# Free Applications of Trigonometry 01 Practice Test - 10th Grade

### Question 1

State whether the given statement is true or false:

The angle of depression is the angle between the line of sight of an observer below his horizontal line of sight.

True

False

#### SOLUTION

Solution :A

The angle of depression is the angle between the line of sight of an observer below his horizontal line of sight.

### Question 2

State whether the given statement is true or false:

The angle of elevation is the angle between the line of sight of an observer below his horizontal line of sight.

True

False

#### SOLUTION

Solution :B

The angle of elevation is the angle between the line of sight of an observer above his horizontal line of sight.

### Question 3

A circus artist is climbing a 30 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. The height of the pole, if the angle made by the rope with the ground level is 30∘

m.

#### SOLUTION

Solution :

The figure above represents the given situation.

In the given figure,

sin30∘=ABAC⇒AB=sin30∘×AC =AC2=302=15m.

### Question 4

A circus artist is climbing a 30 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the distance of the pole to the peg in the ground, if the angle made by the rope with the ground level is 30∘.

5 m

15√3 m

18 m

20 m

#### SOLUTION

Solution :B

The figure above represents the given situation.

From the given firgure,

cos30∘=BCAC⇒BC=AC×cos30∘=30√32

=15√3m

### Question 5

A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 2m and is inclined at an angle of 30∘ to the ground. What should be the length of the slide?

3 m

1.5 m

2 m

4 m

#### SOLUTION

Solution :D

The given situation can be represented by the figure below

In right-angled triangle ABC,

sin∠ABC=ACAB=12⇒sin30∘=2AB⇒AB=212⇒AB=4m∴Length of the slide is 4m.

### Question 6

A contractor plans to install two slides for the children to play in a park for elder children. She wants to have a steep slide at a height of 6 m, and inclined at an angle of 60∘ to the ground. What should be the length of the slide (in m)?

3√3 m

5 m

4√3 m

6 m

#### SOLUTION

Solution :C

The above figure represents the given situation.

In the given right-agled triangle,

sin(∠ACB)=ABAC⇒sin60∘=6AC⇒AC=6√32=12√3=4√3 m∴length of slide=4√3 m

### Question 7

A kite is flying at a height of 30 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60∘. Find the length of the string, assuming that there is no slack in the string.

20√3 m

30 m

30√3 m

60 m

#### SOLUTION

Solution :A

The situation can be represented by the figure below:

In the given right-angled triangle:

sin(∠ACB)=ABAC

⇒sin60∘=ABAC

⇒AC=ABsin60∘=30√32=60√3=20√3

∴ Length of the string is 20√3 m.

### Question 8

A kite is flying at a height of 30 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 30∘. Find the length of the string(in meters), assuming that there is no slack in the string.

60 m

40 m

30 m

20√3 m

#### SOLUTION

Solution :A

The given situation can be represented by the figure below

In the given right-angled triangle,

sin30∘=ABAC⇒AC=ABsin30∘=3012=60m

Therefore length of the string is 60 m.

### Question 9

A 1.8 m tall man is standing at some distance from a 31.8 m tall building. The angle of elevation from his eyes to the top of the building increases from 30∘ to 60∘ as he walks towards the building. Find the distance he walked towards the building.

30 m

20√3 m

20 m

30√3 m

#### SOLUTION

Solution :B

The situation can be represented by the figure above,

tan(∠ACB)=ABBC⇒tan60∘=30BC⇒BC=30tan60∘=10√3m⇒tan(∠ADB)=ABBD⇒tan30∘=30BD⇒BD=301√3=30√3m⇒Distance walked=DC=BD−BC=30√3−10√3=20√3m

### Question 10

A 1.8 m tall man is standing at some distance from a building. The angle of elevation from his eyes to the top of the building increases from 30∘ to 60∘ as he walks 20 m towards the building. Find the height of the building (in m).

10 m

20√3 m

20 m

(10√3 + 1.8) m

#### SOLUTION

Solution :D

The situation can be represented by the figure above.

tan(∠ACB)=ABBC⇒BC=ABtan60∘.....(1)tan(∠ADB)=ABBD⇒tan30∘=ABBC+CD⇒BC+CD=ABtan30∘.....(2) on subtracting equation(1) from (2)⇒CD=ABtan30∘−ABtan60∘⇒AB=10√3

Hence, the height of the building is

= AB + BG = (10√3+1.8)m.