# MCQ Questions for Class 10 Maths Application of Trigonometry with Answers

Free PDF Download of CBSE Class 10 Maths Chapter 9 Application of Trigonometry Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Application of Trigonometry MCQs with Answers to know their preparation level.

## Class 10 Maths MCQs Chapter 9 Application of Trigonometry

1. The shadow of a tower is equal to its height at 10-45 a.m. The sun’s altitude is
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Explaination: Reason: Let the height of tower BC = rm and sun’s altitude = θ
Then Length of its shadow, AB = x m

In rt. ∆ABC, tan θ = $$\frac{BC}{AB}$$ = $$\frac{x}{x}$$ = 1
⇒ tan θ = tan 450
∴ θ = 45°

2. In given figure, the value of CE is

(a) 12 cm
(b) 6 cm
(c) 9 cm
(d) 6√3 cm

Explaination: Reason: In rt. ∆EBC, cos 60° = $$\frac{BC}{CE}$$
⇒ $$\frac{1}{2}$$ = $$\frac{6}{CE}$$
⇒ CE = 12 cm

3. In given figure, the value of ZC is

(a) 90°
(b) 45°
(c) 30°
(d) 60°

Explaination: Reason: In rt. ∆ABC, cos C = $$\frac{BC}{AB}$$ = $$\frac{7}{14}$$ = $$\frac{1}{2}$$
⇒ cos C = cos 60°
∴ C = 60°

4. In given Fig., the angle of depression from the observing position D and E of the object at A are

(a) 60°, 60°
(b) 30°, 30°
(c) 30°, 60°
(d) 60°, 30°

Explaination:

Reason: ∵ APD, ∠1 = 90° – 60° = 30°
∴ APE, ∠2 = ∠EAB …[alt Zs]
∴ ∠2 = 60°
Hence the angles of depression at D and E are 30° and 60° respectively.

5. In given figure, the length of AP is

Explaination:

6. In given figure, the value of AE is

Explaination: Reason: ∠AED = ∠EAB = 30°

In rt. ∆AED, sin 30° = $$\frac{AD}{AP}$$
⇒ $$\frac{1}{2}$$ = $$\frac{45}{AE}$$
⇒ AE = 90 cm

7. In given figure, AD = 4 cm, BD = 3 cm and CB = 12 cm. The value of tan θ is

AB² = AD² + BD² = (4)² + (3)² = 16 + 9 = 25
∴ AB = √25 = 5
∴ In rt ∆ABC, tan θ $$\frac{AB}{BC}$$ = $$\frac{5}{12}$$

8. In figure given ABCD is a rectangle, the value of CE is

(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm

Explaination: Reason: Since ABCD is a rectangle
∴ BC = AD = 8 cm and B = 90°
In rt ∆CBE, cos 60° = $$\frac{CE}{BC}$$
⇒ $$\frac{1}{2}$$ = $$\frac{CE}{8}$$
∴ CE = $$\frac{8}{2}$$ = 4 cm

9. In given figure, ABCD is a || gm. The lenght of AP is

(a) 2 cm
(b) 4 cm
(c) 6 cm
(d) 8 cm

Explaination: Reason: Since ABCD is a || gm
∴ AD = BC = 4√3
In rt ∆APD, sin 60° = $$\frac{AP}{AD}$$
⇒ $$\frac{\sqrt{3}}{2}=\frac{\mathrm{AP}}{4 \sqrt{3}}$$
⇒ 2AP = 4 × 3 = 12
∴ AP = 6 cm

10. When the length of shadow of a vertical pole is equal to √3 times of its height, the angle of elevation of the Sun’s altitude is
(a) 30°
(b) 45°
(c) 60°
(d) 15°

Explaination: Reason: Let the height of the vertical pole, BC = h m
∴ Shadow AB = √3 h m and the angle of elevation ZBAC = θ

In rt ∆ABC, tan θ = $$\frac{B C}{A B}=\frac{h}{\sqrt{3} h}=\frac{1}{\sqrt{3}}$$ = tan 30°
∴ θ = 30°
Hence the Sun’s altitude is 30°

11. The angle of elevation of top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. The length of the tower is
(a) √3 m
(b) 2√3 m
(c) 5√3m
(d) 10√3 m

Explaination:

12. A plane is observed to be approaching the airport. It is at a distance of 12 km from the point of observation and makes an angle of elevation of 60°. The height above the ground of the plane is
(a) 6√3 m
(b) 4√3 m
(c) 3√3 m
(d) 2√3 m

Explaination:

13. The upper part of a tree is broken by the wind and makes an angle of 30° with the ground. The distance from the foot of the tree to the point where the top touches the ground is 5 m. The height of the tree is
(a) 10√33 m
(b) 5√33 m
(c) √3 m
(d) √3/5 m

Explaination:

14. The angles of elevation of the top of a rock from the top and foot of 100 m high tower are respectively 30° and 45°. The height of the rock is
(a) 50 m
(b) 150 m
(c) 5o√3m
(d) 50(3 + √3)

15. The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with horizontal, the length of the wire is
(a) 6 m
(b) 10 m
(c) 12 m
(d) 20 m

16. The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30°. The distance of the car from the base of the tower (in m) is:
(a) 25√3
(b) 50√3
(c) 75√3
(d) 150

Explaination:

17. A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall is

Explaination:

18. The line drawn from the eye of an observer to the point in the object viewed by the observer is known as
(a) horizontal line
(b) vertical line
(c) line of sight
(d) transversal line

19. The tops of two poles of heights 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with the horizontal, then the length of the wire is
(a) 8 m
(b) 10 m
(c) 12 m
(d) 14 m

Explaination:

20. If two towers of heights h1 and h2 subtend angles of 60° and 30° respectively at the mid-point of the line joining their feet, then h1 : h2 =
(a) 1 : 2
(b) 1 : 3
(c) 2 : 1
(d) 3 : 1

Explaination:

21. The angle of elevation of the top of a tower from a point 20 metres away from its base is 45°. The height of the tower is
(a) 10 m
(b) 20 m
(c) 30 m
(d) 20√3 m

Explaination:

22. Two poles are 25 m and 15 m high and the line joining their tops makes an angle of 45° with the horizontal. The distance between these poles is
(a) 5 m
(b) 8 m
(c) 9 m
(d) 10 m

Explaination:

23. A portion of a 60 m long tree is broken by tornado and the top struck up the ground making an angle of 30° with the ground level. The height of the point where the tree is broken is equal to
(a) 30 m
(b) 35 m
(c) 40 m
(d) 20 m

Explaination:

24. The angle of elevation of the top of a 15m high tower at a point 15m away from the base of the tower is ____ .

Explaination:
Hints:
∵ Height of tower = distance of point from the base
∴ Angle of elevation = 45°.

25. The ratio of the height of a tower and the length of its shadow on the ground is √3 : 1. What is the angle of elevation of the sun? [Delhi 2017]

Explaination:

26. An observer, 1.5 m tall, is 28.5 m away from a 30 m high tower. Determine the angle of elevation of the top of the tower from the eye of the observer. [Delhi 2017 (C)]

Explaination:

Here, AB = DM = 1.5 m
CM = CD – DM
= 30 – 1.5 = 28.5 m
Let 0 be the angle of elevation of the top of the tower from the eye of the observer.
∴ In ∆ACM
tan θ = $$\frac{C M}{A M}=\frac{28.5}{28.5}$$
tan θ = 1
tan θ = tan 45°
θ = 45°

27. If a tower 30 m high, casts a shadow 10√3 m long on the ground, then what is the angle of elevation of the sun?
[AI 2017]

Explaination:

28. If two towers of height h1 and h2 subtends angles of 60° and 30° respectively at the mid points of line joining their feet, find h1 : h2