RD Sharma Solutions Chapter 5 Trigonometric Ratios Exercise 5.2 Class 10 Maths

 Chapter Name RD Sharma Chapter 5 Trigonometric Ratios Book Name RD Sharma Mathematics for Class 10 Other Exercises Exercise 5.1Exercise 5.3 Related Study NCERT Solutions for Class 10 Maths

Exercise 5.2 Solutions

Evaluate each of the following (1 - 19) :

1. sin 45° sin 30° + cos 45° cos30°

Solution

2. Sin 60° cos 30° + cos 60° sin 30°

Solution

3. Cos 60° cos45° - sin 60°. sin 45°
Solution

4. sin2 30° + sin2 45° + sin2 60° + sin2 90°
Solution

5. cos2 30° + cos2 45° + cos2 60° + cos2 90°
Solution

6. tan2 30° + tan2 60° + tan2 45°
Solution

7. 2sin2 30° - 3cos2 45° + tan2 60°
Solution

8. sin2 30° cos2 45° + 4tan2 30° + (1/2)sin2 90° - 2 cos2 90° + (1/24)cos2 0°
Solution

9. 4(sin2 60° + cos4 30°) - 3(tan2 60° - tan2 45°) + 5cos2 45°
Solution

10. (cosec2 45° sec2 30°)(sin2 30° + 4 cot2 45° - sec2 60°)
Solution

11. cosec2 30° cos60° tan3 45° sin2 90° sec2 45° cot30°
Solution

12. cot2 30° - 2cos2 60° - (3/4) sec2 45° - 4sec2 30°
Solution

13. (cos0° + sin 45° + sin30°)(sin 90° - cos45° + cos60°)
Solution

14. (sin 30°- sin 90° + 2 cos 0°)/(tan° 30 tan 60°)
Solution

15. 4/cot2 30° + 1/sin2 60° - cos2 45°
Solution

16. 4(sin2 30° + cos2 60°) - 3(cos2 45° - sin2 90°) - sin2 60°
Solution

17. (tan2 60° + 4cos2 45° + 3 sec2 30° + 5cos2 90°)/(cosec 30° + sec 60° - cot2 30°
Solution

18. sin 30°/sin 45° + tan 45°/sec 60° - sin 60°/cot 45° - cos 30°/sin 90°
Solution

19. tan 45°/cosec 30° + sec 60°/cot 45° - 5sin 90°/2cos 0°
Solution

20. 2sin 3x = √3 s = ?
Solution
Sin 3x = √3/2
Sin 3x = sin 60°
Equating angles we get,
3x = 60°
x = 20°

21. 2 sin x/2 = 1   x = ?
Solution
Sin x/2 = 1/2
Sin x/2 = sin 30°
x/2 = 30°
x = 60°

22. √3 sin x = cos x
Solution
√3 tan x = 1
tan x = 1/√3
∴ tan x = tan 30°
x = 30°

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

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

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

26. If Î¸ = 30° verify
(i) tan 2Î¸ = 2tanÎ¸/(1 - tan2 Î¸)
Solution
(i)

27. If A = B = 60°. Verify
(i) cos(A - B) = cosA cosB + sin A + sin B
Solution

28. If A = 30° B = 60° verify
(i) sin (A + B) = sin A cos B + cos A sin B
Solution

29. sin (A - B) = sin A cos B - cos A sin B
cos (A - B) = cos A cos B - sin A sin B
Find sin 15° cos 15°
Solution

30. In right angled triangle ABC. ∠C = 90°, ∠B = 60° . AB = 15 units. Find remaining angles and sides.
Solution

31. In Î”ABC is a right triangle such that ∠C = 90° , ∠A = 45° ,  BC = 7 units find ∠B, AB and AC.
Solution

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

33. If sin(A + B) = 1 and cos(A - B) = 1, 0° < A + B ≤ 90° A ≥ B. Fin A & B
Solution

34. If tan(A - B) = 1/√3 and tan(A + B) = √3, 0° < A + B ≤ 90° , A ≥ B, Find A & B.
Solution

35. If sin(A - B) = 1/2 and cos(A +B) = 1/2, 0° < A + B ≤ 90° , A > B, Find A & B
Solution

36. In right angled triangle Î”ABC at B, ∠A = ∠C. Find the values of
(i) sin A cos C + cos A sin C
Solution

37. Find acute angles A & B, if sin (A + 2B) = √3/2 cos(A + 4B) = 0, A > B>
Solution

38.  If A and B are acute angles such that tan A = 1/2 , tan B = 1/3 and tan (A + B) = (tanA + tan B)/(1 - tan A tan B)  , A + B = ?
Solution

39. In Î”PQR, right angled at Q, PQ = 3cm PR = 6cm. Determine ∠P = ? ∠R = ?
Solution