# Free Applications of Trigonometry 03 Practice Test - 10th Grade

A kite is attached to a string of length 203m is tied to a pole on the level ground. If the string of the kite makes an angle of elevation of 60 with the ground, then the kite is flying at a height of 20 m.

A.

True

B.

False

#### SOLUTION

Solution : B sinACB=ABACsin60=ABACAB=AC × sin60=203×32Height of the kite is 30m.

A tower is 1003 m high. Find the angle of elevation of its top from a point 100 m away from its foot.

A. 30
B. 60
C. 45
D. 15

#### SOLUTION

Solution : B Let the angle of elevation be θ.
tan θ = PerpendicularBase

tan θ = 1003100 = 3

​Therefore, θ = 60

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. Find the height of the tower.

A.

103 m

B.

203 m

C.

303 m

D.

20 m

#### SOLUTION

Solution : A Angle of elevation = θ = 30
tanθ = PerpendicularBase
tanθ = h30
tan30 = 13
h30 = 13
Therefore, h=103m.

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

A.

1003

B.

2003

C.

503

D.

1003

#### SOLUTION

Solution : C We know, angle of elevation = 60
sin θ = perpendicularhypotenuse
sin 60 = 32h100
h = 503 m.

Find the angle of elevation of the sun when the shadow of a 10m long pole is 103 m.

A.

30

B.

60

C.

45

D.

15

#### SOLUTION

Solution : A

Let, angle of elevation of sun = θ
Tanθ = Perpendicularbase10103
Tanθ = 13
θ = 30 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 .

#### SOLUTION

Solution : Angle of elevation θ =  30

Sinθ = perpendicularhypotenuse

Sin30  = 12 = height of pole20

Thus, height of pole = 10 m.

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. From the top of tower B, the angle of depression of the foot of tower A is found to be 60. If the height of tower B is ‘h’ m then the height of tower A in terms of ‘h’ is _____ m

A.

h2m

B.

h3m

C.

3hm

D.

h3m

#### SOLUTION

Solution : B Let the height of tower A be = AB =  H.

And the height of tower B = CD = h

In triangle ABC

tan30 = ABAC = HAC ...... (1)

tan60 = CDAC = hAC...... (2)

Dividing (1) by (2) we get, tan30tan60 = Hh

H= h3

The angle of elevation of the top of a tree from a point A on the ground is 60 . On walking 20 m away from point A, to a point B, the angle of elevation changes to 30. Find the height of the tree.

A.

103 m

B.

203 m

C.

303 m

D.

403 m

#### SOLUTION

Solution : A In triangle ACD,

tanθ=DCCA=hx

tan60=3=hx ..................... 1

In triangle CDB,

tan30=CDCB

13=hx+20....................... 2

Using (1) and (2), we get

13=x3x+20

x+20=3x

2x=20

x=10

h=x3=103

Thus, the height of the tree is 103 m.

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.

A.

6m

B.

10m

C.

8m

D.

12m

#### SOLUTION

Solution : D In triangle ABC,

BC = y = 20-14 = 6m;

Let AB = x

Sin30= BCAB

12 = 6x

x = 12

Thus, the length of the string is 12 m.

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 ___.

A.

25m

B.

75m

C.

50m

D.

100m

#### SOLUTION

Solution : C In triangle ABE,

tanθ2 = ABBE = 25BE

We know, θ1 θ2

tanθ1 = tanθ2