Some Applications of Trigonometry Test

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Some Applications of Trigonometry Test - 2

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Question 1
1 / -0

Two buildings of unequal heights are standing opposite to each other on either side of a road. From a point between them somewhere on the road, the angle of elevation of the top of the smaller building is 45° and that of the top of the other building is 60°. Which of the following is the correct diagram for this situation?

A

B

C

D

Solution
Question 2
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The shadow of a tower, standing on level ground, is found to be 50 m longer when the sun's altitude is 30° than when it is 60°. Which of the following is the correct diagram representing this situation?

A

B

C

D

Solution
Question 3
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From the top of a 10 m high building, the angle of elevation of the top of a tower is 45° and the angle of depression of its foot is 60°. How will you represent this situation diagrammatically?

A

B

C

D

Solution

The observer is at point 'D'.
Question 4
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In the given figure, the angle of elevation of a cloud from a height 'h' above the water level in a lake is . What is the angle of depression of its image in the lake as observed from the same point?

A

B

C

D
90° –

Solution

In BEC:
CBE = and CEB = 90° and angle of depression of the image in the lake = ECB
= 180° – (90° + )
= 90° –
Question 5
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A man standing on the ground is looking at a hill of height 3000 feet. If the angle of elevation from the eye of the man to the top of the hill is 30°, find the distance of the man from the foot of the hill.

A
3000 feet

B
2000 feet

C
1500 feet

D
1000 feet

Solution

Given:
Height of the hill, i.e. AC = 3000 feet
Angle of elevation = 30°

We have to find the distance of the man from the foot of the hill, i.e. BC.
tan 30° =
tan 30° = BC = 3000
Hence, distance of the man from the foot of the hill = 3000 feet
Question 6
1 / -0

The top of a conical tent is 200 feet above the ground. The vertical angle of the tent is 60°. What is its slant height?

A
400 feet

B
100 feet

C
feet

D
feet

Solution

Given the height of the conical tent, h = 200 feet and vertical angle = 60°.

So, the semi-vertical angle is 30°.
We have to find the slant height AB of the cone.
In right-angled triangle ABD,
cos 30° = feet [cos q = ]
AB = feet
Thus, slant height AB = feet
Question 7
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The angle of elevation of the top of a tower from a point at a distance of 50 m from its foot is 30°. Find the height of the tower.

A
m

B
m

C
m

D
25 m

Solution

As, tan = tan 30° =

^_(1/m)
AC = ^_m
Hence, the height of the tower is AC = .
Question 8
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The angle of elevation of the top of a building from a point on the ground is 45°. If 'a' is the height of the building and 'x' is the distance of the point from the foot of the building, then which of the following relations is true?

A
x = 2a

B
x = a

C
x = a

D
x = 2a

Solution

Now, tan q =

tan 45° =
Or, x = a
Question 9
1 / -0

If two towers of heights 'a' and 'b' subtend angles 60° and 30°, respectively, at the midpoint of the line segment joining their feet, then find the value of a : b.

A
3 : 1

B
1 : 1

C

D
3 : 2

Solution

D is the midpoint of BC.

BD = CD

Now, we have to find a : b.

tan q =
In ΔABD,
tan 60° =

BD = … (1)
Similarly, in ΔCDE,
tan 30° = (CD = BD)
BD = b ... (2)
Thus, from equations (1) and (2), we get:
= b
Hence, a : b : : 3 : 1
Question 10
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The angle of elevation of the top of a 60 m high tower from a point on the ground is 60°. Find the distance between the foot of the tower and that point.

A
60 m

B
40 m

C
20 m

D
30 m

Solution

Given: Height of the tower = 60 m

Let the distance between the foot of the tower and the point be x m.

Also, tan =
tan 60° = x = 20m
So, the required distance is 20m.
Question 11
1 / -0

If the length of the shadow of a pole is equal to the height of the pole, then find the angle of elevation of the sun's altitude.

A
0°

B
90°

C
45°

D
60°

Solution

The angle of elevation of the sun's altitude is the angle of elevation of the top of the tower from the tip of the shadow.

As tan q = ,
so in right-angled ABC,
tan q =
tan = h/h ()
tan = 1
= 45^o [As tan 45^o = 1]
Question 12
1 / -0

The shadow of an 8 inches long pencil standing vertically on a table is 8 inches long. Find the angle of elevation of the top of the pencil from the end point of its shadow.

A
60^o

B
30^o

C
45^o

D
90^o

Solution

As tan q =

tan q =
Also, tan 30^o =
Hence, q = 30^o
Question 13
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A tower stands vertically on the ground. From a point on the ground, which is 20 m away from the foot of the tower, the angle of elevation of the top of the tower is 60°. Find the height of the tower.

A
m

B
m

C
10 m

D
10 m

Solution

Given angle of elevation = 60^oand BC = 20 m.

We have to find the height (AB) of the tower.
As tan q = ^,
tan 60^o =
= 20m
Question 14
1 / -0

From the top of a building of height k, the angle of depression of a car on the ground is x. What is the distance of the car from the foot of the building?

A
k tan x

B
k sin x

C
k cot x

D
k cos x

Solution

The angle of depression is the angle between the line of sight and the horizontal line as shown below:

Angle of depression, DAC = ACB ( alternate angles are equal)

So, in DABC, tan x =
tan x =
Hence, BC =
Or BC = k cot x [ tan x = ]
Question 15
1 / -0

The angle of elevation of the top of a tower from a point at a distance of 60 m from its foot is 30°. Find the height of the tower.

A
60 m

B
20 m

C
30 m

D
30 m

Solution

Given: Angle of elevation = 30°

We have to find the height of the tower, i.e. AB.
Now, tan =
tan 30° =

AB = 20m