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

10th Grade > Mathematics

APPLICATIONS OF TRIGONOMETRY MCQs

Total Questions : 58 | Page 1 of 6 pages

Question 1. 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√3 m
30 m
30√3 m
60 m

Discuss Question

Answer: Option A. -> 20√3 m
:
A
The 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 (∠ 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 2. 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∘
60∘
45∘
15∘

Discuss Question

Answer: Option B. -> 60∘
:
B

A Tower Is 100√3 m High. Find The Angle Of Elevation Of I...

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}

Question 3. Find the angle of elevation of the sun when the shadow of a 10m long pole is 10

\sqrt{3}

30∘
60∘
45∘
15∘

Discuss Question

Answer: Option A. -> 30∘
:
A
Let, angle of elevation of sun =

θ

\Rightarrow

Tan

θ

\frac{P e r p e n d i c u l a r}{b a s e}

\frac{10}{10 \sqrt{3}}

\Rightarrow

Tan

θ

\frac{1}{\sqrt{3}}

\Rightarrow

θ

30^{\circ}

Find The Angle Of elevation Of The Sun When The Shadow Of...

Question 4. From the top of a cliff 25m high the angle of elevation of a tower is found to be equal to the angle of depression to the foot of the tower. The height of the tower is ___.

25m
75m
50m
100m

Discuss Question

Answer: Option C. -> 50m
:
C

From The Top Of A Cliff 25m High The Angle Of Elevation Of A...

In triangle ABE,

t a n θ_{2}

\frac{A B}{B E}

\frac{25}{B E}

In triangle ADC,

t a n θ_{1}

\frac{C D}{A D}

We know,

θ_{1}

θ_{2}

\Rightarrow

t a n θ_{1}

t a n θ_{2}

\frac{C D}{A D}

\frac{25}{B E}

\Rightarrow

CD = 25 [Since AD = BE]
DE =AB = 25m

\Rightarrow

Height of tower = CD+ DE
= 25+25 = 50m

Question 5. The angle of elevation of the top of a tree from a point A on the ground is

60^{\circ}

. On walking 20 m away from point A, to a point B, the angle of elevation changes to

30^{\circ}

. Find the height of the tree.

10√3 m
20√3 m
30√3 m
40√3 m

Discuss Question

Answer: Option A. -> 10√3 m
:
A

The Angle Of Elevation Of The Top Of A Tree From A Point A O...

In triangle ACD,

tan θ = \frac{D C}{C A} = \frac{h}{x}

\Rightarrow tan 60^{\circ} = \sqrt{3} = \frac{h}{x}

..................... 1
In triangle CDB,

\Rightarrow tan 30^{\circ} = \frac{C D}{C B}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 20}

....................... 2
Using (1) and (2), we get

\Rightarrow \frac{1}{\sqrt{3}} = \frac{x \sqrt{3}}{x + 20}

\Rightarrow x + 20 = 3 x

\Rightarrow 2 x = 20

\Rightarrow x = 10

∴ h = x \sqrt{3} = 10 \sqrt{3}

Thus, the height of the tree is

10 \sqrt{3} m

Question 6. 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:

An Observer 2.25 M Tall Is 42.75 M Away From A Chimney. The ...

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 thatBC = ED.
Thus, AC = AB + ED
= 42.75 + 2.25
= 45 m.

Question 7. 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√3 m
20 m
60 m

Discuss Question

Answer: Option B. -> 20√3 m
:
B

A person Observes The Angle Of Elevation Of The Top Of A 60...

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}

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