Chapter 5 Trigonometric Ratios R.D. Sharma Solutions for Class 10th Math Exercise 5.2
Exercise 5.2Evaluate each of the following (1 – 19):
Solution
4. Sin2 30° + sin2 45° + sin2 60° + sin2 60° + sin2 90°
Solution
5. cos2 30° + cos2 45° + cos2 60° + cos2 90°
Solution
6. tan2 30° + tan2 60° + tan2 45°
7. 2 sin2 30° - 3 cos2 45° + tan2 60°
8. sin2 30° cos2 45° + 4tan2 30° + 1/2 sin2 90° - 2 cos2 90° + 1/24 cos2 0° .
9. 4(sin4 60° + cos4 30°) + 3(tan2 60° - tan2 45°) + 5 cos2 45°
Solution
10. (cosec2 45° sec2 30°)(sin2 30° + 4 cot2 45° - sec2 60°)
Solution
11. cosec3 30° cos 60° tan3 45° + sin2 90° sec2 45° cot 30°
Solution
12. cot2 30° - 2cos2 60° - 3/4 sec2 45° - 4 sec2 30°
13. ( cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°)
Solution
14. sin 30° - sin 90° + 2 cos 0°/tan 30° tan 60°
Solution
15. 4/cot2 30° + 1/ sin2 60° - cos2 45°
16. 4(sin4 30° + cos2 60°) - 3(cos2 45° - sin2 90°) - sin2 60°
17. tan2 60° + 4 cos2 45° + 3 sec2 30° + 5 cos2 90°/cosec 30° + sec 60° - cot2 30°
18. sin 30°/sin 45° + tan 45°/sec 60° - sin 60°/cot 45° - cos 30° sin 90°
Solution
19. tan 45°/sec 60° - sin 60°/cot 45° - 5 sin 90°/ 2 cos 0°
Solution
20. 2sin 3x = √3/2 = ?
sin 3x = √3/2
sin 3x = sin60°
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°
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 θ = 30° verify
(i) Tan 2 θ = 2 tan θ/1-tan2 θ
(ii) Sin θ = 2 tan θ/1-tan2 θ
(iii) Cos 2 θ = 1-tan2θ/1+tan2θ
(iv) Cos 3 θ = 4 cos3 θ - 3 cos θ
Solution
(i)
Solution
29. If sin (A+B) = 1 and cos (A-B) = 1,0°< A+B ≤ 90°, find A and B.
(i)
(ii)
(iii)
(iv)
27. If A = B = 60° verify
(i) cos (A-B) = Cos A cos B + sin A sin B
(ii) sin (A-B) = sin A cos B - cos A sin B
(iii) tan (A-B) = tan A - tan B/1 + tan A tan B
Solution
(i)
(ii)
(iii)
28. If A = 30° B = 60° verify
(i) Sin(A+B) = Sin A Cos B + cos A sin B
(ii) Cos(A+B) = cos A cos B - sin A sin B
(i)
(ii)
Given :
Sin(A+B) = 1…(1)
Cos(A-B) = 1…(2)
We know that ,
Sin 90° = 1…(3)
Cos 0° = 1 …(4)
Now by comparing equation (1) and (3)
We get,
A+B = 90 …(5)
Now by comparing equation (2) and (4)
We get,
A-B=0 …(6)
Now we get the values of A and B, let us solve equation (5) and (6) simultaneously
Therefore by adding equation (5) and (6)
We get ,
A+B = 90
+A-B = 0
2A + 0 = 90
Therefore,
2A = 90
⇒ A = 90/2
⇒ A = 45°
Hence A = 45°
Now by subtracting equation (6) from equation (5)
We get,
A+B = 90
-A-B = 0
(-)(+)(-)
0+2B = 90
Therefore,
2B = 90
⇒ B = 90/2
⇒ B = 45°
Hence B = 45°
Therefore the values of A and B are as follows
A = 45° and B = 45°
30. If tan (A-B) 1/√3 and tan (A+B) = √3, 0° < A+B ≤ 90°, A≥B, Find A&B.
Solution
31. If sin (A-B) = 1/2 and cos (A+B) = 1/2 , 0°< A + B ≤ 90°, A < B find A and B .
Solution
Solution
Sin(A-B) = 1/2 …(1)
Cos(A+B) = 1/2 …(2)
We know that ,
sin 30° = 1/2 …(3)
cos 60° = 1/2 …(4)
Now by comparing equation (1) and (3)
We get,
A-B = 30 …(5)
Now by comparing equation (2) and (4)
We get,
A+B = 60 …(6)
Now to get the values of A and B, let us solve equation (5) and (6) simultaneously
Therefore by adding equation (5) and (6)
We get,
A-B = 30
+ A+B = 60
2A + 0 = 90
Therefore,
2A = 90
⇒ A = 90/2
⇒ A = 45°
Hence A = 45°
Now by subtracting equation (5) from equation (6)
We get,
A+B = 60
-A-B = 30
(-)(+)(-)
0 + 2B = 30
Therefore ,
2B = 30
⇒ B = 30/2
⇒ B = 15°
Hence B = 15°
Therefore the values of A and B are as follows
A = 45° and B = 15°
(i) sin A cos C + cos A sin C
(ii) sin A sin B + cos A cos B
Solution
Solution
34. In ∆PQR, right angled at Q, PQ = 3cm PR = 6cm. Determine ∠P = ? ∠R = ?
Solution
35. If sin (A-B) = sin A - 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
Solution
37 . If △ABC is a right triangle such that ∠C = 90° ∠A = 45°, BC = 7 units find ∠B , AB and AC .
33. Find the acute angles A&B, if sin (A+2B) = √3/2 cos(A+4B) = 0, A>B.
34. In ∆PQR, right angled at Q, PQ = 3cm PR = 6cm. Determine ∠P = ? ∠R = ?
Solution
35. If sin (A-B) = sin A - 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
36. In a right triangle ABC, right angled at C , if ∠B = 60° and AB = 15 units . Find the remaining angles and sides .
Solution
Solution
38. In rectangle ABCD AB = 20 cm ∠BAC = 60° BC, calculate side BC and diagonals AC and BD.
39. If A and B are acute angles such that tan A = 1/2 , tan B = 1/3 and tan (A+B) = tan A + tan B/1-tan A tan B , find A + B .
Solution
40. Prove that (√3+1) (3-cot 30°) = tan3 60° - 2 sin 60°.
Solution
