ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18

More Exercises

Question 1.
If A is an acute angle and sin A = \(\\ \frac { 3 }{ 5 } \) find all other trigonometric ratios of angle A (using trigonometric identities).
Solution:
sin A = \(\\ \frac { 3 }{ 5 } \)
In ∆ABC, ∠B = 90°
AC = 5 and BC = 3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q1.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q1.2

Question 2.
If A is an acute angle and sec A = \(\\ \frac { 17 }{ 8 } \), find all other trigonometric ratios of angle A (using trigonometric identities).
Solution:
sec A = \(\\ \frac { 17 }{ 8 } \) (A is an acute angle)
In right ∆ABC
sec A = \(\\ \frac { AC }{ AB } \) = \(\\ \frac { 17 }{ 8 } \)
AC = 17, AB = 8
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q2.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q2.2

Question 3.
Express the ratios cos A, tan A and sec A in terms of sin A.
Solution:
cos A = \(\sqrt { { 1-sin }^{ 2 }A } \)
tan A = \(\frac { SinA }{ CosA } =\frac { sinA }{ \sqrt { { 1-sin }^{ 2 }A } } \)
sec A = \(\frac { 1 }{ cosA } =\frac { 1 }{ \sqrt { { 1-sin }^{ 2 }A } } \)

Question 4.
If tan A = \(\frac { 1 }{ \sqrt { 3 } } \), find all other trigonometric ratios of angle A.
Solution:
tan A = \(\frac { 1 }{ \sqrt { 3 } } \)
In right ∆ABC,
tan A = \(\\ \frac { BC }{ AB } \) = \(\frac { 1 }{ \sqrt { 3 } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q4.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q4.2

Question 5.
If 12 cosec θ = 13, find the value of \(\frac { 2sin\theta -3cos\theta }{ 4sin\theta -9cos\theta } \)
Solution:
12 cosec θ = 13
⇒ cosec θ = \(\\ \frac { 13 }{ 12 } \)
In right ∆ABC,
∠A = θ
cosec θ = \(\\ \frac { AC }{ BC } \) = \(\\ \frac { 13 }{ 12 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q5.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q5.2

Without using trigonometric tables, evaluate the following (6 to 10) :

Question 6.
(i) cos² 26° + cos 64° sin 26° + \(\frac { tan{ 36 }^{ O } }{ { cot54 }^{ O } } \)
(ii) \(\frac { sec{ 17 }^{ O } }{ { cosec73 }^{ O } } +\frac { tan68^{ O } }{ cot22^{ O } } \) + cos² 44° + cos² 46°
Solution:
Given that
(i) cos² 26° + cos 64° sin 26° + \(\frac { tan{ 36 }^{ O } }{ { cot54 }^{ O } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q6.1

Question 7.
(i) \(\frac { sin65^{ O } }{ { cos25 }^{ O } } +\frac { cos32^{ O } }{ sin58^{ O } } \) – sin 28° sec 62° + cosec² 30° (2015)
(ii) \(\frac { sin29^{ O } }{ { cosec61 }^{ O } } \) + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3(sin² 38° + sin² 52°).
Solution:
given that
(i) \(\frac { sin65^{ O } }{ { cos25 }^{ O } } +\frac { cos32^{ O } }{ sin58^{ O } } \) – sin 28° sec 62° + cosec² 30°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q7.1

Question 8.
(i) \(\frac { { sin }35^{ O }{ cos55 }^{ O }+{ cos35 }^{ O }{ sin }55^{ O } }{ { cosec }^{ 2 }{ 10 }^{ O }-{ tan }^{ 2 }{ 80 }^{ O } } \)
(ii) \({ sin }^{ 2 }{ 34 }^{ O }+{ sin }^{ 2 }{ 56 }^{ O }+2tan{ 18 }^{ O }{ tan72 }^{ O }-{ cot }^{ 2 }{ 30 }^{ O }\)
Solution:
Given that
(i) \(\frac { { sin }35^{ O }{ cos55 }^{ O }+{ cos35 }^{ O }{ sin }55^{ O } }{ { cosec }^{ 2 }{ 10 }^{ O }-{ tan }^{ 2 }{ 80 }^{ O } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q8.1

Question 9.
(i) \({ \left( \frac { { tan25 }^{ O } }{ { cosec }65^{ O } } \right) }^{ 2 }+{ \left( \frac { { cot25 }^{ O } }{ { sec65 }^{ O } } \right) }^{ 2 }+{ 2tan18 }^{ O }{ tan }45^{ O }{ tan72 }^{ O } \)
(ii) \(\left( { cos }^{ 2 }25+{ cos }^{ 2 }65 \right) +cosec\theta sec\left( { 90 }^{ O }-\theta \right) -cot\theta tan\left( { 90 }^{ O }-\theta \right) \)
Solution:
(i) \({ \left( \frac { { tan25 }^{ O } }{ { cosec }65^{ O } } \right) }^{ 2 }+{ \left( \frac { { cot25 }^{ O } }{ { sec65 }^{ O } } \right) }^{ 2 }+{ 2tan18 }^{ O }{ tan }45^{ O }{ tan72 }^{ O } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q9.1

Question 10.
(i) 2(sec² 35° – cot² 55°) – \(\frac { { cos28 }^{ O }cosec{ 62 }^{ O } }{ { tan18 }^{ O }tan{ 36 }^{ O }{ tan30 }^{ O }{ tan54 }^{ O }{ tan72 }^{ O } } \)
(ii) \(\frac { { cosec }^{ 2 }(90-\theta )-{ tan }^{ 2 }\theta }{ 2({ cos }^{ 2 }{ 48 }^{ O }+{ cos }^{ 2 }{ 42 }^{ O }) } -\frac { { 2tan }^{ 2 }{ 30 }^{ O }{ sec }^{ 2 }{ 52 }^{ O }{ sin }^{ 2 }{ 38 }^{ O } }{ { cosec }^{ 2 }{ 70 }^{ O }-{ tan }^{ 2 }{ 20 }^{ O } } \)
Solution:
(i) 2(sec² 35° – cot² 55°) – \(\frac { { cos28 }^{ O }cosec{ 62 }^{ O } }{ { tan18 }^{ O }tan{ 36 }^{ O }{ tan30 }^{ O }{ tan54 }^{ O }{ tan72 }^{ O } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q10.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q10.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q10.3

Question 11.
Prove that following:
(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1
(ii) \(\frac { tan\theta }{ tan({ 90 }^{ O }-\theta ) } +\frac { sin({ 90 }^{ O }-\theta ) }{ cos\theta } ={ sec }^{ 2 }\theta \)
(iii) \(\frac { cos({ 90 }^{ O }-\theta )cos\theta }{ tan\theta } +{ cos }^{ 2 }({ 90 }^{ O }-\theta )=1\)
(iv) sin (90° – θ) cos (90° – θ) = \(\frac { tan\theta }{ { 1+tan }^{ 2 }\theta } \)
Solution:
(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1
L.H.S. = cos θ sin (90° – θ) + sin θ cos (90° – θ)
= cos θ . cos θ + sin θ . sin θ
= cos2 θ + sin2 θ = 1 = R.H.S.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q11.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q11.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q11.3

Prove that following (12 to 30) identities, where the angles involved are acute angles for which the trigonometric ratios as defined:

Question 12.
(i) (sec A + tan A) (1 – sin A) = cos A
(ii) (1 + tan2 A) (1 – sin A) (1 + sin A) = 1.
Solution:
(i) (sec A + tan A) (1 – sin A) = cos A
L.H.S. = (sec A + tan A) (1 – sin A)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q12.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q12.2

Question 13.
(i) tan A + cot A = sec A cosec A
(ii) (1 – cos A)(1 + sec A) = tan A sin A.
Solution:
(i) tan A + cot A = sec A cosec A
L.H.S. = tan A + cot A
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q13.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q13.2

Question 14.
(i) \(\frac { 1 }{ 1+cosA } +\frac { 1 }{ 1-cosA } =2{ cosec }^{ 2 }A\)
(ii) \(\frac { 1 }{ secA+tanA } +\frac { 1 }{ secA-tanA } =2{ sec }A\)
Solution:
(i) \(\frac { 1 }{ 1+cosA } +\frac { 1 }{ 1-cosA } =2{ cosec }^{ 2 }A\)
L.H.S = \(\frac { 1 }{ 1+cosA } +\frac { 1 }{ 1-cosA }\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q14.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q14.2

Question 15.
(i) \(\frac { sinA }{ 1+cosA } =\frac { 1-cosA }{ sinA } \)
(ii) \(\frac { 1-{ tan }^{ 2 }A }{ { cot }^{ 2 }A-1 } ={ tan }^{ 2 }A\)
(iii) \(\frac { sinA }{ 1+cosA } =cosecA-cotA\)
Solution:
(i) \(\frac { sinA }{ 1+cosA } =\frac { 1-cosA }{ sinA } \)
L.H.S = \(\frac { sinA }{ 1+cosA } \)
(multiplying and dividing by (1 – cosA))
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q15.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q15.2

Question 16.
(i) \(\frac { secA-1 }{ secA+1 } =\frac { 1-cosA }{ 1+cosA } \)
(ii) \(\frac { { tan }^{ 2 }\theta }{ { (sec\theta -1) }^{ 2 } } =\frac { 1+cos\theta }{ 1-cos\theta } \)
(iii) \({ (1+tanA) }^{ 2 }+{ (1-tanA) }^{ 2 }=2{ sec }^{ 2 }A\)
(iv) \({ sec }^{ 2 }A+{ cosec }^{ 2 }A={ sec }^{ 2 }A{ .cosec }^{ 2 }A\)
Solution:
(i) \(\frac { secA-1 }{ secA+1 } =\frac { 1-cosA }{ 1+cosA } \)
L.H.S = \(\frac { secA-1 }{ secA+1 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q16.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q16.2

Question 17.
(i) \(\frac { 1+sinA }{ cosA } +\frac { cosA }{ 1+sinA } =2secA \)
(ii) \(\frac { tanA }{ secA-1 } +\frac { tanA }{ secA+1 } =2cosecA\)
Solution:
(i) \(\frac { 1+sinA }{ cosA } +\frac { cosA }{ 1+sinA } =2secA \)
L.H.S = \(\frac { 1+sinA }{ cosA } +\frac { cosA }{ 1+sinA } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q17.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q17.2

Question 18.
(i) \(\frac { cosecA }{ cosecA-1 } +\frac { cosecA }{ cosecA+1 } =2{ sec }^{ 2 }A\)
(ii) \(cotA-tanA=\frac { { 2cos }^{ 2 }A-1 }{ sinA-cosA } \)
(iii) \(\frac { cotA-1 }{ 2-{ sec }^{ 2 }A } =\frac { cotA }{ 1+tanA } \)
Solution:
(i) \(\frac { cosecA }{ cosecA-1 } +\frac { cosecA }{ cosecA+1 } =2{ sec }^{ 2 }A\)
L.H.S = \(\frac { cosecA }{ cosecA-1 } +\frac { cosecA }{ cosecA+1 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q18.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q18.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q18.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q18.4

Question 19.
(i) \({ tan }^{ 2 }\theta -{ sin }^{ 2 }\theta ={ tan }^{ 2 }\theta { sin }^{ 2 }\theta \)
(ii) \(\frac { cos\theta }{ 1-tan\theta } -\frac { { sin }^{ 2 }\theta }{ cos\theta -sin\theta } =cos\theta +sin\theta \)
Solution:
(i) \({ tan }^{ 2 }\theta -{ sin }^{ 2 }\theta ={ tan }^{ 2 }\theta { sin }^{ 2 }\theta \)
L.H.S = \({ tan }^{ 2 }\theta -{ sin }^{ 2 }\theta \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q19.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q19.2

Question 20.
(i) cosec4 θ – cosec2 θ = cot4 θ + cot2 θ
(ii) 2 sec2 θ – sec4 θ – 2 cosec2 θ + cosec4 θ = cot4 θ – tan4 θ.
Solution:
(i) cosec4 θ – cosec2 θ = cot4 θ + cot2 θ
L.H.S = cosec4 θ – cosec2 θ
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q20.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q20.2

Question 21.
(i) \(\frac { 1+cos\theta -{ sin }^{ 2 }\theta }{ sin\theta (1+cos\theta ) } =cot\theta \)
(ii) \(\frac { { tan }^{ 3 }\theta -1 }{ tan\theta -1 } ={ sec }^{ 2 }\theta +tan\theta \)
Solution:
(i) \(\frac { 1+cos\theta -{ sin }^{ 2 }\theta }{ sin\theta (1+cos\theta ) } =cot\theta \)
L.H.S = \(\frac { 1+cos\theta -{ sin }^{ 2 }\theta }{ sin\theta (1+cos\theta ) }\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q21.1

Question 22.
(i) \(\frac { 1+cosecA }{ cosecA } =\frac { { cos }^{ 2 }A }{ 1-sinA } \)
(ii) \(\sqrt { \frac { 1-cosA }{ 1+cosA } } =\frac { sinA }{ 1+cosA } \)
Solution:
(i) \(\frac { 1+cosecA }{ cosecA } =\frac { { cos }^{ 2 }A }{ 1-sinA } \)
L.H.S = \(\frac { 1+cosecA }{ cosecA }\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q22.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q22.2

Question 23.
(i) \(\sqrt { \frac { 1+sinA }{ 1-sinA } } =tanA+secA\)
(ii) \(\sqrt { \frac { 1-cosA }{ 1+cosA } } =cosecA-cotA\)
Solution:
(i) \(\sqrt { \frac { 1+sinA }{ 1-sinA } } =tanA+secA\)
L.H.S = \(\sqrt { \frac { 1+sinA }{ 1-sinA } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q23.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q23.2

Question 24.
(i) \(\sqrt { \frac { secA-1 }{ secA+1 } } +\sqrt { \frac { secA+1 }{ secA-1 } } =2cosecA\)
(ii) \(\frac { cotAcotA }{ 1-sinA } =1+cosecA \)
Solution:
(i) \(\sqrt { \frac { secA-1 }{ secA+1 } } +\sqrt { \frac { secA+1 }{ secA-1 } } =2cosecA\)
L.H.S = \(\sqrt { \frac { secA-1 }{ secA+1 } } +\sqrt { \frac { secA+1 }{ secA-1 } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q24.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q24.2

Question 25.
(i) \(\frac { 1+tanA }{ sinA } +\frac { 1+cotA }{ cosA } =2(secA+cosecA)\)
(ii) \({ sec }^{ 4 }A-{ tan }^{ 4 }A=1+2{ tan }^{ 2 }A \)
Solution:
(i) \(\frac { 1+tanA }{ sinA } +\frac { 1+cotA }{ cosA } =2(secA+cosecA)\)
L.H.S = \(\frac { 1+tanA }{ sinA } +\frac { 1+cotA }{ cosA } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q25.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q25.2

Question 26.
(i) cosec6 A – cot6 A = 3 cot2 A cosec2 A + 1
(ii) sec6 A – tan6 A = 1 + 3 tan2 A + 3 tan4 A
Solution:
(i) cosec6 A – cot6 A = 3 cot2 A cosec2 A + 1
L.H.S = cosec6 A – cot6 A
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q26.1

Question 27.
(i) \(\frac { cot\theta -cosec\theta -1 }{ cot\theta -cosec\theta +1 } =\frac { 1+cos\theta }{ sin\theta } \)
(ii) \(\frac { sin\theta }{ cot\theta +cosec\theta } =2+\frac { sin\theta }{ cot\theta -cosec\theta } \)
Solution:
(i) \(\frac { cot\theta -cosec\theta -1 }{ cot\theta -cosec\theta +1 } =\frac { 1+cos\theta }{ sin\theta } \)
L.H.S = \(\frac { cot\theta -cosec\theta -1 }{ cot\theta -cosec\theta +1 }\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q27.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q27.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q27.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q27.4

Question 28.
(i) (sinθ + cosθ)(secθ + cosecθ) = 2 + secθ cosecθ
(ii) (cosecA – sinA)(secA – cosA) sec2A = tanA
(iii) (cosecθ – sinθ)(secθ – cosθ)(tan θ + cotθ) = 1
Solution:
(i) (sinθ + cosθ)(secθ + cosecθ) = 2 + secθ cosecθ
L.H.S = (sinθ + cosθ)(secθ + cosecθ)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q28.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q28.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q28.3

Question 29.
(i) \(\frac { { sin }^{ 3 }A+{ cos }^{ 3 }A }{ sinA+cosA } +\frac { { sin }^{ 3 }A-{ cos }^{ 3 }A }{ sinA-cosA } =2\)
(ii) \(\frac { { tan }^{ 2 }A }{ { 1+tan }^{ 2 }A } +\frac { cot^{ 2 }A }{ 1+{ cot }^{ 2 }A } =1\)
Solution:
(i) \(\frac { { sin }^{ 3 }A+{ cos }^{ 3 }A }{ sinA+cosA } +\frac { { sin }^{ 3 }A-{ cos }^{ 3 }A }{ sinA-cosA } =2\)
L.H.S = \(\frac { { sin }^{ 3 }A+{ cos }^{ 3 }A }{ sinA+cosA } +\frac { { sin }^{ 3 }A-{ cos }^{ 3 }A }{ sinA-cosA } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q29.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q29.2

Question 30.
(i) \(\frac { 1 }{ secA+tanA } -\frac { 1 }{ cosA } =\frac { 1 }{ cosA } -\frac { 1 }{ secA-tanA } \)
(ii) \({ (sinA+secA) }^{ 2 }+{ (cosA+cosecA) }^{ 2 }={ (1+secA\quad cosecA) }^{ 2 }\)
(iii) \(\frac { tanA+sinA }{ tanA-sinA } =\frac { secA+1 }{ secA-1 } \)
Solution:
(i) \(\frac { 1 }{ secA+tanA } -\frac { 1 }{ cosA } =\frac { 1 }{ cosA } -\frac { 1 }{ secA-tanA } \)
L.H.S = \(\frac { 1 }{ secA+tanA } -\frac { 1 }{ cosA } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q30.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q30.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q30.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q30.4

Question 31.
If sin θ + cos θ = √2 sin (90° – θ), show that cot θ = √2 + 1
Solution:
sin θ + cos θ = √2 sin (90° – θ)
sin θ + cos θ = √2 cos θ
dividing by sin θ
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q31.1

Question 32.
If 7 sin2 θ + 3 cos2 θ = 4, 0° ≤ θ ≤ 90°, then find the value of θ.
Solution:
7 sin2 θ + 3 cos2 θ = 4, 0° ≤ θ ≤ 90°
3 sin2 θ + 3 cos2 θ + 4 sin2 θ = 4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q32.1

Question 33.
If sec θ + tan θ = m and sec θ – tan θ = n, prove that mn = 1.
Solution:
sec θ + tan θ = m and sec θ – tan θ = n
mn = (sec θ + tan θ) (sec θ – tan θ) = sec2 θ – tan2 θ = 1
(∴ sec2 θ – tan2 θ = 1)
Hence proved.

Question 34.
If x – a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 – y2 = a2 – b2.
Solution:
x – a sec θ + b tan θ and y = a tan θ + b sec θ
To prove that x2 – y2 = a2 – b2.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q34.1

Question 35.
If x = h + a cos θ and y = k + a sin θ, prove that (x – h)2 + (y – k)2 = a2.
Solution:
x = h + a cos θ and y = k + a sin θ
To prove that (x – h)2 + (y – k)2 = a2.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q35.1

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20

More Exercises

Question 1.
An electric pole is 10 metres high. If its shadow is 10√3 metres in length, find the elevation of the sun.
Solution:
Let AB be the pole and
OB is its shadow.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q1.1

Question 2.
The angle of elevation of the top of a tower from a point on the ground and at a distance of 150 m from its foot is 30°. Find the height of the tower correct to one place of decimal
Solution:
Let BC be the tower and
A is the point on the ground such that
∠A= 30° and AC = 150 m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q2.1

Question 3.
A ladder is placed against a wall such that it just reaches the top of the wall. The foot of the ladder is 1.5 metres away from the wall and the ladder is inclined at an angle of 60° with the ground. Find the height of the wall.
Solution:
Let AB be the wall and AC be the ladder
whose foot C is 1.5 m away from B
Let AB = x m and angle of inclination is 60°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q3.1

Question 4.
What is the angle of elevation of the sun when the length of the shadow of a vertical pole is equal to its height.
Solution:
Let AB be the pole and CB be its shadow
and θ is the angle of elevation of the sun.
Let AB = x m, then BC = x m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q4.1

Question 5.
A river is 60 m wide. A tree of unknown height is on one bank. The angle of elevation of the top of the tree from the point exactly opposite to the foot of the tree on the other bank is 30°. Find the height of the tree.
Solution:
Let AB be the tree and BC is the width of the river
and C is the point exactly opposite to B on the other bank
and angle of elevation is 30°.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q5.1

Question 6.
From a point P on level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100 m high, how far is P from the foot of the tower ?
Solution:
Let AB be the tower and P is at a distance of x m from B, the foot of the tower.
While the height of the tower AB = 100 m
and angle of elevation = 30°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q6.1

Question 7.
From the top of a cliff 92 m high, the angle of depression of a buoy is 20°. Calculate to the nearest metre, the distance of the buoy from the foot of the cliff. (2005)
Solution:
Let AB be cliff whose height is 92 m
and C is buoy making depression angle of 20°.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q7.1

Question 8.
A boy is flying a kite with a string of length 100 m. If the string is tight and the angle of elevation of the kite is 26°32′, find the height of the kite correct to one decimal place, (ignore the height of the boy).
Solution:
Let AB be the height of the kite A and AC is the string
and angle of elevation of the kite is 26°32′
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q8.1

Question 9.
An electric pole is 10 m high A steel wire tied to the top of the pole is affixed at a point on the ground to keep the pole upright. If the wire makes an angle of 45° with the horizontal through the foot of the pole, find the length of the wire.
Solution:
Let AB be the pole and AC be the wire
which makes an angle of 45° with the ground.
Height of the pole AB = 10 m
and let the length of wire AC = x m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q9.1

Question 10.
A bridge across a river makes an angle of 45° with the river bank. If the length of the bridge across the river is 200 metres, what is the breadth of the river.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q10.1
Solution:
Let AB be the width of river = xm
Length of the bridge AC = 200 m
and angle with the river bank = 45°
sin θ = \(\\ \frac { AB }{ AC } \)
⇒ sin 45° = \(\\ \frac { x }{ 200 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q10.2

Question 11.
A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower ? (2001)
Solution:
Let AB be the tower and
let a man C stands at a distance from the foot of the tower = x m
and cos θ = 0.53
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q11.1

Question 12.
The upper part of a tree broken by wind, falls to the ground without being detached. The top of the broken part touches the ground at an angle of 38°30′ at a point 6 m from the foot of the tree. Calculate.
(i) the height at which the tree is broken.
(ii) the original height of the tree correct to two decimal places.
Solution:
Let TR be the total height of the tree
and TP is the broken part which touches the ground
at the distance of 6 m from the foot of the tree
making an angle of 38°30′ with the ground.
Let PR = x and TR = x + y
PQ = PT = y
In right ∆PQR
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q12.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q12.2
Height of the tree = 4.7724 + 7.6665 = 12.4389 = 12.44 m
and height of the tree at which it is broken = 4.77 m

Question 13.
An observer 1.5 m tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Solution:
In the figure, AB is tower and CD is an observer.
θ is the angle of observation from
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q13.1

Question 14.
From a boat 300 metres away from a vertical cliff, the angles of elevation of the top and the foot of a vertical concrete pillar at the edge of the cliff are 55°40′ and 54°20′ respectively. Find the height of the pillar correct to the nearest metre.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q14.1
Solution:
Let CB be the cliff and AC be the pillar
and D be the boat which is 300 m away from
the foot of the cliff i.e. BD = 300 m.
Angles of elevation of the top and foot of the pillar
are 55°40′ and 54°20′ respectively.
Let CB = x and AC = y
In right ∆CBD,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q14.2

Question 15.
From a point P on the ground, the angle of elevation of the top of a 10 m tall building and a helicopter hovering over the top of the building are 30° and 60° respectively. Find the height of the helicopter above the ground.
Solution:
let AB be the building and H is the helicopter hovering over it.
P is a point on the ground,
the angle of elevation of the top of building and helicopter are 30° and 60°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q15.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q15.2

Question 16.
An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the distance between the two planes at the instant.
Solution:
Let the distance between the two planes = h m
Given that, AD = 3125 m and ∠ACB = 60° and ∠ACD = 30°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q16.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q16.2

Question 17.
A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60° ; when he retires 20 m from the bank, he finds the angle to be 30°. Find the height of the tree and the breadth of the river. .
Solution:
Let TR be the tree and PR be the width of the river.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q17.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q17.2

Question 18.
The shadow of a vertical tower on a level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. Find the height of the tower, correct to two decimal places. (2006)
Solution:
In the figure, AB is the tower,
BD and BC are the shadow of the tower in two situations.
Let BD = x m and AB = h m
In ∆ABD,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q18.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q18.2

Question 19.
From the top of a hill, the angles of depression of two consecutive kilometer stones, due east are found to be 30° and 45° respectively. Find the distance of two stones from the foot of the hill.
Solution:
Let A and B be the position of two consecutive kilometre stones.
Then AB = 1 km = 1000m
Let the dIstance BC = x m
∴ Distance AC = (1000 + x) m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q19.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q19.2

Question 20.
A man observes the angles of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60 m the angle of elevation changes to 60°. Find the height of the building correct to the nearest me he.
Solution:
Given that
AB is a building CD = 60 m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q20.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q20.2

Question 21.
At a point on level ground, the angle,of elevation of a vertical lower is found to be such that its tangent is \(\\ \frac { 5 }{ 12 } \). On walking 192 m towards the tower,the tangent of the angle is found to be \(\\ \frac { 3 }{ 4 } \). Find the height of the tower. (1990)
Solution:
Let TR be the tower and P is the point on the
ground such that tan θ = \(\\ \frac { 5 }{ 12 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q21.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q21.2

Question 22.
In the figure, not drawn to scale, TF is a tower. The elevation of T from A is x° where tan x = \(\\ \frac { 2 }{ 5 } \) and AF = 200 m. The elevation of T from B, where AB = 80 m, is y°. Calculate :
(i) The height of the tower TF.
(ii) The angle y, correct to the nearest degree. (1997)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q22.1
Solution:
Let height of the tower TF = x
tan x = \(\\ \frac { 2 }{ 5 } \), AF = 200 m, AB = 80 m
(i) In right ∆ATF,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q22.2

Question 23.
From the top of a church spire 96 m high, the angles of depression of two vehicles on a road, at the same level as the base of the spire and on the same side of it are x° and y°, where tan x° = \(\\ \frac { 1 }{ 4 } \) and tan y° = \(\\ \frac { 1 }{ 7 } \). Calculate the distance between the vehicles. (1994)
Solution:
Height of the church CH.
Let A and B are two vehicles which make the angle of depression
from C are x° and y° respectively.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q23.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q23.2

Question 24.
In the adjoining figure, not drawn to the scale, AB is a tower and two objects C and D are located on the ground, on the same side of AB. When observed from the top A of the tower, their angles of depression are 45° and 60°. Find the distance between the two objects. If the height of the tower is 300 m. Give your answer to the nearest metre. (1998)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q24.1
Solution:
Let CB = x and
DB = y
AB = 300 m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q24.2

Question 25.
The horizontal distance between two towers is 140 m. The angle of elevation of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 60 m, find the height of the first tower.
Solution:
Let the height of first tower TR = x
height of second tower PQ = 60 m
Distance between the two towers QR = 140 m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q25.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q25.2

Question 26.
As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships on the same side of the ,lighthouse in horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to the nearest metre.
Solution:
Let AB be the lighthouse and C and D be the two ships.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q26.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q26.2

Question 27.
The angle of elevation of a pillar from a point A on the ground is 45° and from a point B diametrically opposite to A and on the other side of the pillar is 60°. Find the height of the pillar, given that the distance between A and B is 15 m.
Solution:
Let CD be the pillar and let CD = x
Angles of elevation of points A and B are 45° and 60° respectively.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q27.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q27.2

Question 28.
From two points A and B on the same side of a building, the angles of elevation of the top of the building are 30° and 60° respectively. If the height of the building is 10 m, find the distance between A and B correct to two decimal places
Solution:
In ∆DBC, tan 60° = \(\\ \frac { 10 }{ BC } \)
⇒ √3= \(\\ \frac { 10 }{ BC } \)
⇒ BC = \(\frac { 10 }{ \sqrt { 3 } } \)
∆DBC ,tan 30° = \(\\ \frac { 10 }{ BC+AB } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q28.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q28.2

Question 29.
(i) The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 60° and 45° respectively. If the; two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number. (2017)
(ii) An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number. (2014)
Solution:
(i) Let AD be the height of the lighthouse CD = 60 m
Let AD = x m, BD = y m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q29.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q29.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q29.3

Question 30.
From a tower 126 m high, the angles of depression of two rocks which are in a horizontal line through the base of the tower are 16° and 12°20′ Find the distance between the rocks if they are on
(i) the same side of the tower
(ii) the opposite sides of the tower.
Solution:
Let CD be the tower and CD = 126 m
Let A and B be the two rocks on the same line
and angles of depression are 16° and 12°20′ respectively,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q30.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q30.2

Question 31.
A man 1.8 m high stands at a distance of 3.6 m from a lamp post and casts a shadow of 5.4 m on the ground. Find the height of the lamp post.
Solution:
AB is the lamp post CD is the height of man.
BD is the distance of man from the foot of the lamp
and FD is the shadow of man.
CE || DB.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q31.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q31.2

Question 32.
The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45° respectively. Find the height of tire multi-storeyed building and the distance between the two buildings, correct to two decimal places.
Solution:
Let AB be the CD be the building
The angles of depression in from A, to C
and D are 30° and 45° respectively
∠ACE = 30° and ∠ADB = 45°
CD = 8 m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q32.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q32.2

Question 33.
A pole of height 5 m is fixed on the top of a tower. The angle of elevation of the top of the pole as observed from a point A on the ground is 60° and the angle of depression of the point A from the top of the tower is 45°. Find the height of the tower. (Take √3 = 1.732).
Solution:
Let QR be the tower and PQ be the pole on it
Angle of elevation from P to a point A is ∠PAR = 60°
and angle of depression from Q to A = 45°
∠QAR = 45° (alternate angle)
PQ = 5 m,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q33.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q33.2

Question 34.
A vertical pole and a vertical tower are on the same level ground. From the top of the pole the angle of elevation of the top of the tower is 60° and the angle of depression of the foot of the tower is 30°. Find the height of the tower if the height of the pole is 20 m.
Solution:
Let TR is tower and
PL is the pole on the same level, ground PL = 20m
From P, draw PQ || LR
then ∠ TPQ = 60° and ∠ QPR = 30°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q34.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q34.2

Question 35.
From the top of a building 20 m high, the angle of elevation of the top of a monumenti is 45° and the angle of depression of its foot is 15°. Find the height of the monument.
Solution:
Let AB be the building and AB = 20 m and
let CD be the monument and let CD = x
The distance between the building and the monument be y,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q35.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q35.2

Question 36.
The angle of elevation of the top of an unfinished tower at a point distant 120 m from its base is 45°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60°?
Solution:
Let AB be the unfinished tower and AB = 120 m
and angle of elevation = 45°
Let x be higher raised so that
the angle of elevation becomes 60°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q36.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q36.2

Question 37.
In the adjoining figure, the shadow of a vertical tower on the level ground increases by 10 m, when the altitude of the sun changes from 45° to 30°. Find the height of the tower and give your answer, correct to \(\\ \frac { 1 }{ 10 } \) of a metre.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q37.1
[Remark. Altitude of the sun means angle of elevation of the sun.]
Solution:
Let TR be the tower and TR = h ;
Let BR = x,
AB = 10 m
Angles of elevation from the top of the tower
at A and B are 30° and 45° respectively.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q37.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q37.3

Question 38.
An aircraft is flying at a constant height with a speed of 360 km/h. From a point on the ground, the angle of elevation of the aircraft at an instant was observed to be 45°. After 20 seconds, the angle of elevation was observed to be 30°. Determine the height at which the aircraft is flying (use √3 = 1.732)
Solution:
Speed of aircraft = 360 km/h
Distance covered in 20 seconds = \(\\ \frac { 360X20 }{ 60X60 } \) = 2 km
E is the fixed point on the ground
and CD is the position of AB in height of aircraft
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q38.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 Q38.2

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test

More Exercises

Question 1.
Using trigonometrical tables, find the values of :
(i) sin 48° 52′
(ii) cos 37° 34′
(iii) tan 18° 21′.
Solution:
Using tables, we find that
(i) sin 48° 52′ = .7524 + .0008 = .7532
(ii) cos 37° 34′ = .7934 – .0007 = .7927
(iii) tan 18° 21′ = .3307 + .0010 = .3317.

Question 2.
Use tables to find the acute angle θ, given that
(i) sin θ = 0.5766
(ii) cos θ = 0.2495
(iii) tan θ = 2.4523.
Solution:
Using table, we find that
(i) sin θ = 0.5766 = 0.5764 + 0.0002
= sin (35° 12’+ 1′)
= sin 35° 13′
θ = 35° 13′
(ii) cos θ = 0.2495 = 0.2487 + 0.0008
= cos (75° 36′ – 3′)
= cos 75° 33′
θ = 75° 33′
(iii) tan θ = 2.4523 = 2.4504 + 0.0019
= tan (67° 48′ + 1′)
= tan 67° 49′

Question 3.
If θ is acute and cos θ = 0.53, find the value of tan θ.
Solution:
From the table, we find that
cos θ = 0.53 = .5299 + .0001 = cos 58°
θ = 58°
and tan 58° = 1.6003

Question 4.
Find the value of: sin 22° 11′ + cos 57° 20′ – 2 tan 9° 9′.
Solution:
Using the tables, we find that
sin 22° 11′ = 0.3762 + 0.0014 = 0.3776
cos 57° 20′ = 0.5402 – 0.0005 = 0.5397
tan 9° 9′ = 0.1602 + 0.0009 = 0.1611
∴ sin 22° 11′ + cos 57° 20′ – 2 tan 9° 9′
= 0.3376 + 0.5397 – 0.1611 × 2
= 0.3776 + 0.5397 – 0.3222
= 0.9173 – 0.3222
= .5951.

Question 5.
If θ is acute and sin θ = 0.7547, find the value of: (i) θ (ii) cos θ (iii) 2 cos θ – 3 tan θ.
Solution:
Using the tables, we find that
(i) sin θ = 0.7547 = sin 49°
θ = 49°.
(ii) cos θ = cos 49° = 0.6561.
(iii) tan θ = tan 49° = 1.1504
2 cos θ – 3 tan θ
= 2 × θ.6561 – 3 × 1.1504
= 1.3122 – 3.4512
= – 2.1390

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test

More Exercise

Question 1.
A cylindrical container is to be made of tin sheet. The height of the container is 1 m and its diameter is 70 cm. If the container is open at the top and the tin sheet costs Rs 300 per m2, find the cost of the tin for making the container.
Solution:
Height of container opened at the top (h) = 1 m = 100 cm
and diameter = 70 cm
∴Radius (r) = \(\\ \frac { 70 }{ 2 } \) = 35 cm
∴Total surface area = 2πrh + πr2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q1.1

Question 2.
A cylinder of maximum volume is cut out from a wooden cuboid of length 30 cm and cross-section of square of side 14 cm. Find the volume of the cylinder and the volume of wood wasted.
Solution:
Dimensions of the wooden cuboid = 30 cm × 14 cm × 14 cm
Volume = 30 × 14 × 14 = 5880 cm3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q2.1

Question 3.
Find the volume and the total surface area of a cone having slant height 17 cm and base diameter 30 cm. Take π = 3.14.
Solution:
Slant height of a cone (l) = 17 cm
Diameter of base = 30 cm
Radius (r) = \(\\ \frac { 30 }{ 2 } \) = 15 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q3.1

Question 4.
Find the volume of a cone given that its height is 8 cm and the area of base 156 cm2.
Solution:
Height of a cone = 8 cm
Area of base = 156 cm
.’. Volume = \(\\ \frac { 1 }{ 3 } \) × area of base × height
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q4.1

Question 5.
The circumference of the edge of a hemispherical bowl is 132 cm. Find the capacity of the bowl.
Solution:
Circumference of the edge of bowl = 132 cm
Radius of a hemispherical bowl
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q5.1

Question 6.
The volume of a hemisphere is \(2425 \frac { 1 }{ 2 } \) cm2. Find the curved surface area.
Solution:
Volume of a hemisphere = \(2425 \frac { 1 }{ 2 } \) cm3
= \(\\ \frac { 4851 }{ 2 } \) cm3
Let radius = r, then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q6.1

Question 7.
A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the toy
Solution:
A wooden solid toy is of a shape of a right circular cone
mounted on a hemisphere.
Radius of hemisphere (r) = 4.2 cm
Total height = 10.2 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q7.1

Question 8.
A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of the entire capsule is 2 cm. Find the capacity of the capsule.
Solution:
Diameter of cylindrical part = 0.5 cm
Total length of the capsule = 2 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q8.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q8.2

Question 9.
A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and the total surface area of the solid.
Solution:
Radius of cylinder = \(\\ \frac { 7 }{ 2 } \)cm
and height of cylinder = 19 – 2 × \(\\ \frac { 7 }{ 2 } \) cm
= 19 – 7 = 12 cm
and radius of hemisphere = \(\\ \frac { 7 }{ 2 } \) cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q9.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q9.2

Question 10.
The radius and height of a right circular cone are in the ratio 5 : 12. If its volume is 2512 cm , find its slant height. (Take π = 3.14).
Solution:
Let radius of cone (r) = 5x
then height (h) = 12x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q10.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q10.2

Question 11.
A cone and a cylinder are of the same height. If diameters of their bases are in the ratio 3 : 2, find the ratio of their volumes.
Solution:
Let height of cone and cylinder = h
Diameter of the base of cone = 3x
Diameter of base of cylinder = 2x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q11.1

Question 12.
A solid cone of base radius 9 cm and height 10 cm is lowered into a cylindrical jar of radius 10 cm, which contains water sufficient to submerge the cone completely. Find the rise in water level in the jar.
Solution:
Radius of the cone (r) = 9 cm
Height of the cone (h) = 10 cm
Volume of water filled in cone
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q12.1

Question 13.
An iron pillar has some part in the form of a right circular cylinder and the remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if one cu. cm of iron weighs 7.8 grams.
Solution:
Radius of the base of cone = 8 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q13.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q13.2

Question 14.
A circus tent is made of canvas and is in the form of right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 126 m and 5 m respectively. The total height of the tent is 21 m. Find the total cost of the tent if the canvas used costs Rs 36 per square metre.
Solution:
Diameter of the cylindrical part = 126 m
Radius (r) = \(\\ \frac { 126 }{ 2 } \) = 63m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q14.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q14.2

Question 15.
The entire surface of a solid cone of base radius 3 cm and height 4 cm is equal to the entire surface of a solid right circular cylinder of diameter 4 cm. Find the ratio of their
(i) curved surfaces
(ii) volumes.
Solution:
Radius of the base of a cone (r) = 3 cm
Height (h) = 4 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q15.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q15.2
Question 16.
A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. Find the radius of the sphere.
Solution:
Radius of base of a cone (r) = 2. 1 cm
and height (h) = 8.4 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q16.1

Question 17.
How many lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm and 21 cm.
Solution:
Dimensions of a solid rectangular lead piece
= 66 cm × 42 cm × 21 cm
.’. Volume = 66 × 42 × 21 cm3
Diameter of a lead shot = 4.2 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q17.1

Question 18.
Find the least number of coins of diameter 2.5 cm and height 3 mm which are to be melted to form a solid cylinder of radius 3 cm and height 5 cm.
Solution:
Radius of a cylinder (r) = 3 cm
Height (h) = 5 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q18.1

Question 19.
A hemisphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Determine the height of the cone correct to 2 places of decimal.
Solution:
Radius of hemisphere = 8 cm
Volume = \(\frac { 2 }{ 3 } \pi { r }^{ 3 }{ cm }^{ 3 }\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q19.1

Question 20.
A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of the water in the cylinder.
Solution:
Radius of hemispherical bowl = 6 cm
.’. Volume of the water in the bowl
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q20.1

Question 21.
The diameter of a metallic sphere is 42 cm. It is metled and drawn into a cylindrical wire of 28 cm diameter. Find the length of the wire.
Solution:
Diameter of sphere = 42 cm
Radius of sphere =\(\\ \frac { 42 }{ 2 } \) = 21 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q21.1

Question 22.
A sphere of diameter 6 cm is dropped into a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?
Solution:
Radius of sphere = \(\\ \frac { 6 }{ 2 } \) = 3 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q22.1

Question 23.
A solid sphere of radius 6 cm is metled into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.
Solution:
Radius of solid sphere = 6 cm
Volume of solid sphere = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q23.1

Question 24.
A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 3.5 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical vessel, full of water, in such a Way that the whole solid is submerged in water. If the radius of the cylindrical vessel is 5 cm and its height is 10.5 cm, find the volume of water left in the cylindrical vessel.
Solution:
Radius of hemisphere (r) = 3.5 cm
Height of cone (h1) = 4 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q24.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test Q24.2

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19

More Exercises

Question 1.
Find the value of the following:
(i) sin 35° 22′
(ii) sin 71° 31′
(iii) sin 65° 20′
(iv) sin 23° 56′.
Solution:
(i) sin 35° 22′
Using the table of natural sines,
we see 35° in the horizontal line and for 18′,
in the vertical column, the value is 0.5779.
Now read 22′ – 18′ = 4′ in the difference column, the value is 10.
Adding 10 in 0.5779 + 10 = 0.5789,
we find sin 35° 22′ = 0.5789.
(ii) sin 71° 31′
Using the table of natural sines, we see 71° in the horizontal line
and for 30′ in the vertical column, the value is 0.9483 and for 31′ – 30′ = 1′,
we see in the mean difference column, the value is 1.
∴ sin 71° 31′ = 0.9483 + 1 = 0.9484.
(iii) sin 65° 20′
Using the table of natural sines, we see 65° in the horizontal line
and for 18′ in the vertical column, the value is .9085 and for 20′ – 18′ = 2′,
we see in the mean difference column. We find 2.
∴ sin 65° 20′ = 0.9085 + 2 = 0.9087 Ans.
(iv) sin 23° 56′
Using the table of natural sines, we see 23° in the horizontal line
and for 54′, we see in vertical column, the value is 0.4051
and for 56′ – 54′ = 2′ in the mean difference. It is 5.
∴ sin 23° 56′ = 0.4051 + 5 = 0.4056

Question 2.
Find the value of the following:
(i) cos 62° 27′
(ii) cos 3° 11′
(iii) cos 86° 40′
(iv) cos 45° 58′.
Solution:
(i) cos 62° 27′
From the table of natural cosines,
we see 62° in the horizontal line and 24′ in the vertical column, the value is .4633
and 27′ – 24′ = 3′ in the mean difference. Its value is 8.
∴ cos 62° 27′ = 0.4633 – 8 = 0.4625 Ans.
(ii) cos 3° 11′
From the table of natural cosines, we see 3° in the horizontal line
and 6′ in the vertical column, its value is 0.9985
and 11′ – 6′ = 5′ in the mean difference, its value is 1.
∴ cos 3° 11′ = 0.9985 – 1 = 0.9984 Ans.
(iii) cos 86° 40′
From the table of natural cosines, we see 86° in the horizontal line
and 36′ in the vertical column, its value is 0.0593
and for 40′ – 36′ = 4′ in the mean difference, it is 12.
cos 86° 40’= 0.0593 – 12 = 0 0581 Ans.
(iv) cos 45° 58′
From the table of natural cosines, we see 45° in the horizontal column
and 54′ in the vertical column, its value is 0.6959
and for 58′ – 54′ = 4′, in the mean difference, it is 8.’
cos 45° 58′ = 0.6959 – 8 = 0.6951

Question 3.
Find the value of the following :
(i) tan 15° 2′
(ii) tan 53° 14′
(iii) tan 82° 18′
(iv) tan 6° 9′.
Solution:
(i) tan 15° 2′
From the table of natural tangents, we see 15° in the horizontal line,
its value is 0.2679 and for 2′, in the mean difference, it is 6.
tan 15° 2′ = 0.2679 + 6 = 0.2685.
(ii) tan 53° 14′
From the table of natural tangents, we see 53° in the horizontal line
and 12′ in the vertical column, its value is 1.3367
and 14′ – 12′ = 2′ in the mean difference, it is 16.
∴ tan 53° 14′ = 1.3367 + 16 = 1 .3383 Ans.
(iii) tan 82° 18′
From the table of natural tangents, we see 82° in the horizontal line
and 18′ in the vertical column, its value is 7.3962.
∴ tan 82° 18’= 7.3962.
(iv) tan 6° 9′
From the table of natural tangents, we see 6° in the horizontal line
and 6′ in the vertical column, its value is .1069
and 9′ – 6′ = 3′, in the mean difference, it is 9.
tan 6°9′ = .1069 + 9 = .1078.

Question 4.
Use tables to find the acute angle θ, given that:
(i) sin θ = – 5789
(ii) sin θ = – 9484
(iii) sin θ = – 2357
(iv) sin θ = – 6371.
Solution:
(i) sin θ = – 5789
From the table of natural sines,
we look for the value (≤ 5789), which must be very close to it,
we find the value .5779 in the column 35° 18′ and in mean difference,
we see .5789 – .5779 = .0010 in the column of 4′.
θ = 35° 18’+ 4’= 35° 22′ Ans.
(ii) sin θ = . 9484
From the table of natural sines,
we look for the value (≤ 9484) which must be very close to it,
we find the value .9483 in the column 71° 30′
and in the mean differences,
we see .9484 – 9483 = 0001, in the column of 1′.
θ = 71° 30′ + 1′ = 71° 31′ Ans.
(iii) sin θ = – 2357
From the table of natural sines,
we look for the value (≤ 2357) which must be very close to it,
we find the value .2351 in the column 13° 36′ and in the mean difference,
we see .2357 – 2351 = .0006, in the column of 2′.
θ = 13° 36′ +2’= 13° 38′ Ans.
(iv) sin θ = .6371
From the table of natural sines,
we look for the value (≤ 6371) which must be very close to it,
we find the value .6361 in the column 39° 30′ and in the mean difference,
we see .6371 – .6361 = .0010 in the column of 4′.
θ = 39° 30′ + 4′ = 39° 34′

Question 5.
Use the tables to find the acute angle θ, given that:
(i) cos θ = .4625
(ii) cos θ = .9906
(iii) cos θ = .6951
(iv) cos θ = .3412.
Solution:
(i) cos θ = .4625
From the table of natural cosines,
we look for the value (≤ .4625) which must be very close to it,
we find the value .4617 in the column of 62° 30′ and in the mean difference,
we see .4625 – .4617 = .0008 which is in column of 3′.
θ = 62° 30′ – 3’= 62° 27′.
(ii) cos θ = .9906
From the table of cosines,
we look for the value (≤ .9906) which must be very close to it,
we find the value of .9905 in the column of 7° 54′ and in mean difference,
we see .9906 – 9905 = .0001 which is in column of 3′.
θ = 70 54′ – 3’= 7° 51′
(iii) cos θ = .6951
From the tables of cosines,
we look for the value (≤ 6951) which must be very close to it,
we find the value .6947 in the column of 46° and in mean difference,
.6951 – .6947 = 0.0004 which in the column of 2′.
θ = 46° – 2′ = 45° 58′ Ans.
(iv) cos θ = .3412
From the table of cosines,
we look for the value of (≤ .3412) which must be very close to it,
we find the value .3404 in the column of 70° 6′ and in the mean difference,
.3412 – 3404 = .0008 which is in the column of 3′.
θ = 70° 6′ – 3′ = 70° 3′

Question 6.
Use tables to find the acute angle θ, given that:
(i) tan θ = .2685
(ii) tan θ = 1.7451
(iii) tan θ = 3.1749
(iv) tan θ = .9347
Solution:
(i) tan θ = .2685
From the table of natural tangent,
we look for the value of (≤ .2685) which must be very close to it,
we find the value .2679 in the column of 15° and in the mean difference,
.2685 – .2679 which is in the column of 2′.
θ = 15° +2′ = 15° 2′ Ans.
(ii) tan θ = 1.7451
From the tables of natural tangents,
we look for the value of (≤ 1.7451) which must be very close to it,
we find the value 1.7391 in the column of 60°’ 6′
and in the mean difference 1.7451 + 1.7391 = 0.0060 which is in the column of 5′.
θ = 60° 6’+ 5’= 60° 11’Ans.
(iii) tan θ = 3.1749
From the tables of natural tangents,
we look for the value of (≤ 3.1749) which must be very close to it,
we find the value 3.1716 in the column of 72° 30′
and in the mean difference 3.1749 – 3.1716 = 0.0033 which is in the column of 1′.
θ = 720 30′ + 1′ = 72° 31′ Ans.
(iv) tan θ = .9347
From the tables of natural tangents,
we look for the value of (≤ .9347 which must be very close to it,
we find the value .9325 in the column of 43°
and in the mean difference .9347 – .9325 = 0.0022 which is in the column of 4′.
θ = 43° + 4′ = 43° 4′

Question 7.
Using trigonometric table, find the measure of the angle A when sin A = 0.1822.
Solution:
sin A = 0.1822
From the tables of natural sines,
we look for the value (≤ .1822) which must be very close to it,
we find the value .1822 in column 10° 30′.
A = 10° 30′

Question 8.
Using tables, find the value of 2 sin θ – cos θ when (i) θ = 35° (ii) tan θ = .2679.
Solution:
(i) θ = 35°
2 sin θ – cos θ = 2 sin 35° – cos 35°
= 2 x .5736 – .8192
(From the tables)
= 1.1472 – .8192 = 0.3280.
(ii) tan θ = .2679
From the tables of natural tangents,
we look for the value of ≤ .2679,
we find the value of the column 15°.
θ = 15°
Now, 2 sin θ – cos θ = 2 sin 15° – cos 15°
= 2 (.2588) – .9659 = 5136 – .9659
= -0.4483

Question 9.
If sin x° = 0.67, find the value of
(i) cos x°
(ii) cos x° + tan x°.
Solution:
sin x° = 0.67
From the table of natural sines,
we look for the value of (≤ 0.67) which must be very close to it,
we find the value .6691 in the column 42° and in the mean difference,
the value of 0.6700 – 0.6691 = 0.0009 which is in the column 4′.
θ = 42° + 4′ = 42° 4′
Now
(i) cos x° = cos 42° 4′ = .7431 – .0008
= 0.7423 Ans.
(ii) cos x° + tan x° = cos 42° 4′ + tan 42° 4′
= 0.7423 + .9025
= 1.6448

Question 10.
If θ is acute and cos θ = .7258, find the value of (i) θ (ii) 2 tan θ – sin θ.
Solution:
cos θ = .7258
From the table of cosines,
we look for the value of (≤ .7258) which must be very close to it,
we find the value .7254 in the column of 43° 30′
and in the mean differences the value of .7258 – .7254 = 0.0004
which in the column of 2′.
(i) θ = 43° 30′ – 2’= 43° 28′.
(ii) 2 tan θ – sin θ
= 2 tan43°28′ – sin43°28′
= 2 (.9479) – .6879
= 1.8958 – .6879
= 1.2079

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5

More Exercises

Question 1.
Draw an ogive for the following frequency distribution:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5 Q1.1
Solution:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5 Q1.2
Plot the points (160, 8), (170, 11), (180, 15), (190, 25) and (200, 27)
on the graph and join them with the free hand. We get an ogive as shown:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5 Q1.3

Question 2.
Draw an ogive for the following data:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5 Q2.1
Solution:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5 Q2.2
Plot the points (10.5, 3), (20.5, 8), (30.5, 16), (40.5, 23),
(50.5, 29), (60.5, 31) on the graph and join them with a free hand,
we get an ogive as shown:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5 Q2.3

Question 3.
Draw a cumulative frequency curve for the following data:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5 Q3.1
Solution:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5 Q3.2
Plot the points (29, 1), (34, 3), (39, 8), (44, 14), (49, 18),
(54, 21) and (59, 23) on the graph and join them
with a free hand to get an ogive as shown:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5 Q3.3

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2

More Exercises

Question 1.
A student scored the following marks in 11 questions of a question paper : 3, 4, 7, 2, 5, 6, 1, 8, 2, 5, 7 Find the median marks.
Solution:
Arranging in the ascending order, 1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8
Here, n = 11 i.e. odd,
The middle term = \(\\ \frac { n+1 }{ 2 } \) = \(\\ \frac { 11+1 }{ 2 } \) = \(\\ \frac { 12 }{ 2 } \) = 6th term
Median = 5

Question 2.
(a) Find the median of the following set of numbers : 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7 (1990)
(b)For the following set of numbers, find the median: 10, 75, 3, 81, 17, 27, 4, 48, 12, 47, 9, 15.
Solution:
(a) Arranging in ascending order :
0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9
Here, n = 12 which is even
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q2.1

Question 3.
Calculate the mean and the median of the numbers : 2, 1, 0, 3, 1, 2, 3, 4, 3, 5
Solution:
Writing in ascending order 0, 1, 1, 2, 2, 3, 3, 3, 4, 5
Here, n = 10 which is even
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q3.1

Question 4.
The median of the observations 11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.
Solution:
Observation are :
11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47
n = 9
Median = \(\\ \frac { 9+1 }{ 2 } \) th term
i.e, 5th term = x + 4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q4.1

Question 5.
The mean of the numbers 1, 7, 5, 3, 4, 4, is m. The numbers 3, 2, 4, 2, 3, 3, p have mean m – 1 and median q. Find
(i) p
(ii) q
(iii) the mean of p and q.
Solution:
(i) Mean of 1, 7, 5, 3, 4, 4 is m.
Here n = 6
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q5.1

Question 6.
Find the median for the following distribution:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q6.1
Solution:
Writing the distribution in cumulative frequency table:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q6.2

Question 7.
Find the median for the following distribution.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q7.1
Solution:
Writing the distribution in cumulative frequency table :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q7.2

Question 8.
Marks obtained by 70 students are given below :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q8.1
Calculate the median marks.
Solution:
Arranging the variates in ascending order and in c.f. table.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q8.2

Question 9.
Calculate the mean and the median for the following distribution :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q9.1
Solution:
Writing the distribution in c.f. table :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q9.2

Question 10.
The daily wages (in rupees) of 19 workers are
41, 21, 38, 27, 31, 45, 23, 26, 29, 30, 28, 25, 35, 42, 47, 53, 29, 31, 35.
Find
(i) the median
(ii) lower quartile
(iii) upper quartile range,
(iv) interquartile range.
Solution:
Arranging the observations in ascending order
21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53
Here n = 19 which is odd.
(i) Median = \(\\ \frac { n+1 }{ 2 } \) th term = \(\\ \frac { 19+1 }{ 2 } \) = \(\\ \frac { 20 }{ 2 } \) = 10th term = 31
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q10.1

Question 11.
From the following frequency distribution, find :
(i) the median
(ii) lower quartile
(iii) upper quartile
(iv) inter quartile range
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q11.1
Solution:
Writing frequency distribution in c.f. table :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q11.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q11.3

Question 12.
For the following frequency distribution, find :
(i) the median
(ii) lower quartile
(iii) upper quartile
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q12.1
Solution:
Writing the distribution in cumulative frequency (c.f.) table :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q12.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 Q12.3

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS

More Exercise

Choose the correct answer from the given four options (1 to 32) :

Question 1.
In a cylinder, if radius is halved and height is doubled then the volume will be
(a) same
(b) doubled
(c) halved
(d) four times
Solution:
Let radius of cylinder = r
and height = h
then volume = πr²h
If the radius is halved and the height is doubled
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q1.1

Question 2.
In a cylinder, if the radius is doubled &nd height is halved then its curved surface area will be
(a) halved
(b) doubled
(c) same
(d) four times
Solution:
Let radius of a cylinder = r
and height = h
Then curved surface area = 2πrh
Now if radius is doubled and height is halved,
then curved surface area = 2π\(\\ \frac { r }{ 2 } \) × 2h = 2πrh
which is same (c)

Question 3.
If a well of diameter 8 m has been dug to the depth of 14 m, then the volume of the earth dug out is
(a) 352 m3
(b) 704 m3
(c) 1408 m3
(d) 2816 m3
Solution:
Diameter of a well = 8 m
Radius (r) = \(\\ \frac { 8 }{ 2 } \) = 4m
Depth (h) = 14 m
Volume of the earth dug put = πr2h
= \(\\ \frac { 22 }{ 7 } \) × 4 × 4 × 14 m3
= 704 m3 (b)

Question 4.
If two cylinders of the same lateral surface have their radii in the ratio 4 : 9, then the ratio of their heights is
(a) 2 : 3
(b) 3 : 2
(c) 4 : 9
(d) 9 : 4
Solution:
Ratio in two cylinder having same lateral surface area their radii is 4 : 9
Let r1 be the radius of the first and r2 be the second cylinder
and h1, h2 and their heights
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q4.1
Ratio in their heights = 9 : 4 (d)

Question 5.
The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is
(a) 10 : 17
(b) 20 : 27
(c) 17 : 27
(d) 20 : 37
Solution:
Radii of two cylinder are in the ratio = 2 : 3
Let radius (r1) = 2x
and radius (r2) = 3x
Ratio in their height = 5 : 3
Let height of the first cylinder = 5y
and of second = 3y
Now, volume of the first cylinder
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q5.1

Question 6.
The total surface area of a cone whose radius is \(\\ \frac { r }{ 2 } \) and slant height 2l is
(a) 2πr (l + r)
(b) \(\pi r\left( l+\frac { r }{ 4 } \right) \)
(c) πr(l + r)
(d) 2πrl
Solution:
Radius of a cone = \(\\ \frac { r }{ 2 } \)
and slant height = 2l
total surface area of a cone
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q6.1

Question 7.
If the diameter of the base of cone is 10 cm and its height is 12 cm, then its curved surface area is
(a) 60π cm2
(b) 65π cm2
(c) 90π cm2
(d) 120π cm2
Solution:
Diameter of the base of a cone = 10 cm
Radius (r) = \(\\ \frac { 10 }{ 2 } \) = 5 cm
and height (h) = 12 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q7.1

Question 8.
If the diameter of the base of a cone is 12 cm and height is 20 cm, then its volume is ,
(a) 240π cm3
(b) 480π cm3
(c) 720π cm3
(d) 960π cm3
Solution:
Diameter of the base of a cone = 12 cm
Radius (r) = \(\\ \frac { 12 }{ 2 } \) = 6 cm
and height (h) = 20 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q8.1

Question 9.
If the radius of a sphere is 2r, then its volume will be
(a) \(\frac { 4 }{ 3 } \pi { r }^{ 3 } \)
(b) \(4\pi { r }^{ 3 }\)
(c) \(\frac { 8\pi { r }^{ 3 } }{ 3 } \)
(d) \(\frac { 32\pi { r }^{ 3 } }{ 3 } \)
Solution:
Radius of a sphere = 2r
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q9.1

Question 10.
If the diameter of a sphere is 16 cm, then its surface area is
(a) 64π cm2
(b) 256π cm2
(c) 192π cm2
(d) 256 cm2
Solution:
Diameter of a sphere = 16 cm
Radius (r) = \(\\ \frac { 16 }{ 2 } \) = 8 cm
Surface area = 4π2 = 4π × 8 × 8 cm2 = 256π cm2 (b)

Question 11.
If the radius of a hemisphere is 5 cm, then its volume is
(a) \(\frac { 250 }{ 3 } \pi { \quad cm }^{ 3 }\)
(b) \(\frac { 500 }{ 3 } \pi { \quad cm }^{ 3 } \)
(c) \(75\pi { \quad cm }^{ 3 } \)
(d) \(\frac { 125 }{ 3 } \pi { \quad cm }^{ 3 } \)
Solution:
Radius of a hemisphere (r) = 5 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q11.1

Question 12.
If the ratio of the diameters of the two spheres is 3 : 5, then the ratio of their surface areas is
(a) 3 : 5
(b) 5 : 3
(c) 27 : 125
(d) 9 : 25
Solution:
Ratio in the diameters of two spheres = 3 : 5
Let radius of the first sphere = 3x cm
and radius of the second sphere = 5x cm
Ratio in their surface area
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q12.1

Question 13.
The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is
(a) 1 : 4
(b) 1 : 3
(c) 2 : 3
(d) 2 : 1
Solution:
Radius of balloon (hemispherical) in the original position = 6 cm
and in increased position = 12 cm
Ratio in their surface areas
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q13.1

Question 14.
The shape of a Gilli, in the game of Gilli- danda, is a combination of
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q14.1
(a) two cylinders
(b) a cone and a cylinders
(c) two cones and a cylinder
(d) two cylinders and a cone
Solution:
The shape of a Gilli is the combination of
two cones and a cylinder (as shown in the figure). (c)

Question 15.
If two solid hemisphere of same base radius r are joined together along with their bases, then the curved surface of this new solid is
(a) 4πr2
(b) 6πr2
(c) 3πr2
(d) 8πr2
Solution:
Radius of two solid hemispheres = r
These are joined together along with the bases
Curved surface area = 2π2 × 2 = 4πr2 (a)

Question 16.
During conversion of a solid from one shape to another, the volume of the new shape will
(a) increase
(b) decrease
(c) remain unaltered
(d) be doubled
Solution:
During the conversion of a solid into another,
the volume of the new shaper will be the same.
i.e. remain unaltered (c)

Question 17.
If a solid of one shape is converted to another, then the surface area of the new solid
(a) remains same
(b) increases
(c) decreases
(d) can’t say
Solution:
If a solid of one shape has conversed into another then
the surface area of the new solid will same or not same
i.e. can’t say. (d)

Question 18.
If a marble of radius 2.1 cm is put into a cylindrical cup full of water of radius 5 cm and height 6 cm, then the volume of water that flows out of the cylindrical cup is
(a) 38.8 cm3
(b) 55.4 cm3
(c) 19.4 cm3
(d) 471.4 cm3
Solution:
Radius of a marble = 2.1 cm
Volume of marble = \(\frac { 4 }{ 3 } \pi { { r }^{ 3 }\quad cm }^{ 3 }\)
= \(\\ \frac { 4 }{ 3 } \) x \(\\ \frac { 22 }{ 7 } \) × 2.1 × 2.1 × 2.1 cm3
= 38.88 cm3
= 38.8 cm3 (a)

Question 19.
The volume of the largest right circular cone that can be carved out from a cube of edge 4.2 cm is
(a) 9.7 cm3
(b) 77.6 cm3
(c) 58.2 cm3
(d) 19.4 cm3
Solution:
Edge of cube = 4.2 cm
Radius of largest cone cut out = \(\\ \frac { 42.2 }{ 2 } \) = 2.1 cm
and height = 4.2 cm
Volume = \(\frac { 1 }{ 3 } \pi { { r }^{ 2 }h } \)
= \(\\ \frac { 1 }{ 3 } \) x \(\\ \frac { 22 }{ 7 } \) × 2.1 × 2.1 × 4.2 cm3
= 19.404
= 19.4 cm3 (d)

Question 20.
The volume of the greatest sphere cut off from a circular cylindrical wood of base radius 1 cm and height 6 cm is
(a) 288 π cm3
(b) \(\frac { 4 }{ 3 } \pi \) cm3
(c) 6 π cm3
(d) 4 π cm3
Solution:
Radius of cylinder (r) = 1 cm
Height (h) = 6 cm
The largest sphere that can be cut off from the cylinder of radius 1 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q20.1

Question 21.
The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is
(a) 3 : 4
(b) 4 : 3
(c) 9 : 16
(d) 16 : 9
Solution:
Ratio in volumes of two spheres = 64 : 27
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q21.1

Question 22.
If a cone, a hemisphere and a cylinder have equal bases and have same height, then the ratio of their volumes is
(a) 1 : 3 : 2
(b) 2 : 3 : 1
(c) 2 : 1 : 3
(d) 1 : 2 : 3
Solution:
If a cone, a hemisphere and a cylinder have equal bases = r (say)
and height = h in each case and r = h
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q22.1

Question 23.
If a sphere and a cube have equal surface areas, then the ratio of the diameter of the sphere to the edge of the cube is
(a) 1 : 2
(b) 2 : 1
(c) √π : √6
(d) √6 : √π
Solution:
A sphere and a cube have equal surface area
Let a be the edge of a cube and r be the radius of the sphere, then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q23.1

Question 24.
A solid piece of iron in the form of a cuboid of dimensions 49 cm x 33 cm x 24 cm is moulded to form a sphere. The radius of the sphere is
(a) 21 cm
(b) 23 cm
(c) 25 cm
(d) 19 cm
Solution:
Dimension of a cuboid = 49 cm × 33 cm × 24 cm
Volume of a cuboid = 49 × 33 × 24 cm3
⇒ Volume of sphere = Volume of a cuboid
Volume of a sphere = 49 × 33 × 24 cm3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q24.1

Question 25.
If a solid right circular cone of height 24 cm and base radius 6 cm is melted and recast in the shape of a sphere, then the radius of the sphere is
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 12 cm
Solution:
Height of a circular cone (h) = 24 cm
and radius (r) = 6 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q25.1

Question 26.
If a solid circular cylinder of iron whose diameter is 15 cm and height 10 cm is melted and recasted into a sphere, then the radius of the sphere is
(a) 15 cm
(b) 10 cm
(c) 7.5 cm
(d) 5 cm
Solution:
Diameter of a cylinder = 15 cm
Radius = \(\\ \frac { 15 }{ 2 } \) cm
and height = 10 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q26.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q26.2

Question 27.
The number of balls of radius 1 cm that can be made from a sphere of radius 10 cm is
(a) 100
(b) 1000
(c) 10000
(d) 100000
Solution:
Radius of sphere (R) = 10 cm
Volume of sphere = \(\frac { 4 }{ 3 } \pi { R }^{ 3 }=\frac { 4 }{ 3 } \pi { (10) }^{ 3 }{ cm }^{ 3 }\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q27.1

Question 28.
A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form of a cone of base diameter 8 cm. The height of the cone is
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 18 cm
Solution:
The internal diameter of the metallic shell = 4 cm
and external diameter = 8 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q28.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q28.2

Question 29.
A cubical icecream brick of edge 22 cm is to be distributed among some children by filling icecream cones of radius 2 cm and height 7 cm upto its brim. The number of children who will get the icecream cones is
(a) 163
(b) 263
(c) 363
(d) 463
Solution:
Edge of a cubical icecream brick = 22 cm
Volume = a3 = (22)3 = 10648 cm3
Radius (r) of ice cream cone (r) = 2 cm
and height (h) = 7 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q29.1

Question 30.
Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is
(a) 4 cm
(b) 3 cm
(b) 2 cm
(d) 6 cm
Solution:
Diameter of cylinder = 2 cm
Radius = \(\\ \frac { 2 }{ 2 } \) = 1 cm
and height = 16 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q30.1

Question 31.
A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that \(\\ \frac { 1 }{ 8 } \) space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is
(a) 142296
(b) 142396
(c) 142496
(d) 142596
Solution:
Internal edge of a hollow cube = 22 cm
Volume = (side)3 = (22)3 = 22 × 22 × 22 cm3 = 10648 cm3
Diameter of spherical marble = 0.5 cm = \(\\ \frac { 1 }{ 2 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q31.1

Question 32.
In the given figure, the bottom of the glass has a hemispherical raised portion. If the glass is filled with orange juice, the quantity of juice which a person will get is
(a) 135 π cm3
(b) 117 π cm3
(c) 99 π cm3
(d) 36 π cm3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q32.1
Solution:
Radius of base of cylinder (r) = \(\\ \frac { 6 }{ 2 } \) cm = 3 cm
and height (h)= 15 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS Q32.2

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test

More Exercises

Question 1.
(i) If θ is an acute angle and cosec θ = √5 find the value of cot θ – cos θ.
(ii) If θ is an acute angle and tan θ = \(\\ \frac { 8 }{ 15 } \), find the value of sec θ + cosec θ.
Solution:
(i) θ is an acute angle.
cosec θ = √5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q1.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q1.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q1.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q1.4

Question 2.
Evaluate the following:
(i) \(2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right) \) – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°
(ii) \(\frac { { sec }29^{ O } }{ { cosec }61^{ O } } \) + 2 cot 8° cot 17° cot 45° cot 73°0 cot 82° – 3(sin2 38° + sin2 52°)
(iii) \(\frac { { sin }^{ 2 }{ 22 }^{ O }+{ sin }^{ 2 }{ 68 }^{ O } }{ { cos }^{ 2 }{ 22 }^{ O }+{ cos }^{ 2 }{ 68 }^{ O } } \) + sin2 63° + cos 63° sin 27°
Solution:
(i) \(2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right) \) – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q2.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q2.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q2.3

Question 3.
If \(\\ \frac { 4 }{ 3 } \) (sec2 59° – cot2 31°) – \(\\ \frac { 2 }{ 2 } \) sin 90° + 3tan2 56° tan2 34° = \(\\ \frac { x }{ 2 } \), then find the value of x.
Solution:
Given
\(\\ \frac { 4 }{ 3 } \) (sec2 59° – cot2 31°) – \(\\ \frac { 2 }{ 2 } \) sin 90° + 3tan2 56° tan2 34° = \(\\ \frac { x }{ 2 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q3.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q3.2

Prove the following (4 to 11) identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

Question 4.
(i) \(\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA \)
(ii) \(\frac { cosA }{ cosecA+1 } +\frac { cosA }{ cosecA-1 } =2tanA \)
Solution:
(i) \(\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA \)
L.H.S = \(\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q4.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q4.2

Question 5.
(i) \(\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta \)
(ii) (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1.
Solution:
(i) \(\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta \)
L.H.S = \(\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q5.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q5.2

Question 6.
(i) sin2 θ + cos4 θ = cos2 θ + sin4 θ
(ii) \(\frac { cot\theta }{ cosec\theta +1 } +\frac { cosec\theta +1 }{ cot\theta } =2sec\theta \)
Solution:
L.H.S. = sin2 θ + cos4 θ
= (1 – cos2 θ + cos4 θ
= 1 – cos2 θ + cos4 θ
= 1 – cos2 θ (1 – cos2 θ)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q6.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q6.2

Question 7.
(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1
(ii) \(\frac { 1 }{ sinA+cosA+1 } +\frac { 1 }{ sinA+cosA-1 } =secA+cosecA\)
Solution:
(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1
L.H.S = sec4 A (1 – sin4 A) – 2 tan2 A
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q7.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q7.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q7.3

Question 8.
(i) \(\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1\)
(ii) (sec A – tan A)2 (1 + sin A) = 1 – sin A.
Solution:
(i) \(\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1\)
L.H.S = \(\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q8.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q8.2

Question 9.
(i) \(\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA \)
(ii) (sec A – cosec A) (1 + tan A + cot A) = tan A sec A – cot A cosec A
(iii) \(\frac { { tan }^{ 2 }\theta }{ { tan }^{ 2 }\theta -1 } -\frac { { cosec }^{ 2 }\theta }{ { sec }^{ 2 }\theta -{ cosec }^{ 2 }\theta } =\frac { 1 }{ { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta } \)
Solution:
(i) \(\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA \)
L.H.S = \(\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q9.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q9.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q9.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q9.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q9.5

Question 10.
\(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 } \)
Solution:
\(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 } \)
L.H.S = \(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q10.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q10.2

Question 11.
2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = θ
Solution:
2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = θ
L.H.S = 2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q11.1

Question 12.
If cot θ + cos θ = m, cot θ – cos θ = n, then prove that (m2 – n2)2 = 16 run.
Solution:
cot θ + cos θ = m…..(i)
cot θ – cos θ = n……(ii)
Adding (i)&(ii) we get
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q12.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q12.2

Question 13.
If sec θ + tan θ = p, prove that sin θ = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)
Solution:
sec θ + tan θ = p,
prove that sin θ = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)
\(\frac { 1 }{ cos\theta } +\frac { sin\theta }{ cos\theta } =p\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q13.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q13.2

Question 14.
If tan A = n tan B and sin A = m sin B, prove that cos2 A = \(\frac { { m }^{ 2 }-1 }{ { n }^{ 2 }-1 } \)
Solution:
m = \(\\ \frac { sinA }{ sinB } \)
n = \(\\ \frac { tanA }{ tanB } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q14.1

Question 15.
If sec A = \(x+ \frac { 1 }{ 4x } \), then prove that sec A + tan A = 2x or \(\\ \frac { 1 }{ 2x } \)
Solution:
sec A = \(x+ \frac { 1 }{ 4x } \)
To prove that sec A + tan A = 2x or \(\\ \frac { 1 }{ 2x } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q15.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q15.2

Question 16.
When 0° < θ < 90°, solve the following equations:
(i) 2 cos2 θ + sin θ – 2 = 0
(ii) 3 cos θ = 2 sin2 θ
(iii) sec2 θ – 2 tan θ = 0
(iv) tan2 θ = 3 (sec θ – 1).
Solution:
0° < θ < 90°
(i) 2 cos2 θ + sin θ – 2 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q16.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q16.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q16.3

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS

More Exercises

Choose the correct answer from the given four options (1 to 12) :

Question 1.
\({ cot }^{ 2 }\theta -\frac { 1 }{ { sin }^{ 2 }\theta } \) is equal to
(a) 1
(b) -1
(c) sin2 θ
(d) sec2 θ
Solution:
\({ cot }^{ 2 }\theta -\frac { 1 }{ { sin }^{ 2 }\theta } \)
= \(\frac { { cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta } -\frac { 1 }{ { sin }^{ 2 }\theta } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS Q1.1

Question 2.
(sec2 θ – 1) (1 – cosec2 θ) is equal to
(a) – 1
(b) 1
(c) 0
(d) 2
Solution:
(sec2 θ – 1) (1 – cosec2 θ)
= \(\left( \frac { 1 }{ { cos }^{ 2 }\theta } -1 \right) \left( 1-\frac { 1 }{ { sin }^{ 2 }\theta } \right) \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS Q2.1

Question 3.
\(\frac { { tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } \) is equal to
(a) 2 sin2 θ
(b) 2 cos2 θ
(c) sin2 θ
(d) cos2 θ
Solution:
\(\frac { { tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS Q3.1
= \(=\frac { { sin }^{ 2 }\theta }{ 1 } ={ sin }^{ 2 }\theta \) (c)

Question 4.
(cos θ + sin θ)2 + (cos θ – sin θ)2 is equal to
(a) – 2
(b) 0
(c) 1
(d) 2
Solution:
(cos θ + sin θ)2 + (cos θ – sin θ)2
= cos2 θ + sin2 θ + 2 sin θ cos θ + cos2 θ + sin2 θ – 2 sin θ cos θ
= 2(sin2 θ + cos2 θ)
= 2 × 1 = 2 (d)
(∵ sin2 θ + cos2 θ = 1)

Question 5.
(sec A + tan A) (1 – sin A) is equal to
(a) sec A
(b) sin A
(c) cosec A
(d) cos A
Solution:
(sec A + tan A) (1 – sin A)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS Q5.1

Question 6.
\(\frac { { 1+tan }^{ 2 }A }{ { 1+cot }^{ 2 }A } \) is equal to
(a) sec2 A
(b) – 1
(c) cot2 A
(d) tan2 A
Solution:
\(\frac { { 1+tan }^{ 2 }A }{ { 1+cot }^{ 2 }A } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS Q6.1

Question 7.
If sec θ – tan θ = k, then the value of sec θ + tan θ is
(a) \(1-\frac { 1 }{ k } \)
(b) 1 – k
(c) 1 + k
(d) \(\\ \frac { 1 }{ k } \)
Solution:
sec θ – tan θ = k
\(\frac { 1 }{ cos\theta } -\frac { sin\theta }{ cos\theta } =k\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS Q7.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS Q7.2

Question 8.
Which of the following is true for all values of θ (0° < θ < 90°):
(a) cos2 θ – sin2 θ = 1
(b) cosec2 θ – sec2 θ = 1
(c) sec2 θ – tan2 θ = 1
(d) cot2 θ – tan2 θ = 1
Solution:
∴ sec2 θ – tan2 θ = 1 is true for all values of θ as it is an identity.
(0° < θ < 90°) (c)

Question 9.
If θ is an acute angle of a right triangle, then the value of sin θ cos (90° – θ) + cos θ sin (90° – θ) is
(a) 0
(b) 2 sin θ cos θ
(c) 1
(d) 2 sin2 θ
Solution:
sin θ cos (90° – θ) + cos θ sin (90° – θ)
= sin θ sin θ + cos θ cos θ
{ ∵ sin(90° – θ) = cosθ, cos (90° – θ) = sin θ }
= sin2 θ + cos2 θ = 1 (c)

Question 10.
The value of cos 65° sin 25° + sin 65° cos 25° is
(a) 0
(b) 1
(b) 2
(d) 4
Solution:
cos 65° sin 25° + sin 65° cos 25°
= cos (90° – 25°) sin 25° + sin (90° – 25°) cos 25°
= sin 25° . sin 25° + cos 25° . cos 25°
= sin2 25° + cos2 25°
( ∵ sin2 θ + cos2 θ = 1)
= 1 (b)

Question 11.
The value of 3 tan2 26° – 3 cosec2 64° is
(a) 0
(b) 3
(c) – 3
(d) – 1
Solution:
3 tan2 26° – 3 cosec2 64°
= 3 tan2 26° – 3 cosec (90° – 26°)
= 3 tan2 26° – 3 sec2 26°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS Q11.1

Question 12.
The value of \(\frac { sin({ 90 }^{ O }-\theta )sin\theta }{ tan\theta } -1 \) is
(a) – cot θ
(b) – sin2 θ
(c) – cos2 θ
(d) – cosec2 θ
Solution:
\(\frac { sin({ 90 }^{ O }-\theta )sin\theta }{ tan\theta } -1 \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS Q12.1

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4

More Exercises

Question 1.
Draw a histogram for the following frequency distribution and find the mode from the graph :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q1.1
Solution:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q1.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q1.3
Mode = 14

Question 2.
Find the modal height of the following distribution by drawing a histogram :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q2.1
Solution:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q2.2
Now present the Height on x-axis and No. of students (frequency) on the y-axis
and draw a histogram as shown. In the histogram join AB and CD intersecting at M.
From M, draw MN to the x-axis. N shows the mode.
Hence mode = 174 cm

Question 3.
A Mathematics aptitude test of 50 students was recorded as follows :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q3.1
Draw a histogram for the above data using a graph paper and locate the mode. (2011)
Solution:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q3.2
Hence, the required mode is 82.5.

Question 4.
Draw a histogram and estimate the mode for the following frequency distribution :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q4.1
Solution:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q4.2
Representing classes on x-axis and frequency on the y-axis,
we draw a histogram as shown.
In the histogram, join AB and CD intersecting at M.
From M, draw ML perpendicular to the x-axis. L shows the mode
Hence Mode = 23

Question 5.
IQ of 50 students was recorded as follows
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q5.1
Draw a histogram for the above data and estimate the mode.
Solution:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q5.2
Representing the IQ scores on x-axis and number of students on the y-axis,
we draw a histogram as shown. Join AB and CD intersecting each other at M.
From M draw ML ⊥ x-axis. L is the mode which is 107

Question 6.
Use a graph paper for this question. The daily pocket expenses of 200 students in a school are given below:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q6.1
Draw a histogram representing the above distribution and estimate the mode from the graph.
Solution:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q6.2
Mark the upper comers of the highest rectangle
and the corners of the adjacent rectangles as A, B, C, D as shown.
Join AC and BD to intersect at P. Draw PM ⊥ x-axis.
Then the abscissa of M is 21, which is the required mode.
Hence, mode = 21

Question 7.
Draw a histogram for the following distribution :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q7.1
Hence estimate the modal weight.
Solution:
We write the given distribution in the continuous form :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q7.2
Representing the weight (in kg) on x-axis
and No. of students on y-axis. We draw a histogram as shown.
Now join AB and CD intersecting each other at M.
From M, draw ML perpendicular to x-axis.
L is the mode which is 51.5 kg

Question 8.
Find the mode of the following distribution by drawing a histogram
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q8.1
Also state the modal class.
Solution:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 Q8.2
Representing class on x-axis and frequency on the y-axis,
we draw a histogram as shown.
Join AB and CD intersecting each other at M.
From M, draw ML perpendicular to the x-axis.
L shows the mode which is 30.5 and class is 27-33.

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4, drop a comment below and we will get back to you at the earliest.