Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 14

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Question 1
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An observer $$1.6\ m$$ tall is $$20\sqrt{3}m$$ away from a tower. The angle of elevation from his eye to the top of the tower is $$30^{\circ}$$. The heights of the tower is:

A
$$21.6\ m$$

B
$$23.2\ m$$

C
$$24.72\ m$$

D
None of these

Solution

Let $$AB$$ be the observer and $$CD$$ be the tower.
Draw $$BE \perp CD$$.
Then, $$CE = AB = 1.6\ m$$,
$$BE = AC = 20\sqrt {3}m$$.
$$\dfrac {DE}{BE} = \tan 30^{\circ} = \dfrac {1}{\sqrt {3}}$$
$$\Rightarrow DE = \dfrac {20\sqrt {3}}{\sqrt {3}} m = 20\ m$$.
$$\therefore CD = CE + DE = (1.6 + 20) m = 21.6\ m$$.
Question 2
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A ladder '$$x$$' meters long is laid against a wall making an angle '$$\theta$$' with the ground. If we want to directly find the distance between the foot of the ladder and the foot of the wall, which trignometrical ratio should be considered?

A
$$\sin { \theta } $$

B
$$\cos { \theta } $$

C
$$\tan { \theta } $$

D
$$\cot { \theta } $$

Solution

Distance to be found is adjacent side to the given angle which means base. The length of base in a right angled triangle, when the length of hypotenuse $$(x)$$ can be found using the $$\cos$$ function.
So, trigonometric ratio to be used is $$\cos \theta$$.
As, $$\cos \theta=\dfrac{\mathrm{Base}}{\mathrm{Hypotenuse}}$$
Question 3
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If the height of a tower and the length of its shadow is equal, then the value of the angle of elevation of the sun is-

A
$$30^{\circ}$$

B
$$45^{\circ}$$

C
$$60^{\circ}$$

D
None of these

Solution

$$ \tan \theta = \dfrac{height}{shadow} $$
Here the height is equal to the shadow
$$\tan { \theta } =\dfrac { height }{ height } =1$$
$$ \tan \theta = 1 $$
$$ \theta = 45^{o} $$
Question 4
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A $$20 m$$ pole casts a $$5 m$$ long shadow. If at the same time of the day, a building casts a shadow of $$20 m$$, how high is the building?

A
$$400$$ m

B
$$4$$ m

C
$$80$$ m

D
$$100$$ m

Solution

Let the angle subtended between pole and shadow be $$\theta$$
$$\Rightarrow$$ $$\tan\theta = \dfrac{20}{5} = 4$$

Similarly,
At same time of the day, angle subtended will remain same as position of sun is fixed.
Given that the shadow is $$20m$$ long.
Let height of the building be $$x$$
$$\Rightarrow$$ $$\tan\theta = \dfrac{x}{20} = 4$$

$$\Rightarrow x = 80$$ m
Question 5
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If the length of the shadow of a pole is equal to the height of the pole, then the angle of elevation of the sun is

A
$${ 30 }^{ o }\quad $$

B
$${ 75 }^{ o }$$

C
$${ 60 }^{ o }\quad $$

D
$${ 45 }^{ o }$$

Solution

$$AC=BC$$
$$tan(B)=\dfrac{AC}{BC}$$
$$=\dfrac { AC }{ AC } =1$$
$$\angle B =tan^{-1}1$$
$$\angle B={ 45 }^{ 0 }$$
Question 6
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The heights of two poles are $$80$$ m and $$65$$ m. If the line joining their tops makes an angle of $$45^{\circ}$$ with the horizontal, then the distance between the poles is :

A
$$15$$ m

B
$$22.5$$ m

C
$$30$$ m

D
$$7.5$$ m

Solution

Let, $$AB$$ and $$CD$$ be the two poles of height $$80$$ m and $$65$$ m respectively.

According to the picture,
$$AE=AB-CD=$$$$(80-65)\ m=15\ m$$ [Since $$CD=EB$$]

Now, from $$\Delta$$ $$ACE$$ we have,
$$\tan 45^o=\dfrac{AE}{EC}$$
$$\Rightarrow 1=\dfrac{15}{EC}$$
$$\Rightarrow EC=15$$

Now, $$BD=EC=15\ m$$.

Hence, the distance between the two poles is $$15\ m$$.
Question 7
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If the ratio of the height of tower and the length of its shadow is $$1: \sqrt{3}$$, then the angle of elevation of the sum has measure_________.

A
$$60$$

B
$$45$$

C
$$30$$

D
$$90$$

Solution

Let $$\overline{AB}$$ is a tower and $$\overline{BC}$$ is its shadow.
Given that , $$\dfrac{AB}{BC}=\dfrac{1}{\sqrt{3}}$$
Let the angle of elevation of the Sun is $$\theta$$
$$\therefore m\angle ACB = \theta$$
Now, $$\tan \theta =\dfrac{AB}{BC} $$
$$\therefore \tan \theta =\dfrac{1}{\sqrt{3}}=\tan 30^o$$
$$\therefore \theta =30^o$$
$$\therefore$$ the angle of elevation of the Sun $$=30^o$$
Question 8
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At an instant, the length of the shadow of a pole is $$\sqrt{3}$$ times the height of the pole. The angle of elevation of the sun is

A
$$30^{\circ}$$

B
$$45^{\circ}$$

C
$$60^{\circ}$$

D
$$75^{\circ}$$

Solution

Let the height of pole be $$h$$ metres and Angle of elevation be $$\theta.$$
According to question,
Shadow of pole $$=\sqrt{3}h$$ metres
We know that,
$$\tan \, \theta \, = \, \dfrac{perpendicular}{base}$$
$$\Rightarrow\tan \, \theta \, = \, \dfrac{h}{\sqrt{3}h}$$
$$\Rightarrow \displaystyle \tan \, \theta \, = \, \frac{1}{\sqrt{3}} $$
$$\Rightarrow \tan\theta = \, \tan \, 30^{\circ}$$
$$\Rightarrow\theta \, = \, 30^{\circ}$$

Hence, the elevation of the sun is $$30^o$$
Question 9
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Upper part of a vertical tree which is broken over by the winds just touches the ground and makes an angle of $${30}^{o}$$ with the ground. If the length of the broken part is $$20$$ metres, then the remaining part of the tree is of length.

A
$$20$$ metres

B
$$10\sqrt { 3 } $$ metres

C
$$10$$ metres

D
$$10\sqrt { 2 } $$ metres

Solution

$$\cfrac { AB }{ 20m } =\sin { { 30 }^{ o } } $$
$$AB=10m$$
Question 10
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A man of height $$6$$ ft. observes the top of a tower and the foot of the tower at angles of $$45^\circ $$ and $$30^\circ $$ of elevation and depression respectively. Assuming the man to stand on the level ground, the height of the tower is :-

A
$$6\left( {\sqrt 3 - 1} \right)ft.$$

B
$$6\left( {\sqrt 3 + 1} \right)ft.$$

C
$$6\left( {\sqrt 2 - 1} \right)ft.$$

D
None of these

Solution

$$In\,ABC$$

$$\tan 30^{0} = {\dfrac{AB} {BC}}$$
$${\dfrac 1 {\sqrt 3 }} = {\dfrac 6 {BC}}$$
$$BC = 6\sqrt 3 $$
$$AD = 6\sqrt 3 \,\,\,\,\,\,\,\,\left( {BC = AD} \right)$$

$$Now\,in\,\,ADE$$
$$\tan 45^{0} = {\dfrac{ED} {AD}}$$
$$1 = {\dfrac{ED} {6\sqrt 3 }}$$
$$ED = 6\sqrt 3 $$
$$\,h = ED + DC$$
$$=6\sqrt 3 + 6$$
$$=\left( {6\sqrt 3 + 1} \right)$$ft