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**Also, check:** Heights and Distances (An application of trigonometry)

## Class 10 Maths Chapter 9 MCQs (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.

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**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

Answer: **(a) Increasing**

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

Answer: **(b) 30/√3 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

Answer: **(c) Do not change**

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°

Answer:** (a) 60°**

Explanation: As per the given question:

Hence,

tan θ = 6/2√3

tan θ = √3

tan θ = tan 60°

⇒ 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

Answer: **(a) 10√3 m**

Explanation: As per the given question:

Hence,

tan 60° = 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

Answer: **(b) Angle of depression**

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

**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

Answer: **(a) Angle of elevation**

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

**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 (in m) standing straight is:**

(a) 15√3

(b) 10√3

(c) 12√3

(d) 20√3

Answer: **(a) 15√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

Answer: **(c) Line of sight**

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

**10. The height or length of an object or the distance between two distant objects can be determined with the help of:**

(a) Trigonometry angles

(b) Trigonometry ratios

(c) Trigonometry identities

(d) None of the above

Answer: **(b) Trigonometry ratios**

The height or length of an object or the distance between two distant objects can be determined with the help of trigonometry ratios.

**11. When the shadow of a pole h metres high is √3h metres long, the angle of elevation of the Sun is**

(a) 30°

(b) 60°

(c) 45°

(d) 15°

Answer: **(a) 30°**

Explanation:

Let AB be the pole and BC be its shadow.

Consider θ is the angle of elevation of the Sun.

In right triangle ABC,

tan θ = AB/BC = h/√3h = 1/√3

tan θ = tan 30°

θ = 30°

**12. A ladder makes an angle of 60° with the ground, when placed along a wall. If the foot of ladder is 8 m away from the wall, the length of ladder is**

(a) 4 m

(b) 8 m

(c) 8√3 m

(d) 16 m

Answer: **(d) 16 m**

Explanation:

Let AB be the wall, AC be the length of the ladder.

In right angled triangle ABC,

cos 60° = BC/AC

½ = 8/AC

AC = 8 × 2 = 16

Therefore, the length of the ladder is 16 m.

**13. If the height and length of a shadow of a tower are the same, then the angle of elevation of Sun is**

(a) 30°

(b) 60°

(c) 45°

(d) 15°

Answer:** (c) 45°**

Explanation:

Let AB be the tower and BC be its shadow.

Consider the angle of elevation of the Sun as θ.

According to the given,

AB = BC

In right triangle ABC,

tan θ = AB/BC

tan θ = AB/AB {since AB = BC}

tan θ = 1

tan θ = tan 45°

θ = 45°

**14. The angle of depression of an object on the ground, from the top of a 25 m high tower is 30°. The distance of the object from the base of tower is**

(a) 25√3 m

(b) 50√3 m

(c) 75√3 m

(d) 50 m

Answer: **(a) 25√3**

Explanation:

Let AB be the tower and BC be the distance of the object (at C) from the base of the tower.

In right triangle ABC,

tan 30° = AB/BC

1/√3 = 25/BC

BC = 25√3 m

**15. The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it was 60°. The height of the tower is **

(a) 40√3 m

(b) 20√3

(c) 20 m

(d) 15√3 m

Answer: **(b) 20√3 m**

Explanation:

When the sun’s altitude is the angle of elevation of the top of the tower from the tip of the shadow.

Let AB be h m and BC be x m. From the question, DC is 40 m longer than BC.

So, BD = (40 + x) m

And two right triangles ABC and ABD are formed as shown in the figure.

In ΔABC,

tan 60° = AB/ BC

√3 = h/x

x = h/√3….(i)

In ΔABD,

tan 30° = AB/ BD

1/ √3 = h/ (x + 40)

x + 40 = √3h

h/√3 + 40 = √3h [using (i)]

h + 40√3 = 3h

2h = 40√3

h = 20√3

Therefore, the height of the tower is 20√3 m.

**16. If the angles of elevation of the top of a tower from two points at the distance of a m and b m from the base of tower and in the same straight line with it are complementary, then the height of the tower (in m) is**

(a) √(a/b)

(b) √ab

(c) √(a + b)

(d) √(a – b)

Answer: **(b) √ab**

Explanation:

If the angles of elevation of the top of a tower from two points at the distance of a m and b m from the base of tower and in the same straight line with it are complementary, then the height of the tower (in m) is √ab.

**17. From a point on a bridge across a river the angle of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at the height of 30 m from the banks, the width of the river is**

(a) 30(1 + √3) m

(b) 30(√3 – 1) m

(c) 30√3 m

(d) 60√3 m

Answer: **(a) 30(1 + √3) m**

Explanation:

The bridge is at a height of 30 m from the banks.

Let, A and B represent the points on the bank on opposite sides of the river. And, AB is the width of the river. P is a point on the bridge which is at the height of 30 m from the banks.

AB = AD + DB

In right ΔAPD,

∠A = 30°

So, tan 30° = PD/ AD

1/√3 = PD/ AD

AD = √3(30)

AD = 30√3 m

Next, in right ΔPBD

∠B = 45°

So, tan 45° = PD/ BD

1 = PD/ BD

BD = PD

BD = 30 m

We know that, AB = AD + DB = 30√3 + 30 = 30(√3 + 1)

Hence, the width of the river = 30(1 + √3)m

**18. The ratio of the height of a tower and the length of its shadow on the ground is √3 : 1. The angle of elevation of the Sun is**

(a) 30°

(b) 45°

(c) 60°

(d) 75°

Answer: **(c) 60°**

Explanation:

Let AB be the height and BC be its shadow.

θ be the angle of elevation of the Sun.

According to the given, AB : BC = √3 : 1

So, AB = √3x and BC = x

In right triangle ABC,

tan θ = AB/BC

tan θ = √3x/x

tan θ = √3

tan θ = tan 60°

θ = 60°

**19. A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with the ground. The distance between the foot of the tree to the point where the top touches the ground is 8 m. The height of the tree is**

(a) 4√3 m

(b) 8√3 m

(c) 6√3 m

(d) 16√3 m

Answer: **(b) 8√3 m**

Explanation:

Let AC be the initial height of the tree.

Due to the storm the tree is broken at B.

Let the bent portion of the tree be AB = x m and the remaining portion BC = h m.

So, the height of the tree AC = (x + h) m

And, given DC = 8m

In right ΔBCD,

tan 30° = BC/DC

1/√3 = h/8

h = 8/√3

Again in right ΔBCD,

cos 30° = DC/BD

√3/2 = 8/x

x = 16/√3 m

So, x + h = 16/√3 + 8/√3

= 24/√3

= (24 . √3)/(√3 . √3)

= (24√3)/3

= 8√3

Therefore, the height of the tree is 8√3 m.

**20. The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will be**

(a) Greater than 60°

(b) Equal to 30°

(c) Less than 60°

(d) Equal to 60°

Answer: **(c) Less than 60°**

Explanation:

Let AB = h be the height and BC = x.

In this case, the angle of elevation is 30°.

When height is doubled, the height of the tower will be PQ = 2h.

In this case, consider the angle of elevation as θ.

Also, BC = QR = x.

In triangle ABC,

tan 30° = AB/BC = h/x

1/√3 = h/x….(i)

In triangle PQR,

tan θ = PQ/QR

tan θ = 2h/x

tan θ = 2(1/√3) {from (i)}

tan θ = 1.15 (approx)

θ = tan^{-1}(1.15) < 60°.

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