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

10th Grade > Mathematics

APPLICATIONS OF TRIGONOMETRY MCQs

Total Questions : 58 | Page 3 of 6 pages

Question 21. A tower stands vertically on the ground. From a point on the ground, which is 24 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be

60^{\circ}

. Find the height of the tower.

26√3 m
24√3 m
24 m
26 m

Discuss Question

Answer: Option B. -> 24√3 m
:
B

A Tower Stands Vertically On The Ground. From A Point On The...

The above

Δ A B C

represents the given situation.
AB = 24 m
In Triangle ABC,

t a n 60^{\circ} = \sqrt{3} = \frac{C B}{A B} \Rightarrow A B \times \sqrt{3} = C B \Rightarrow C B = 24 \sqrt{3} m .

Hence, the height of the tower =

24 \sqrt{3} m .

Question 22. 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

A Contractor Plans To Install Two Slides For The Children To...

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 23. 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}

60^{\circ}

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

Discuss Question

Answer: Option D. -> (10√3 + 1.8) m
:
D

A 1.8 M Tall Man Is Standing At Some Distance From A Buildin...

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 24. 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

Discuss Question

Answer: Option B. -> False
:
B
The angle of elevation is the angle between the line of sight of an observer above his horizontal line of sight.

Question 25. 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

Discuss Question

Answer: Option A. -> True
:
A

The angle of depression is the angle between the line of sight of anobserverbelow his horizontal line of sight.

Question 26. 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√3 m
5 m
4√3 m
6 m

Discuss Question

Answer: Option C. -> 4√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 27. 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}

60^{\circ}

as he walks towards the building. Find the distance he walked towards the building.

30 m
20√3 m
20 m
30√3 m

Discuss Question

Answer: Option B. -> 20√3 m
:
B

A 1.8 M Tall Man Is Standing At Some Distance From A 31.8 M ...

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 28. 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}

___
m.

Discuss Question

A Circus Artist Is Climbing A 30 M Long Rope, Which Is Tight...

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 29.

A tower stands vertically on the ground. From a point on the ground, which is 24 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be $60^{\circ}$ . Find the height of the tower.

$26 \sqrt{3} m$
$24 \sqrt{3} m$
$24 m$
$26 m$

Discuss Question

Answer: Option B. ->

24 \sqrt{3} m

:
B

The above $Δ A B C$ represents the given situation.
AB = 24 m
In Triangle ABC,
$t a n 60^{\circ} = \sqrt{3} = \frac{C B}{A B} \Rightarrow A B \times \sqrt{3} = C B \Rightarrow C B = 24 \sqrt{3} m .$
Hence, the height of the tower = $24 \sqrt{3} m .$

Question 30.

A Technician has to repair a light on a pole of height 10 m. She needs to reach a point 2 m below the top of the pole to undertake the repair work. The length of the ramp that she should use which, when inclined at an angle of $30^{\circ}$ to the horizontal, would enable her to reach the required position, is

___ m

(You may take $\sqrt{3}$ = 1.73)

Discuss Question

Answer: Option B. ->