Some Applications of Trigonometry Test

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Some Applications of Trigonometry Test - 6

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Question 1
1 / -0

AB is a vertical pole. The end A is on the level ground and C is the middle point of AB. P is a point on the level ground. The portion BC subtends an angle at P. If AP = nAB, then tan equals

A

B

C

D
None of these

Solution
Question 2
1 / -0

What is the measure of in the given figure?

A
15°

B
30°

C
45°

D
60°

Solution

In ABC,

In ABD,

Now, h =

tan = =
= 30°
Question 3
1 / -0

A 40 m tall building and a tower are standing in front of each other. Angles of elevation to the top of the tower from the top and base of the building are 45° and 60°, respectively. How high is the tower? ()

A
76.9 m

B
80.4 m

C
88.5 m

D
94.6 m

Solution
Question 4
1 / -0

From the foot of a 50 m tall building, the angle of elevation of the top of a hill is 60°and the angle of elevation of the top of the building from the foot of the hill is 30°. How high is the hill (length DC in the figure)?

A
50 m

B
m

C
100 m

D
150 m

Solution
Question 5
1 / -0

The angle of elevation of the top of an unfinished tower at a point of distance 120 m from its base is 45°. How much height of the tower must be increased so that the angle of elevation becomes 60°?

A
120() m

B
() m

C
120() m

D
None of these

Solution

Let BC be the unfinished tower.
Let AB be the required increase in height of the tower.

In ADC,

AC = m

In BDC,

BC = 120 m
AB = AC - BC = 120m
Question 6
1 / -0

When the elevation of sun changes from 45°to 30°, the shadow of a tower increases by 60 meters. The height of the tower is

A
30(- 1) metres

B
30(+ 1) metres

C
30(- 1) metres

D
30( + 1) metres

Solution

As tan 45°= 1,
h = AC
Also, tan 30° =
=
Or, h =
=
= 30 (+ 1) metres
Question 7
1 / -0

If the angle of elevation of a cloud from a height h above the level of water in a lake is and the angle of depression of its image is , then which of the following cannot be the height of the cloud above the level of water in the lake?

A

B

C

D

Solution

Let BC = BD = x

Also, BM = AP = h
MC = x - h
and DM = x + h
Now, in CPM;
PM = MC cot …(i)
In DPM;
PM = DM cot …(ii)
From eq. (i) and (ii), we get:

(x - h) cot

=

= h
Question 8
1 / -0

In the given figure, AB is the height of a tree. C and D are the two points in a straight line on the ground at b units and a units respectively away from the tree. What is the height of the tree (AB)?

A
units

B
units

C
units

D
units

Solution

In ABC
= tan 60°
AB = units

In ABD
= tan 30°
AB = units
Now AB AB = units²
AB² = ab units²AB = units
Question 9
1 / -0

The angle of elevation of a cloud from a point 60 m above a lake is 30^o and from the same point, the angle of depression of its image in the lake is 60^o. The height of the cloud is

A
100 m

B
80 m

C
40 m

D
120 m

Solution

h =

=

=

= 120 m
Question 10
1 / -0

The angle of elevation of the top of a vertical tower from a point P on the ground is 60°. At a point Q, 40 m vertically above P, the angle of elevation is 45°. Find the height of the tower.

A

B

C

D

Solution

h = = =
Question 11
1 / -0

If two towers of heights h₁ and h₂ subtend angles 30° and 60°, respectively at the mid-point of the line joining their feet, then h₁ : h₂ is

A
1 : 2

B
1 : 3

C
2 : 1

D
3 : 1

Solution

In ABE,
tan 30° =
x = 2 …(i)
and in BCD,

tan 60° =
x = …(ii)

From Eqs. (i) and (ii),
2
h₁ : h₂ = 1 : 3
Question 12
1 / -0

A light house is constructed on a cliff on the bank of a river. Height of the cliff and the light house is 110 m and 70 m, respectively. From the top of the light house, James sees two boats coming towards the bank in a straight line such that both of them make an angle of depression as 30° and 60°, respectively. What is the distance between the two boats?

A
207.6 m

B
190.9 m

C
184.6 m

D
169.3 m

Solution
Question 13
1 / -0

A 1.6 m tall observer is 45 meters away from a tower. If the angle of elevation from his eye to the top of the tower is 30°, then the height of the tower in meters is
(Take = 1.732)

A
25.98

B
26.58

C
27.58

D
27.98

Solution

Given height of the observer, DE = 1.6 m
Hence, BC = 1.6 m
AC is the height of the tower.
Let AB = h.
Given distance between the observer and the tower, CD = 45 m
⇒ BE = 45 m
In right ΔABE,

tan 30° = (h/45)

45 = h
h = 45 ÷ 1.732 = 25.98 m

Thus, height of tower = AC = AB + BC = 25.98 m + 1.6 m = 27.58 m
Hence, answer option 3 is correct.
Question 14
1 / -0

Two ships are sailing in the sea on the two opposite sides of a light house. The angles of elevation of the top of the light house as observed from the two ships are 60° and 45°, respectively. If the light house is 100 m high, the distance between the two ships is (take = 1.73)

A
173 m

B
273 m

C
158 m

D
300 m

Solution

In the diagram, AC represents the height of light house = 100 m.
Distance between the two ships is BD.
In triangle ABC,
(AC/BC) = tan 60°= 1.73
BC = 57.8 m ~ 58 m
Similarly,
In triangle ACD,
(AC/CD) = tan 45° = 1
DC = 100 m
Distance between the two ships = BD = BC + CD = 58 m + 100 m = 158 m
Question 15
1 / -0

A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot is pulled away from the wall through a distance p, such that its upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal.
What is the value of ?

A

B

C

D

Solution

Length of the ladder will remain the same.
∴AC = ED

(AC = ED)
cos - cos =
=
Now,

sin - sin=
(∵ ED = AC)
Question 16
1 / -0

A 50 m high pole stands on a 250 m high building. To an observer at a height of 300 m, the building and the pole subtend equal angles. The distance of the observer from the top of the pole is

A
25 m

B
50 m

C
25 m

D
25 m

Solution

From the figure, we find that,
tanθ = (From triangle OMP) ... (1)
Also, tan 2θ = (From triangle OBM) ... (2)
=
= (from equation 1)
3(x² - 2500) = x²
x²=3x² - 7500
2x² = 7500
x² = 3750 = 625 × 6
x = 25
Hence, distance of the observer from the top of the pole is 25.
Question 17
1 / -0

A man on the top of a tower observes a car moving towards the base of the tower at an angle of depression . Ten minutes later, the angle of depression of the car is found to be . If the tangent of angle is equal to, co-tangent of angle is equal to and the car is moving with a uniform speed, find the total time taken by the car to reach the base of the tower.

A
10 minutes

B
5 minutes

C
2.5 minutes

D
None of these

Solution

Distance travelled in 10 minutes = CD = x metres
tan =
tan =
Now, AB = … (i)
AB = CB … (ii)
Solving (i) and (ii), we get

CB + x = 5CB
x = 4CB
Or, CB = metres
Total time taken to cover a distance of metres = 10 = = 2.5 minutes
Question 18
1 / -0

From a point on a horizontal plane, the elevation of the top of a hill is 45°. Now, at an elevation of 15°, he again starts walking 500 m towards its summit where he has the elevation of 75° towards the hill's summit. Then the height of the hill will be

A
500m

B
500m

C
250m

D
250m

Solution
Question 19
1 / -0

The upper three-quarters of a vertical pole subtend an angle tan^-1(3/5) at a point in the horizontal plane through its foot at a distance of 40 m from it. What is the height of the pole?

A
80 m

B
100 m

C
160 m

D
200 m

Solution

Hence, height of the vertical pole = 160 m
Question 20
1 / -0

An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. If after 10 seconds, the elevation is observed to be 30°, then the uniform speed of the aeroplane is

A
120 km/hr

B
240 km/hr

C
_km/hr

D
_km/hr

Solution