RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2

RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

Evaluate each of the following (1-19) : 1.
Question 1.
sin45° sin30° + cos45° cos30°.
Solution:
sin 45° sin 30° + cos 45° cos 30°

Question 2.
sin60° cos30° + cos60° sin30°.
Solution:

Question 3.
cos60° cos45° – sin60° sin45°.
Solution:

Question 4.
sin²30° + sin²45° + sin²60° + sin²290°.
Solution:

Question 5.
cos²30° + cos²45° + cos²60° + cos²90°.
Solution:

Question 6.
tan²30° + tan²60° + tan²45°.
Solution:

Question 7.
2sin²30° – 3cos²45° + tan²60°.
Solution:

Question 8.
sin²30°cos²4S° + 4tan²30° + \(\frac { 1 }{ 2 }\) sin²90° -2cos²90° + \(\frac { 1 }{ 24 }\) cos20°.
Solution:

Question 9.
4 (sin⁴ 60° + cos⁴ 30°) – 3 (tan² 60° – tan² 45°) + 5cos² 45°
Solution:

Question 10.
(cosecc² 45° sec² 30°) (sin² 30° + 4cot² 45° – sec² 60°).
Solution:

Question 11.
cosec³ 30° cos 60° tan³ 45° sin2 90° sec² 45° cot 30°.
Solution:

Question 12.
cot² 30° – 2cocs² 60° – \(\frac { 3 }{ 4 }\)sec2 45° – 4sec² 30°.
Solution:

Question 13.
(cos0° + sin45° + sin30°) (sin90° – cos45° + cos60°)
Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.
4 (sin⁴ 30° + cos² 60°) – 3 (cos² 45° – sin² 90°) – sin² 60°
Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Find the value of x in each of the following : (20-25)

Question 20.
2sin 3x = √3
Solution:

Question 21.
2sin \(\frac { x }{ 2 }\) = 1
Solution:

Question 22.
√3 sin x=cos x
Solution:

Question 23.
tan x = sin 45° cos 45° + sin 30°
Solution:

Question 24.
√3 tan 2x = cos 60° +sin 45° cos 45°
Solution:

Question 25.
cos 2x = cos 60° cos 30° + sin 60° sin 30°
Solution:

Question 26.
If θ = 30°, verify that :

Solution:

Question 27.
If A = B = 60°, verify that:
(i) cos (A – B) = cos A cos B + sin A sin B
(ii) sin (A – B) = sin A cos B – cos A sin B tan A – tan B
(iii) tan (A – B) = \(\frac { tanA-tanB }{ 1+tanA-tanB }\)
Solution:

Question 28.
If A = 30° and B = 60°, verify that :
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) cos (A + B) = cos A cos B – sin A sin B
Solution:

Question 29.
If sin (A + B) = 1 and cos (A,-B) = 1,0° < A + B < 90°, A > B find A and B.
Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.
In a ∆ABC right angle at B, ∠A = ∠C. Find the values of
(i) sin A cos C + cos A sin C
(ii) sin A sin B + cos A cos B
Solution:

Question 33.
Find acute angles A and B, if sin (A + 2B)=\(\frac { \sqrt { 3 } }{ 2 }\) and cos (A + 4B) = 0, A > B.
Solution:

Question 34.
In ΔPQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠P and ∠R.
Solution:

Question 35.
If sin (A – B) = sin A cos B – cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of sin 15° and cos 15°.
Solution:

Question 36.
In a right triangle ABC, right angled at ∠C if ∠B = 60° and AB – 15 units. Find the remaining angles and sides.
Solution:

Question 37.
If ΔABC is a right triangle such that ∠C = 90°, ∠A = 45° and BC = 7 units. Find ∠B, AB and AC.
Solution:

Question 38.
In a rectangle ABCD, AB = 20 cm, ∠BAC = 60°, calculate side BC and diagonals AC and BD.
Solution:

Question 39.
If A and B are acute angles such that

Solution:

Question 40.
Prove that (√3 + 1) (3 – cot 30°) = tan³ 60° – 2sin 60°. [NCERT Exemplar]
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2 are helpful to complete your math homework.

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