Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 42

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Question 1
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On walking $$x$$ meters, making an angle of $${30}^{o}$$ with the ground, to find a ball fallen in a valley, one can reach a depth of '$$y$$' meters below the ground, then

A
$$x=y$$

B
$$x=2y$$

C
$$2x=\sqrt { 3 } y$$

D
$$2x=y$$

Solution

According to the given fig.
$$\sin{30^{o}} =\cfrac{1}{2}$$
$$\cfrac{y}{x}=\cfrac{1}{2}$$
$$\therefore x=2y$$
Question 2
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An observer 1.5 m tall is 28.5 m away from a tower. The angle of elevation of the top of the tower from his/her eyes has measure 45. What is the height of the tower?

A
28.5 m

B
30 m

C
27 m

D
1.5 m

Solution

We can form the above figure by given data
$$\implies BD = CE = 1.5$$
In $$\Delta ABC, \angle CBA=90^{o}$$
$$\implies \tan 45^{o} = \dfrac{AB}{BC}$$
$$\implies 1 = \dfrac{AB}{28.5}$$
$$\implies AB = 28.5$$
Now, $$h = AB + BD = 28.5 + 1.5 = 30$$
Hence, height of the tower is $$30\ m$$
Question 3
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A series of steps lead to a temple. The number of steps is 30. The height of each step is 20 cm. Then find the height of the temple from the base.

A
5.80 m

B
6 m

C
30 m

D
6.20 m

Solution

Number of steps $$= 30$$
Height of each step $$= 20\ cm$$
Thus, height of the temple $$= (30 \times 20) cm = 600\ cm = 6\ m$$
Hence, the correct answer is $$6\ m$$
Question 4
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When the length of the shadow of a pole is equal to the height of the pole, the angle of elevation of the Sun has measure of ................

A
$${ 30 }^{ o }$$

B
$${ 45 }^{ o }$$

C
$${ 60 }^{ o }$$

D
$${ 75 }^{ o }$$

Solution

Let $$AB=h$$ be the height of the pole.
The shadow of a pole will form base $$BC$$
Let $$\theta$$ be the angle of elevation.
According to the fig.
$$\tan{\theta} =\cfrac{AB}{BC} $$
$$\implies \tan{\theta}=\cfrac{h}{h} =1$$
$$\implies \theta =\tan^{-1}(1)$$
$$\implies \theta=45^{o}$$
Hence, the angle elevation is $$45^{o}$$.
Question 5
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In given figure, the minimum distance to reach from point "C" to point "A" will be ....................

A
$$a^2$$

B
$$\sqrt 2$$

C
2

D
2a

Solution

$$tan A = \dfrac{BC}{AB}$$

$$tan 60^o = \dfrac{BC}{a}$$

$$BC = a tan 60^o$$

$$BC = \sqrt 3 a$$

In $$\Delta ABC$$,

Using Pythagoras theorem we get,

$$AC^2 = AB^2 + BC^2$$

$$AC^2 = a^2 + (\sqrt 3a)^2$$

$$AC^2 = 4a^2$$

$$AC = 2a$$
Question 6
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From the top of a building of height $$h,$$ the angle of depression of an object on the ground is $$\theta.$$ What is the distance of the object from the foot of the building in terms of $$h$$ and $$\theta?$$

A
$$h \cot \theta$$

B
$$h \tan\theta$$

C
$$h \cos\theta$$

D
$$h \sin\theta$$

Solution

Let $$x$$ be the distance of the object from the foot of the building, we have

$$\tan\theta =\dfrac { h }{ x }$$

$$\Rightarrow x=h\cot\theta$$

Hence, option $$A$$ is correct.
Question 7
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The angle of elevation of a tower at a level ground is $$30^o$$ . The angle of elevation becomes $$\theta$$ when moved 10 m towards the tower. If the height of tower is $$ 5 \sqrt{3} m $$, then what is $$\theta$$ equal to ?

A
$$45^o$$

B
$$60^o$$

C
$$75^o$$

D
None of the above

Solution

In $$\triangle ABC,$$
$$\tan(30^{o})=\dfrac { 5\sqrt { 3 } }{ 10+x } $$

$$\Rightarrow \dfrac { 1 }{ \sqrt { 3 } } =\dfrac { 5\sqrt { 3 } }{ 10+x } $$
$$\Rightarrow x+10=15$$
$$\Rightarrow x=5$$

In $$\triangle ADC,$$
$$\tan(\theta )=\dfrac{5\sqrt{3}}{x}=\dfrac { 5\sqrt { 3 } }{ 5 } =\sqrt { 3 }$$
$$\Rightarrow \theta =60^{o}$$
Hence, option B is correct.
Question 8
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Two poles are 10 m and 20 m high. The line joining their tips makes an angle of $$15^o$$ with the horizontal. What is the distance between the poles ?

A
$$10 ( \sqrt{3} - 1 ) m$$

B
$$ 5 ( 4 + 2 \sqrt{3} -1 ) m$$

C
$$20 ( \sqrt{3} + 1 ) m$$

D
$$10 ( \sqrt{3} +2 ) m$$

Solution

Now AC will be the distance between the poles
Here,
$$\tan(\frac { \theta }{ 2 } )=\pm\sqrt { \dfrac { 1-cos\theta }{ 1+cos\theta } } $$

Now, $$\tan(15^{o})=\dfrac { 10 }{ AC } $$

$$\Rightarrow AC=\dfrac { 10 }{\tan(15^{o})}$$

But, $$\tan(15^{o})=\sqrt { \dfrac { 1-\cos(30^{o}) }{ 1+\cos(30^{o})}}$$

$$=\sqrt { \dfrac { 1-\dfrac { \sqrt { 3 } }{ 2 } }{ 1+\dfrac { \sqrt { 3 } }{ 2 } } } =\sqrt { \dfrac { 2-\sqrt { 3 } }{ 2+\sqrt { 3 } } } =\sqrt { \dfrac { 2-\sqrt { 3 } }{ 2+\sqrt { 3 } } \times \dfrac { 2-\sqrt { 3 } }{ 2-\sqrt { 3 } } } =2-\sqrt { 3 } $$

So, $$AC=\dfrac { 10 }{ 2-\sqrt { 3 } } =\dfrac { 10 }{ 2-\sqrt { 3 } } \times \dfrac { 2+\sqrt { 3 } }{ 2+\sqrt { 3 } } =20+10\sqrt { 3 } $$
Question 9
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From an aeroplane above a straight road the angles of depression of two positions at a distance 20 m apart on the road are observed to be $$30^o$$ and $$45^o$$. The height of the aeroplane above the ground is :

A
$$10 \sqrt{3} m$$

B
$$ 10 (\sqrt{3}-1) m$$

C
$$ 10 (\sqrt{3}+1) m$$

D
$$20 \, m$$

Solution

From the figure
$$ \cfrac { h }{ 20-x } =\tan { { 30 }^{ 0 } }$$
$$ \cfrac { h }{ 20-x } =\cfrac { 1 }{ \sqrt { 3 } }$$
$$\sqrt { 3 } h=\quad 20-x$$ ....... $$(1)$$
and $$\cfrac { h }{ x } =\tan { { 45 }^{ 0 } }$$
$$h=x$$ ....... $$(2)$$
From $$(1)$$ and $$(2)$$
$$ \sqrt { 3 } h=20-h\\ \Rightarrow \cfrac { 20 }{ \sqrt { 3 } +1 } =h$$
$$\Rightarrow h=10(\sqrt { 3 } -1)$$
Question 10
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The angle of elevation of the top of a tower standing on a horizontal plane from two points on a line passing through the foot of the tower at distances $$49\ m$$ and $$36\ m$$ are $$43^{\circ}$$ and $$47^{\circ}$$ respectively. What is the height of the tower?

A
$$40\ m$$

B
$$42\ m$$

C
$$45\ m$$

D
$$47\ m$$

Solution

Let $$BD $$ be the tower
In triangle $$ABD$$
$$\tan A=\dfrac { BD }{ AB } \\ \Longrightarrow \tan 43^o=\dfrac { BD }{ 49} \\ \Longrightarrow BD=49 \tan43^o$$

In triangle $$BDC$$

$$\tan C=\dfrac { BD }{ BC } \\ \Longrightarrow \tan 47^o=\dfrac { BD }{ 36 } \\ \Longrightarrow BD= 36 \tan 47^o$$

$$\Longrightarrow 36 \tan 47^o=49 \tan 43^o\\ \Longrightarrow \dfrac {\tan 47 ^o}{\tan 43^o } =\dfrac { 49 }{ 36 } \\ \Longrightarrow \dfrac {\tan 47^o}{\cot 47^o } =\dfrac { 49 }{ 36 } \\ \Longrightarrow {\tan }^{ 2 }47^o=\dfrac { 49 }{ 36 } \\ \Longrightarrow \tan 47^o=\dfrac { 7 }{ 6 }$$
But $$BD=36\tan 47^o$$
$$\Longrightarrow BD=36\times \dfrac{7}{6}\\=42m$$