## RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles Ex 6

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles Ex 6.

Other Exercises

Question 1.
Solution:
On substituting the value of various T-ratios, we get
sin60° cos30° + cos60° sin30° Question 2.
Solution:
On substituting the value of various T-ratios, we get
cos60° cos30° – sin60° sin30° Question 3.
Solution:
On substituting the value of various Tratios, we get
cos45° cos30° + sin45° sin30° Question 4.
Solution:
On substituting the value of various Tratios, we get Question 5.
Solution: Question 6.
Solution:
On substituting the value of various Tratios, we get Question 7.
Solution:
On substituting the value of various Tratios, we get Question 8.
Solution:
On substituting the value of various Tratios, we get Question 9.
Solution:
On substituting the value of various Tratios, we get Question 10.
Solution:
(i) (ii) Question 11.
Solution:
(i) R.H.S. = L.H.S.
Hence, sin60° cos30° – cos60° sin30° = sin30°

(ii)
L.H.S. = cos60° cos30° + sin60° sin30°  (iii) R.H.S. = L.H.S.
Hence,2sin30° cos30° = sin60°

(iv) R.H.S. = sin90° = 1
R.H.S. = L.H.S.
Hence, 2 sin 45° cos45° = sin90°

Question 12.
Solution:
A = 45° 2 A = 90°

(i)Sin 2A = sin90° = 1 (ii) cos2A = cos90° = 0 Question 13.
Solution:
A = 30 ⇒ 2A = 60

(i) (ii) (iii) Question 14.
Solution:
(i) (ii) Question 15.
Solution:
(i) (ii) (iii) Question 16.
Solution: Hence, (A + B) = 45

Question 17.
Solution:
Putting A = 30° 2 A = 60° Question 18.
Solution:
Putting A = 30° 2 A = 60° Question 19.
Solution:
Putting A = 30° 2 A = 60° Question 20.
Solution:
From right angled ∆ABC, Question 21.
Solution:
From right angled ∆ABC, Question 22.
Solution:
From right angled  ∆ABC, (i) (ii)
By Pythagoras theorem Hence, (i) BC = 3cm and (ii) AB = 3cm.

Question 23.
Solution:
sin (A + B)= 1  sin (A + B) = sin90° Adding (1) and (2), we get
2A = 90° ⇒ A = 45°
Putting A = 45° in (1) we get
45° + B = 90° ⇒ B = 45°
Hence, A = 45° and B = 45°.

Question 24.
Solution: Solving (1) and (2), we get
2A = 90° ⇒ A = 45°
Putting A = 45° in (1), we get
45° – B = 30° ⇒ B = 45 – 30° = 15°
Hence, A = 45°, B = 15°.

Question 25.
Solution: Solving (1) and (2), we get
2A = 90° ⇒ A = 45°
Putting A = 45° in (1), we get
45° – B = 30° ⇒ B = 45° – 30°  = 15°
A = 45°, B = 15°

Question 26.
Solution: Question 27.
Solution: We hope the RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles Ex 6 help you. If you have any query regarding RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles Ex 6, drop a comment below and we will get back to you at the earliest.