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

10th Grade > Mathematics

APPLICATIONS OF TRIGONOMETRY MCQs

Total Questions : 58 | Page 4 of 6 pages

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $30^{\circ}$ with it. The distance between the foot of the tree to the point where the top touches the ground is 12 m. Find the height of the tree.

12 m
4 m
4 $\sqrt{3}$ m
12 $\sqrt{3}$ m

Discuss Question

Answer: Option D. -> 12

\sqrt{3}

m
:
D

The given situation can be represented by the figure below

$c o s ∠ A C B = \frac{B C}{A C} \Rightarrow c o s 30^{\circ} = \frac{B C}{A C} \Rightarrow \frac{\sqrt{3}}{2} = \frac{12}{A C} \Rightarrow A C = \frac{24}{\sqrt{3}} = 8 \times \frac{3}{\sqrt{3}} = 8 \sqrt{3} m t a n (∠ A C B) = \frac{A B}{B C} \Rightarrow t a n 30^{\circ} = \frac{A B}{12} \Rightarrow A B = \frac{12}{\sqrt{3}} = 4 \sqrt{3} m$
$∴$ The height of tree was originally AB+AC = $4 \sqrt{3} + 8 \sqrt{3} = 12 \sqrt{3} m .$

Question 32.

An observer 2.25 m tall is 42.75 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is $45^{\circ}$ . What is the height of the chimney?

40 m
35 m
45 m
50 m

Discuss Question

Answer: Option C. -> 45 m
:
C

The given situation is represented by the figure below:

In triangle ABE,
$tan 45^{\circ} = \frac{A B}{E B} A l s o, E B = D C ∴ t a n 45^{\circ} = \frac{A B}{D C} \Rightarrow A B = D C \times t a n 45^{\circ} \Rightarrow A B = 1 \times 42.75$
Hence, the height of the chimney = AC = AB + BC
We can observe that BC = ED.
Thus, AC = AB + ED
= 42.75 + 2.25
= 45 m.

Question 33.

A kite is flying on a string of 40 m. 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 height at which the kite is flying.

10 m
30 m
20 m
40 m

Discuss Question

Answer: Option C. -> 20 m
:
C

The situation can be represented by the figure below:

AB is the height at which the kite is flying.
$s i n (∠ A C B) = \frac{A B}{A C} \Rightarrow s i n 30^{\circ} = \frac{A B}{A C} \Rightarrow \frac{1}{2} \times 40 = A B \Rightarrow A B = 20 m$

Question 34.

A person observes the angle of elevation of the top of a 60m tower from a point on the level ground to be 60 $^{\circ}$ . Find the distance between the person and the foot of the tower.

30 m
20 $\sqrt{3}$ m
20 m
60 m

Discuss Question

Answer: Option B. -> 20

\sqrt{3}

m
:
B

In $Δ A B C$ , tan(60 $^{\circ}$ ) = $\frac{B C}{A B}$

$\Rightarrow$ $\sqrt{3} = \frac{60}{x}$

$\Rightarrow$ x = 20 $\sqrt{3}$

$\Rightarrow$ Distance between the person and the tower = x = 20 $\sqrt{3}$ m.

Question 35.

A vertical stick 10 cm long casts a shadow 8 cm long. At the same time, a tower casts a shadow 30 cm long. Determine the height of the tower?

65 cm
75 cm
37.5 cm
100 cm

Discuss Question

Answer: Option C. -> 37.5 cm
:
C

$tan θ = \frac{10}{8} = \frac{5}{4}$

As both the shadows are formed at the same time, the sun rays forms the same angle in both the cases.
$tan θ = \frac{h}{30}$

$\frac{5}{4} = \frac{h}{30}$

h = 37.5 cm

Question 36.

The angles of depression of two objects from the top of a 100 m hill lying to its east are found to be $45^{\circ} a n d 30^{\circ}$ . Find the distance between the two objects. (Take $\sqrt{3} = 1.732$ )

73.2 metres
107.5 meters
150 metres
200 metres

Discuss Question

Answer: Option A. -> 73.2 metres
:
A

Let C and D be the objects and CD be the distance between the objects.

In $Δ A B C$ , $t a n 45^{\circ}$ = $\frac{A B}{A C}$ = 1

$A B = A C = 100 m$

In $Δ A B D$ , $t a n 30^{\circ}$ = $\frac{A B}{A D}$
$A D \times \frac{1}{\sqrt{3}} = 100$
$A D = 100 \times \sqrt{3} = 173.2 m$
$C D = A D - A C = 173.2 - 100 = 73.2 m e t r e s$

Question 37.

If the angle of elevation of a cloud from a point 200 m above a lake is $30^{\circ}$ and the angle of depression of the reflection of the cloud in the lake from the same point is $60^{\circ}$ , then the height of the cloud above the lake is:

400 m
500 m
30 m
200 m

Discuss Question

Answer: Option A. -> 400 m
:
A

In $△$ ABC',
$t a n 60^{\circ} = \frac{x + 400}{A B}$
$\sqrt{3} = \frac{x + 400}{A B}$
$A B = \frac{x + 400}{\sqrt{3}}$ ___(i)
In $△$ ABC,
$t a n 30^{\circ} = \frac{x}{A B}$
$\frac{1}{\sqrt{3}} = \frac{x}{A B}$
$A B = x \sqrt{3}$ _______(ii)
Plugging the value of AB from equation (ii) in equation (i), we get
$\Rightarrow x \sqrt{3} = \frac{x + 400}{\sqrt{3}}$
$\Rightarrow 3 x = x + 400$
$\Rightarrow x = 200 m$
Hence, the height of the cloud above the lake is
200 + 200 = 400 m

Question 38.

Two ships are on either side of a 100 m lighthouse. The angles of elevation from the two ships to the top of the lighthouse are 30° and 45°. Find the distance between the ships.

300 m
173 m
273 m
200 m

Discuss Question

Answer: Option C. -> 273 m
:
C

Let, BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m, $∠$ BAD = 30 $^{\circ}$ , $∠$ BCD = 45 $^{\circ}$
$t a n 30^{\circ} = \frac{B D}{B A} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{B A} \Rightarrow B A = 100 \sqrt{3} t a n 45^{\circ} = \frac{B D}{B C} \Rightarrow 1 = \frac{100}{B C} \Rightarrow B C = 100$
Distance between the two ships
$= A C = B A + B C = 100 \sqrt{3} + 100 = 100 (\sqrt{3} + 1) = 100 (1.73 + 1) = 100 \times 2.73 = 273 m$