RD Sharma Class 10 Solutions Chapter 5 Trigonometric Ratios Ex 5.3
RD Sharma Class 10 Solutions Trigonometric Ratios Exercise 5.3
Question 1.
Evaluate the following :
Solution:
Question 2.
Evaluate the following :
Solution:
Question 3.
Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°
(i) sin 59° + cos 56°
(ii) tan 65° + cot 49“
(iii) sec 76° + cosec 52°
(iv) cos 78° + sec 78°
(v) cosec 54° + sin 72°
(vi) cot 85″ + cos 75°
(vii) sin 67° + cos 75°
Solution:
(i) sin 59° + cos 56°
= sin (90° – 31°) + cos (90° – 34°)
= cos 31° +sin 34°
(ii) tan 65° + cot 49°
= tan (90° – 25°) + cot (90° – 41°)
= cot 25° + tan 41°
(iii) sec 76° + cosec 52°
= sec (90° – 14°) + cosec (90 0 – 38°)
= cosec 14° + sec 38°
(iv) cos 78° + sec 78°
= cos (90° – 12°) + sec (90°- 12°)
= sin 12° + cosec 12°
(v) cosec 54° + sin 72°
= cosec (90° – 36°) + sin (90°-18°)
= sec 36° + cos 18°
(vi) cot 85° + cos 75°
= cot (90° – 5°) + cos (90° – 15°)
= tan 5° + sin 15°
(vii) sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°
Question 4.
Express cos 75° + cot 75° in terms of angles between 0° and 30°.
Solution:
cos 75° + cot 75° = cos (90° – 15°) + cot (90°-15°)
= sin 15° + tan 15°
Question 5.
If sin 3A = cos (A – 26°), where 3A is an acute angle, And the value of A.
Solution:
sin 3A = cos (A – 26°)
⇒ cos (90° – 3A) = cos (A – 26°)
Comparing,
90° – 3A = A – 26°
⇒ 90° + 26° = A + 3A ⇒ 4A = 116°
Question 6.
If A, B, C are the interior angles of a triangle ABC, prove
Solution:
Question 7.
Prove that :
Solution:
Question 8.
Prove the following :
Solution:
Question 9.
Evaluate :
Solution:
Question 10.
If sin θ= cos (θ – 45°), where θ and (θ – 45°) are acute angles, find the degree measure of θ.
Solution:
Question 11.
If A, B, C are the interior angles of a AABC, show that :
(i) \(sin\frac { B+C }{ 2 } cos\frac { A }{ 2 }\)
(ii) \(cos\frac { B+C }{ 2 } sin\frac { A }{ 2 }\)
Solution:
Question 12.
If 2θ + 45° and 30° – θ are acute angles, find the degree measures of θ satisfying sin (20 + 45°) = cos (30° – θ).
Solution:
Question 13.
If θ is a positive acute angle such that sec θ = cosec 60°, And the value of 2 cos2 θ-1.
Solution:
Question 14.
If cos 2 θ – sin 4 θ, where 2 θ and 4 θ are acute angles, find the value of θ.
Solution:
Question 15.
If sin 3 θ = cos (θ – 6°), where 3 θ and θ – 6° are acute angles, find the value of θ.
Solution:
Question 16.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
Question 17.
If sec 2A = cosec (A – 42°), where 2A is an acute angle, find the value of A. (C.B.S.E. 2008)
Solution:
RD Sharma Class 10 Solutions Chapter 5 Trigonometric Ratios Ex 5.3
RD Sharma Class 10 Solutions Chapter 5 Trigonometric Ratios Ex 5.3
RD Sharma Class 10 Solutions Chapter 5 Trigonometric Ratios Ex 5.3
RD Sharma Class 10 Solutions Chapter 5 Trigonometric Ratios Ex 5.3
