 # Class 10 Maths Chapter 9 Some Applications of Trigonometry MCQs

Class 10 Maths MCQs Chapter 9 (Some applications of trigonometry) are available online here at BYJU’S with solved answers. All the objective questions are prepared as per NCERT guidelines and the CBSE syllabus. Practising these questions will help students to score good marks in the board exam. Also, click here to get chapter-wise multiple-choice questions.

## Class 10 Maths MCQs for Some Applications of Trigonometry

Solve the questions below for Chapter 9 (Some applications of trigonometry) and increase your chances of scoring high in this chapter. Also, get important questions for class 10 Maths here to practice more.

1. If the length of the shadow of a tree is decreasing then the angle of elevation is:

(a)Increasing

(b)Decreasing

(c)Remains the same

(d)None of the above

Explanation: See the following figure: As the shadow reaches from point D to C towards the direction of the tree, the angle of elevation increase from 30 to 60.

2. The angle of elevation of the top of a building from a point on the ground, which is 30 m away from the foot of the building, is 30°. The height of the building is:

(a)10 m

(b)30/√3 m

(c)√3/10 m

(d)30 m

Explanation: Say x is the height of the building.

a is a point 30 m away from the foot of the building.

Here, height is the perpendicular and distance between point a and foot of building is the base.

The angle of elevation formed is 30.

Hence, tan 30 = perpendicular/base = x/30

1/√3 = x/30

x=30/√3

3. If the height of the building and distance from the building foot’s to a point is increased by 20%, then the angle of elevation on the top of the building:

(a)Increases

(b)Decreases

(c)Do not change

(d)None of the above

Explanation: We know, for an angle of elevation θ,

Tan θ = Height of building/Distance from the point

If we increase both the value of the angle of elevation remains unchanged.

4. If a tower 6m high casts a shadow of 2√3 m long on the ground, then the sun’s elevation is:

(a)60°

(b)45°

(c)30°

(d)90°

Explanation: As per the given question: Hence,

tan θ = 6/2√3

tan θ = √3

tan θ = tan60°

θ = 60°

5. The angle of elevation of the top of a building 30 m high from the foot of another building in the same plane is 60°, and also the angle of elevation of the top of the second tower from the foot of the first tower is 30°, then the distance between the two buildings is:

(a)10√3 m

(b)15√3 m

(c)12√3 m

(d)36 m

Explanation: As per the given question: Hence,

tan60° = 30/x

√3 = 30/x

x = 30/√3

x = 10√3m

6. The angle formed by the line of sight with the horizontal when the point is below the horizontal level is called:

(a)Angle of elevation

(b)Angle of depression

(c)No such angle is formed

(d)None of the above

7. The angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level is called:

(a)Angle of elevation

(b)Angle of depression

(c)No such angle is formed

(d)None of the above

8. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. The height of the tower standing straight is:

(a)15√3

(b)10√3

(c)12√3

(d)20√3

Explanation: We know:

Tan (angle of elevation) = height of tower/its distance from the point

Tan 60 = h/15

√3 = h/15

h=15√3

9. The line drawn from the eye of an observer to the point in the object viewed by the observer is said to be

(a)Angle of elevation

(b)Angle of depression

(c)Line of sight

(d)None of the above