Free Applications of Trigonometry 03 Practice Test - 10th Grade

Question 1

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

SOLUTION

Solution : 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 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^{\circ}

60^{\circ}

45^{\circ}

15^{\circ}

SOLUTION

Solution : 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, $θ$ = $60^{\circ}$

Question 3

The angle of elevation of the top of a tower from a point on the ground, which is 30m away from the foot of the tower is $30^{\circ}$ . Find the height of the tower.

10 $\sqrt{3}$ m

20 $\sqrt{3}$ m

30 $\sqrt{3}$ m

20 m

SOLUTION

Solution : A

Angle of elevation = $θ$ = $30^{\circ}$
$\Rightarrow$ tan $θ$ = $\frac{P e r p e n d i c u l a r}{B a s e}$
$\Rightarrow$ tan $θ$ = $\frac{h}{30}$
$\Rightarrow$ tan $30^{\circ}$ = $\frac{1}{\sqrt{3}}$
$\Rightarrow$ $\frac{h}{30}$ = $\frac{1}{\sqrt{3}}$
Therefore, $h = 10 \sqrt{3} m$ .

Question 4

The string of a kite is 100m long and it makes an angle of $60^{\circ}$ with the horizontal. Find the height of the kite from the ground, assuming that there is no slack in the string.

100 $\sqrt{3}$

$\frac{200}{\sqrt{3}}$

50 $\sqrt{3}$

$\frac{100}{\sqrt{3}}$

SOLUTION

Solution : C

We know, angle of elevation = $60^{\circ}$
$\Rightarrow$ $sin$ $θ$ = $\frac{p e r p e n d i c u l a r}{h y p o t e n u s e}$
$\Rightarrow$ $sin$ $60^{\circ}$ = $\frac{\sqrt{3}}{2}$ = $\frac{h}{100}$
$\Rightarrow$ h = 50 $\sqrt{3} m$ .

Question 5

Find the angle of elevation of the sun when the shadow of a 10m long pole is 10 $\sqrt{3}$ m.

$30^{\circ}$

$60^{\circ}$

$45^{\circ}$

$15^{\circ}$

SOLUTION

Solution : 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 $θ$ = $\frac{1}{\sqrt{3}}$
$\Rightarrow$ $θ$ = $30^{\circ}$

Question 6

A circus artist is climbing a 20m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. The height of the pole is

___
m if the angle made by the rope with the ground level is

30^{\circ}

SOLUTION

Solution :

Angle of elevation $θ$ = $30^{\circ}$

$\Rightarrow$ Sin $θ$ = $\frac{p e r p e n d i c u l a r}{h y p o t e n u s e}$

$\Rightarrow$ Sin $30^{\circ}$ = $\frac{1}{2}$ = $\frac{h e i g h t o f p o l e}{20}$

Thus, height of pole = 10 m.

Question 7

Two towers A and B are standing at some distance apart. From the top of tower A, the angle of depression of the foot of tower B is found to be $30^{\circ}$ . From the top of tower B, the angle of depression of the foot of tower A is found to be $60^{\circ}$ . If the height of tower B is ‘h’ m then the height of tower A in terms of ‘h’ is _____ m

$\frac{h}{2}$ m

$\frac{h}{3}$ m

$\sqrt{3} h$ m

$\frac{h}{\sqrt{3}}$ m

SOLUTION

Solution : B

Let the height of tower A be = AB = H.

And the height of tower B = CD = h

In triangle ABC

tan $30^{\circ}$ = $\frac{A B}{A C}$ = $\frac{H}{A C}$ ...... (1)

In triangle ADC

tan $60^{\circ}$ = $\frac{C D}{A C}$ = $\frac{h}{A C}$ ...... (2)

Dividing (1) by (2) we get, $\frac{t a n 30^{\circ}}{t a n 60^{\circ}}$ = $\frac{H}{h}$

H= $\frac{h}{3}$

Question 8

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.

$10 \sqrt{3}$ m

$20 \sqrt{3}$ m

$30 \sqrt{3}$ m

$40 \sqrt{3}$ m

SOLUTION

Solution : A

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 9

The tops of two poles of height 14 m and 20 m are connected by a wire which makes an angle of 30° with the horizontal. The length of the wire is _____m.

10m

12m

SOLUTION

Solution : D

In triangle ABC,

BC = y = 20-14 = 6m;

Let AB = $x$

$\Rightarrow$ Sin $30^{\circ}$ = $\frac{B C}{A B}$

$\Rightarrow$ $\frac{1}{2}$ = $\frac{6}{x}$

$\Rightarrow$ $x$ = 12

Thus, the length of the string is 12 m.

Question 10

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

SOLUTION

Solution : C

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

Free Applications of Trigonometry 03 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

Related ExamsPolynomials 03Triangles 03Introduction to Trigonometry 03

Related Exams
Polynomials 03
Triangles 03
Introduction to Trigonometry 03