Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 53

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Question 1
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The angle of elevation of the top of a tower from a point $$20m$$ away from its base is $${60^ \circ }$$. The height of the tower is-

A
$$2\sqrt{3}\,m$$

B
$$-20\sqrt{3}\,m$$

C
$$\sqrt{3}\,m$$

D
$$20\sqrt{3}\,m$$

Solution

Let the distance between point of observation and foot of tower $$BC=20\,m\,$$

Height of tower $$AB=H$$

Angle of elevation of top of tower $$\theta ={{60}^{0}}$$

Now from figure $$\Delta ABC$$

We know that, $$\Delta ABC$$ is a right angle triangle.

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

$$ \tan {{60}^{0}}=\dfrac{h}{20} $$

$$ \Rightarrow \dfrac{h}{20}=\sqrt{3} $$

$$ \Rightarrow h=20\sqrt{3}\,\,m. $$
Therefore, height of tower$$=20\sqrt{3}\,m$$
Question 2
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A vertical to stands on a horizontal plane and is surmounted by a vertical flag staff of height 6 meters. At point on the plane angle of elevation of the bottom and the top of the flag staff are respectively $${ 30 }^{ \circ }and60^{ \circ }$$. Find the height of tower.

A
5 m

B
3m

C
10 m

D
7.5 m

Solution
Question 3
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From the top of a cliff $$25 \ m$$ height, the angle of the elevation of the the top of tower is found to be equal to the angle of depresion of the foot of the tower. The height of the tower is

A
$$25m$$

B
$$50\ m$$

C
$$\dfrac{25}{\sqrt{2}} m$$

D
$$25 \sqrt{2} m$$

Solution

R.E.F image

Consider $$ \Delta DCB $$ right angled at c,

$$ tan x^{\circ} = \dfrac{BC}{CD} = \dfrac{25}{CD}...(1) $$

Now, consider $$ \Delta DCE $$ right angled atc,
$$ tan x^{\circ} = \dfrac{CE}{CD} = ...(2) $$

equating eq$$^{n}$$ (1) & (2),

$$ \Rightarrow \dfrac{25}{CD} = \dfrac{CE}{CD} $$

$$ \Rightarrow CE = 25 $$

$$ \therefore $$ Total ht. of tower is (25+25) m

($$ \because $$ BE is ht.of tower such that BC+CE & BC = AD = 25 $$ \Rightarrow 25+CE $$ )

$$ = 50 m $$ Ans
So Option b is correct
Question 4
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From the top of a tower $$100m$$ high, the angels of depression of the bottom and the top of a building just opposite to it are observed to be $${60^ \circ }$$ and $${45^ \circ }$$ respectively, then height of the building is:

A
$$\dfrac{{100\left( {3 - \sqrt 3 } \right)}}{3}m$$

B
$$\dfrac{{100\left( {3 + \sqrt 3 } \right)}}{3}m$$

C
$$\dfrac{{3 + \sqrt 3 }}{3}m$$

D
$$\dfrac{{50\left( {3 - \sqrt 3 } \right)}}{3}m$$

Solution

$$AC = 100 - h $$
$$\tan 45^{\circ}= \dfrac{100-h}{x}= 1$$
$$\Rightarrow 100-h=x$$
Now, $$\tan 60^{\circ}=\dfrac{100}{x}=\sqrt{3}$$
$$\dfrac{100\sqrt{3}}{3}=x$$
Now $$h= 100-\dfrac{100\sqrt{3}}{3}=100\left ( 1-\dfrac{\sqrt{3}}{3} \right )=\dfrac{100}{3}(3-\sqrt{3})$$
Question 5
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The angle of elevation of the sun, when the length of the shadow of a tree is $$\sqrt{3}$$ times the height of the tree, is

A
$${30}^{\circ}$$

B
$${45}^{\circ}$$

C
$${60}^{\circ}$$

D
$${90}^{\circ}$$

Solution

REF. Image
Let AB be the tree and AC be its
shadow
Let $$\angle ACB =\theta $$
$$\Rightarrow \frac{AC}{AB}=\sqrt{3}$$
$$\Rightarrow cot\theta =\sqrt{3}$$
$$\therefore \theta = 30^{\circ}$$
Question 6
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If a tower $$30\ m$$ high, casts a shadow $$10\sqrt{3}\ m$$ long on the ground, then what is the angle of elevation of the sun?

A
$$60^{o}$$

B
$$30^{o}$$

C
$$45^{o}$$

D
$$90^{o}$$

Solution

Given, Length = 30 m (Tower)
Base = $$ 10 \sqrt 3$$ m (Shadow)
We have to find the angle, lets take that A
So,
tan A = $$\frac{Perpendicular}{Base}$$
tan A = $$\frac{30}{10 \sqrt 3}$$
tan A = $$\frac{3}{ \sqrt 3}$$
tan A = $$\sqrt 3$$
So, tan A = tan $$60^o$$
So, A = $$60^o$$
Question 7
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The length of a shadow of a vertical tower is $$\dfrac{1}{\sqrt{3}}$$ times its height. The angle of elevation of the sun is:

A
$$30^{o}$$

B
$$45^{o}$$

C
$$60^{o}$$

D
$$90^{o}$$

Solution

$$\because \tan 30^{\circ} = \dfrac{1}{\sqrt{3}}$$

The angle of elevation of the sun is $$30^{\circ}$$

Option A is correct.
Question 8
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From the top of a cliff $$24\mathrm { mt }$$ height, a man observes the angle of depression of a boat is to be $$60 ^ { \circ } .$$ The distance of the boat from the foot of the cliff is

A
$$8 \sqrt { 3 } \mathrm { mt }$$

B
$$8 \sqrt { 2 } \mathrm { mt }$$

C
$$8 \sqrt { 5 } \mathrm { mt }$$

D
$$8 \mathrm { mt }$$

Solution

$$\begin{array}{l} In\, ABC, \\ \tan { 60^{ 0 } } =\cfrac { { AB } }{ { BC } } \\ \sqrt { 3 } =\cfrac { { 24 } }{ { BC } } \\ BC=\cfrac { { 24 } }{ { \sqrt { 3 } } } \times \cfrac { { \sqrt { 3 } } }{ { \sqrt { 3 } } } \\ =8\sqrt { 3 } \end{array}$$

Hence, this is the answer.
Question 9
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A tower is $$50\sqrt{3}$$ m high. The angle of elevation of its top from a point 50 m away from its foot has measure............degree

A
65

B
60

C
30

D
15

Solution
Question 10
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If a pole $$12\ { m }$$ high casts a shadow $$4\sqrt { 3 }\ { m }$$ long on the ground then the angle of elevation is

A
$$30 ^ { \circ }$$

B
$$45 ^ { \circ }$$

C
$$60 ^ { \circ }$$

D
$$90 ^ { \circ }$$

Solution

Here,
Height of pole = $$12$$ $$m$$
Length of shadow = $$4\sqrt{3}$$ $$m$$
Now, angle of elevation, $$\tan \theta = \dfrac{perpendicular}{base}$$

$$\tan \theta = \dfrac{12}{4\sqrt{3}} =\dfrac{3}{\sqrt{3}} =\sqrt{3}$$

$$\tan \theta = \tan 60^o$$
Hence, $$\theta = 60^o$$
Hence, option (c) is correct.