# 10th Maths Chapter 6 Trigonometry Exercise 6.5

10th Standard Maths Chapter 6 Trigonometry Exercise 6.5 Guide. TN SSLC Samacheer Kalvi Guide Chapter 6 Exercise 6.5 Book Back Answers & Solutions. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here.

Multiple choice questions:

1. The value of sin2θ+11+tan2θ is equal to
(1) tan2θ
(2) 1
(3) cot2θ
(4) 0
Solution:
(2) 1
Hint:

2.tan θ cosec2θ – tan θ is equal to
(1) sec θ
(2) cot2θ
(3) sin θ
(4) cot θ
Solution:
(4) cot θ
Hint:

3.If (sin α + cosec α)2 + (cos α + sec α)2 = k + tan2 α + cot2 α, then the value of k is equal to
(1) 9
(2) 7
(3) 5
(4) 3
Solution:
(2) 7
Hint:
(sin α + cosec α)2 + (cos α + sec α)2 = k + tan2 α + cot2 α
sin2 α + cosec2 α + 2 sin α cosec α + cos2 α + sec2 α + 2 cos α sec α = k + tan2 α + cot2 α
sin2 α + cos2 α + cosec2 α + sec2 α + 2 sin α × 1sinα + 2 cos α × 1cosα = k + tan2 α + cot2 α
1 + 1 + cot2 α + 1 + tan2 α + 2 + 2 = k + tan2 α + cot2 α
7 + cot2 α + tan2 α = k + tan2 α + cot2 α
∴ k = 7

4.If sin θ + cos θ = a and sec θ + cosec θ = b, then the value of b(a2 – 1) is equal to
(1) 2a
(2) 3a
(3) 0
(4) 2ab
Solution:
(1) 2a
a = sin θ + cos θ
b = sec θ + cosec θ

#### 5.If 3x = sec θ and 5x = tan θ, then x2−1×2 is (1) 25 (2) 1 / 25 (3) 5 (4) 1Solution:

(2) 1 / 25

6.If sin θ = cos θ , then 2 tan2θ + sin2θ – 1 is equal to
(1) −3/2
(2) 3/2
(3) 2/3
(4) −2/3
Solution:
(2) 3/2
Hint:

7.If x = a tan θ and y = b sec θ then

Solution:

Hint:

8.(1 + tan θ + sec θ) (1 + cot θ – cosec θ) is equal to
(1) 0
(2) 1
(3) 2
(4) -1
Solution:
(3) 2
Hint:

9. a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2– q2is equal to ………….
(1) a2 – b2
(2) b2 – a2
(3) a2 + b2
(4) b – a
(2) b2 – a2
Hint:
p2 – q2 = (p + q) (p – q)
= (a cot θ + b cosec θ + b cot θ + a cosec θ) (a cot θ + b cosec θ – b cot θ – a cosec θ)
= [cot θ (a + b) + cosec θ (a + b)] [cot θ (a – b) + cosec θ (b – a)]
= (a + b) [cot θ + cosec θ] (a – b) [cosec θ (a – b)]
= (a + b) [cot θ + cosec θ] (a – b) [cot θ – cosec θ]
= (a + b) (a – b) (cot2 θ – cosec2 θ)
= (a2 – b2) (-1) = – (a2 – b2)
p2 – q2 = b2 – a2

10.If the ratio of the height of a tower and the length of its shadow is 3–√ : 1, then the angle of elevation of the sun has measure.
(1) 45°
(2) 30°
(3) 90°
(4) 60°
Solution:
(4) 60°
Hint:

11. The electric pole subtends an angle of 30° at a point on the same level as its foot. At a second point ‘b’ metres above the first, the depression of the foot of the tower is 60°. The height of the tower (in metres) is equal to

Solution:                                                                                                                                                                                        (2) b / 3                                                                                                                                                                                              Hint:

12. A tower is 60 m in height. Its shadow is x meters shorter when the sun’s altitude is 45° than when it has been 30°, then x is equal to
(1) 41.92 m
(2) 43.92 m
(3) 43 m
(4) 45.6 m°
Solution:
(2) 43.92 m
Hint:

13. The angle of depression of the top and bottom of a 20 m tall buildings from the top of a multistoried building are 30° and 60° respectively. The height of the multistoried building and the distance between two buildings (in meters) is
(1) 20, 103–√
(2) 30, 53–√
(3) 20, 10
(4) 30, 103–√
Solution:
(4) 30, 103–√
Hint:

∴ Height of tower = 20 + 10 = 30 m
distance = 17.32 m = 103–√

14.Two persons are standing ‘x’ metres apart from each other and the height of the first person is double that of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the shorter person (in metres) is

Solution:

Hint:

15.The angle of elevation of a cloud from a point h metres above a lake is β. The angle of depression of its reflection in the lake is 45°. The height of location of the cloud from the lake is

Solution:
(1) h(1+tanβ)/1−tanβ
Hint:

Solution:

Hence proved.

### 3.If x sin3θ + y cos3θ = sin θ cos θ and x sin θ = y cos θ , then prove that x2 + y2 = 1.

Solution:
x sin3θ +y cos3θ= sinθ cosθ ; x sinθ y cosθ.
x (sinθ) [sin2θ + cos2θ] = sinθ cosθ

Solution:

Hence Proved.

### 5. A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such away that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Determine the speed at which the bird flies. ( √3 = 1.732).

Solution:
Let s be the speed of the bird. In 2 seconds, the bird goes from C to D, it covers a distance ‘d’

### 6. An aeroplane is flying parallel to the Earth’s surface at a speed of 175 m/sec and at a height of 600 m. The angle of elevation of the aeroplane from a point on the Earth’s surface is 37° at a given point. After what period of time does the angle of elevation increase to 53°? (tan 53° = 1.3270, tan 37° = 0.7536)

Solution:
Let Plane’s initial position be A. Plane’s final position = D Plane travels from A ➝ D.

### 7. A bird is flying from A towards B at an angle of 35°, a point 30 km away from A. At B it changes its course of flight and heads towards C on a bearing of 48° and a distance of 32 km away.

• (i) How far is B to the North of A?
• (ii) How far is B to the West of A?
• (iii) How far is C to the North of B?
• (iv) How far is C to the East of B?

(sin 55° = 0.8192, cos 55° = 0.5736,
sin 42° = 0.6691, cos 42° = 0.7431)
Solution:

### 8. Two ships are sailing in the sea on either side of the lighthouse. The angles of depression of two ships as observed from the top of the lighthouse are 60° and 45° respectively. If the distance between the ships is 200(√3+1/√3) metres, find the height of the lighthouse.

Solution:
From the figure AB – height of the light house = h CD – Distance between the ships

∴ The height of the lighthouse is 200 meters.

Solution: