Applications Of Trigonometry(10th Grade > Mathematics ) Questions and answers for exam Preparation

10th Grade > Mathematics

APPLICATIONS OF TRIGONOMETRY MCQs

Total Questions : 58 | Page 5 of 6 pages

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

Discuss Question

Answer: Option B. -> 15

\sqrt{3}

m
:
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 42.

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 $30^{\circ}$

___

Discuss Question

Answer: Option B. -> 15

\sqrt{3}

m
:

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 43.

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

Discuss Question

Answer: Option D. -> 4 m
:
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 44.

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

Discuss Question

Answer: Option C. -> 4

\sqrt{3}

m
:
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 45.

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

Discuss Question

Answer: Option A. ->

20 \sqrt{3}

m
:
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 46.

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

Discuss Question

Answer: Option B. ->

20 \sqrt{3}

m
:
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 47.

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

Discuss Question

Answer: Option A. -> 60 m
:
A
The given situation can be represented by the figure below

A Kite Is Flying At A Height Of 30 M Above The Ground. The S...

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 48.

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

Discuss Question

Answer: Option D. -> (10

\sqrt{3}

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

Question 49.

A kite is attached to a string of length 20 $\sqrt{3}$ m is tied to a pole on the level ground. If the string of the kite makes an angle of elevation of $60^{\circ}$ with the ground, then the kite is flying at a height of 20 m.

True
False

Discuss Question

Answer: Option B. -> False
:
B

$s i n ∠ A C B = \frac{A B}{A C} \Rightarrow s i n 60^{\circ} = \frac{A B}{A C} \Rightarrow A B = A C \times s i n 60^{\circ} = 20 \sqrt{3} \times \frac{\sqrt{3}}{2} ∴ Height of the kite is 30 m .$

Question 50.

A tower is 100 $\sqrt{3}$ m high. Find the angle of elevation of its top from a point 100 m away from its foot.

$30^{\circ}$
$60^{\circ}$
$45^{\circ}$
$15^{\circ}$

Discuss Question

Answer: Option B. ->

60^{\circ}

:
B

Let the angle of elevation be $θ$ .
$\Rightarrow$ $t a n θ$ = $\frac{P e r p e n d i c u l a r}{B a s e}$
$\Rightarrow$ $t a n θ$ = $\frac{100 \sqrt{3}}{100}$ = $\sqrt{3}$
Therefore, $θ$ = $60^{\circ}$