Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 40

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Question 1
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The angle of depression of a car parked on the road from the top of a $$150 $$ m high tower is $$\displaystyle { 30 }^{ \circ }$$. The distance of the car from the tower (in metres) is

A
$$\displaystyle 50\sqrt { 3 } $$

B
$$\displaystyle 150\sqrt { 3 } $$

C
$$\displaystyle 150\sqrt { 2 } $$

D
$$\displaystyle 75$$

Solution

Given: Height $$= 150$$ m

Angle of depression $$= 30^o$$

Consider the diagram,

In $$\Delta ABC$$,

Let, distance of the car from tower $$BC = x$$ m,

In $$\Delta ABC, \tan 30^o = \dfrac{AC}{AB}$$

So, $$\tan 30^o = \dfrac{1}{\sqrt{3}}$$

Therefore, $$AC = 150\sqrt{3}$$

So, distance between the tower and car is $$150\sqrt{3}$$ m
Question 2
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The angle of depression of a car, standing on the ground, from the top of a $$75 \ m$$ high tower, is $$30^o$$. The distance of the car from the base of the tower (in m.) is:

A
$$25\sqrt{3}$$

B
$$50\sqrt{3}$$

C
$$75\sqrt{3}$$

D
$$150$$

Solution

In right angle triangle $$ABC$$ the $$\angle ACB =30^0$$ (Angle of depression of a car) and the tower is $$75\ m$$ high
Let the distance of car from ground is $$x\ m$$
Then $$\tan 30^{0}=\dfrac{AB}{BC}=\dfrac{75}{x}$$
$$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{75}{x}$$
$$\Rightarrow x=75\sqrt{3}$$ m
Question 3
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The shadow of a person $$X$$, when the angle of elevation of the sun is $$\alpha$$, is equal in length to the shadow of another person $$Y$$, when the angle of elevation of the sun is $$\left (\dfrac {\alpha}{2}\right )$$. Which is the correct statement?

A
$$X$$ is shorter than $$Y$$

B
$$X$$ is twice as tall as $$Y$$

C
$$X$$ is taller than $$Y$$ but is not twice as tall as $$Y$$

D
Both $$X$$ and $$Y$$ have equal heights
Question 4
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What is the angle of elevation of a light source when the length of the shadow of a flagpost is equal to its height?

A
$$45^{\circ}$$

B
$$30^{\circ}$$

C
$$60^{\circ}$$

D
$$90^{\circ}$$

Solution

Let $$AB$$ be the height of the flagpost and $$BC$$ be the length of its shadow.
$$AB=BC$$ (given)
Let, the angle of elevation, $$\angle ACB =\theta$$

Now, in $$\triangle ACB$$
$$\Rightarrow \tan\theta=\dfrac{AB}{BC}$$

$$\Rightarrow \tan\theta=\dfrac{AB}{AB}$$

$$\Rightarrow \tan\theta=1$$

$$\Rightarrow \theta=45^{\circ}$$

Hence, the angle of elevation is $$45^{\circ}.$$
Question 5
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The length of shadow of a tower on the plane ground is $$\sqrt3$$ times the height of the tower. The angle of elevation of sun is:

A
$$45^0$$

B
$$30^0$$

C
$$60^0$$

D
$$90^0$$

Solution

Let the height of the tower be h meter.
Then the length of the tower be $$\sqrt{3}h$$ meter
$$\therefore $$ In $$\triangle ABC$$
$$\Rightarrow tan\theta=\dfrac{AB}{BC}$$
$$\Rightarrow tan \theta=\dfrac{h}{\sqrt{3}h}$$
$$\Rightarrow tan\theta=\dfrac{1}{\sqrt{3}}$$
$$\Rightarrow tan\theta=tan30^{\circ}$$ [$$\because tan 30^{\circ}=\dfrac{1}{\sqrt{3}}$$]
$$\Rightarrow \theta=30^{\circ}.$$
Question 6
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A $$20\ \text{m}$$ long ladder rests against a wall. If the foot of the ladder is $$10\ \text{m}$$ from the wall, what is the angle of elevation?

A
$$\dfrac {\pi}{6}$$

B
$$\dfrac {\pi}{4}$$

C
$$\dfrac {\pi}{3}$$

D
$$\dfrac {\pi}{2}$$

Solution

Let $$AC$$ be the ladder and $$AB$$ be the wall,
$$\therefore AC=20\text{ m}$$
$$BC=10\text{ m},$$ distance of the foot of the ladder from the wall.
Let $$\angle ACB=\theta$$.
In $$\triangle ACB,$$
$$\Rightarrow \cos \theta =\dfrac{BC}{AC}=\dfrac{10}{20}=\dfrac{1}{2}$$
$$\Rightarrow \cos \theta =\dfrac{1}{2}$$
$$\Rightarrow \cos\theta =\cos 60°$$
$$\Rightarrow \theta =60°=\dfrac{\pi}{3}$$
Hence, the answer is $$\dfrac{\pi}{3}.$$
Question 7
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The angles of elevation of the top of a vertical tower from points at distance $$a$$ and $$b$$ from the base and in the same line with it are complementary. If $$a > b$$, find the height of the tower.

A
$$\sqrt {ab}$$

B
$$\sqrt {(a + b)}$$

C
$$\sqrt {(a/b)}$$

D
$$\sqrt {(b - a)}$$

Solution

Let the the height of the base be $$=h$$
and the angle of elevation be A and B

Now, $$\tan A=\dfrac{h}{a}\\tanB=\dfrac{h}{b}$$
as the angles A and B are complimentary
$$A+B=90^o\Rightarrow A=90^o-B\ \tan A=\tan(90^o-B)\Rightarrow \tan A=\cot B\Rightarrow \dfrac{h}{a}=\dfrac{b}{h}\Rightarrow h^2=ab\Rightarrow h=\sqrt{ab}$$

Therefore, Answer is $$\sqrt{ab}$$
Question 8
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Two buildings are 140 m apart. The angle of elevation of the top of a building as seen from the top of the other is 30 degrees. If the 2nd building is 60 m tall, how tall is the 1st building?

A
$$83 m$$

B
$$80.83 m$$

C
$$84 m$$

D
$$140.83 m$$

Solution

Let the height of second building be $$'h'$$ meter.
Now, In $$\triangle ABE$$
$$\Rightarrow \tan { { 30 }^{ 0 } } =\dfrac { AB }{ BE } =\dfrac { \left( h-60 \right) }{ 140 } $$
$$\Rightarrow \dfrac { 1 }{ \sqrt { 3 } } =\dfrac { \left( h-60 \right) }{ 140 } $$
$$\Rightarrow h-60=\dfrac { 140 }{ \sqrt { 3 } } =\dfrac { 140\sqrt { 3 } }{ 3 } $$
$$\Rightarrow h=\dfrac { 140\sqrt { 3 } }{ 3 } +60=\dfrac { 140\sqrt { 3 } +180 }{ 3 } $$
$$\Rightarrow h=\dfrac { 242.2+180 }{ 3 } =\dfrac { 422.48 }{ 3 } $$
$$\Rightarrow h=140.83m$$
Hence, the answer is $$140.83m.$$
Question 9
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A person standing on the bank of a river observes that the angle subtended by the tree on the opposite bank is $$60^o$$ and when he retires $$40$$ m from the bank and finds the angle of elevation to be $$30^o$$. What is the breadth of the river?

A
$$24$$ m

B
$$20$$ m

C
$$26$$ m

D
$$15$$ m

Solution

Let $$R$$ br the length of River
and $$T$$ be the length of tree

In $$\triangle ABC$$
$$\tan60^o=\dfrac{T}{R}\Rightarrow T=\sqrt3R.......(1)$$

in $$\triangle ABD$$
$$\tan30^o=\dfrac{T}{R+40}\Rightarrow T=\dfrac{R+40}{\sqrt3}.....(2)$$

From (1) and (2)
$$\sqrt3R=\dfrac{R+40}{\sqrt3}\Rightarrow 3R=R+40\Rightarrow R=20$$

Therefore, Answer is $$20m$$
Question 10
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Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are $$30^{\circ}$$ and $$45^{\circ}$$ respectively. If the lighthouse is $$100\ m$$ high, the distance between the two ships is:

A
$$173\ m$$

B
$$200\ m$$

C
$$273\ m$$

D
$$300\ m$$

Solution

Let $$AB$$ be the lighthouse and C and D be the positions of the ships.
Then, $$AB = 100\ m, \angle ACB = 30^{\circ}$$ and $$\angle ADB = 45^{\circ}$$.
$$\dfrac {AB}{AC} = \tan 30^{\circ} = \dfrac {1}{\sqrt {3}} \Rightarrow AC = AB \times \sqrt {3} = 100\sqrt {3}m$$.
$$\dfrac {AB}{AD} = \tan 45^{\circ} = 1\Rightarrow AD = AB = 100\ m$$.
$$\therefore CD = (AC + AD) = (100\sqrt {3} + 100)m$$
$$= 100(\sqrt {3} + 1)$$
$$= (100\times 2.73)m$$
$$= 273\ m$$.