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 $30^{\circ}$

___

SOLUTION

Solution :

The figure above represents the given situation.
In the given figure,
$s i n 30^{\circ} = \frac{A B}{A C} \Rightarrow A B = s i n 30^{\circ} \times A C = \frac{A C}{2} = \frac{30}{2} = 15 m$ .

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 $30^{\circ}$ .

5 m

15 $\sqrt{3}$ m

18 m

20 m

SOLUTION

Solution : B

The figure above represents the given situation.
From the given firgure,
$c o s 30^{\circ} = \frac{B C}{A C} \Rightarrow B C = A C \times c o s 30^{\circ} = \frac{30 \sqrt{3}}{2}$
$= 15 \sqrt{3} m$

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 $30^{\circ}$ 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,
$s i n ∠ A B C = \frac{A C}{A B} = \frac{1}{2} \Rightarrow s i n 30^{\circ} = \frac{2}{A B} \Rightarrow A B = \frac{2}{\frac{1}{2}} \Rightarrow A B = 4 m ∴ L e n g t h o f t h e s l i d e i s 4 m .$

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 $60^{\circ}$ to the ground. What should be the length of the slide (in m)?

3 $\sqrt{3}$ m

5 m

4 $\sqrt{3}$ m

6 m

SOLUTION

Solution : C

The above figure represents the given situation.
In the given right-agled triangle,
$s i n (∠ A C B) = \frac{A B}{A C} \Rightarrow s i n 60^{\circ} = \frac{6}{A C} \Rightarrow A C = \frac{6}{\frac{\sqrt{3}}{2}} = \frac{12}{\sqrt{3}} = 4 \sqrt{3} m ∴ l e n g t h o f s l i d e = 4 \sqrt{3} m$

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 $60^{\circ}$ . Find the length of the string, assuming that there is no slack in the string.

$20 \sqrt{3}$ m

$30$ m

$30 \sqrt{3}$ m

$60$ m

SOLUTION

Solution : A

The situation can be represented by the figure below:

In the given right-angled triangle:
$s i n (∠ A C B) = \frac{A B}{A C}$
$\Rightarrow s i n 60^{\circ} = \frac{A B}{A C}$
$\Rightarrow A C = \frac{A B}{s i n 60^{\circ}} = \frac{30}{\frac{\sqrt{3}}{2}} = \frac{60}{\sqrt{3}} = 20 \sqrt{3}$
$∴$ Length of the string is $20 \sqrt{3} m .$

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 $30^{\circ}$ . Find the length of the string(in meters), assuming that there is no slack in the string.

60 m

40 m

30 m

20 $\sqrt{3}$ m

SOLUTION

Solution : A

The given situation can be represented by the figure below

In the given right-angled triangle,
$s i n 30^{\circ} = \frac{A B}{A C} \Rightarrow A C = \frac{A B}{s i n 30^{\circ}} = \frac{30}{\frac{1}{2}} = 60 m$
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 $30^{\circ}$ to $60^{\circ}$ as he walks towards the building. Find the distance he walked towards the building.

30 m

$20 \sqrt{3}$ m

$20$ m

$30 \sqrt{3}$ m

SOLUTION

Solution : B

The situation can be represented by the figure above,
$t a n (∠ A C B) = \frac{A B}{B C} \Rightarrow t a n 60^{\circ} = \frac{30}{B C} \Rightarrow B C = \frac{30}{t a n 60^{\circ}} = 10 \sqrt{3} m \Rightarrow t a n (∠ A D B) = \frac{A B}{B D} \Rightarrow t a n 30^{\circ} = \frac{30}{B D} \Rightarrow B D = \frac{30}{\frac{1}{\sqrt{3}}} = 30 \sqrt{3} m \Rightarrow D i s t a n c e w a l k e d = D C = B D - B C = 30 \sqrt{3} - 10 \sqrt{3} = 20 \sqrt{3} m$

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 $30^{\circ}$ to $60^{\circ}$ as he walks 20 m towards the building. Find the height of the building (in m).

10 m

20 $\sqrt{3}$ m

20 m

(10 $\sqrt{3}$ + 1.8) m

SOLUTION

Solution : D

The situation can be represented by the figure above.
$t a n (∠ A C B) = \frac{A B}{B C} \Rightarrow B C = \frac{A B}{t a n 60^{\circ}} . . . . . (1) t a n (∠ A D B) = \frac{A B}{B D} \Rightarrow t a n 30^{\circ} = \frac{A B}{B C + C D} \Rightarrow B C + C D = \frac{A B}{t a n 30^{\circ}} . . . . . (2) on subtracting equation(1) from (2) \Rightarrow C D = \frac{A B}{t a n 30^{\circ}} - \frac{A B}{t a n 60^{\circ}} \Rightarrow A B = 10 \sqrt{3}$
Hence, the height of the building is
= AB + BG = $(10 \sqrt{3} + 1.8) m .$

Free Applications of Trigonometry 01 Practice Test - 10th Grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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Areas Related to Circles 01
Surface Areas and Volumes 01