Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 44

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Question 1
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Two poles are $$10\ m$$ and $$20\,m$$ high. The line joining their tops makes an angle of $$15^\circ$$ with the horizontal. The distance between the poles is approximately equal to

A
$$36.3\ m$$

B
$$37.3\ m$$

C
$$38.3\ m$$

D
$$39.3\ m$$

Solution

Let $$AB$$ and $$CD$$ are two poles

$$AB=10\ m$$ and $$CD=20\ m$$

and $$\angle CAE=15^\circ$$

Here $$CE=CD-ED$$

$$=20-10=10m$$

In $$\triangle ACE,$$

$$\Rightarrow \tan15^\circ=\cfrac{CE}{AE}$$

$$\Rightarrow AE=\cfrac{10}{\tan15^\circ}$$

$$=\cfrac{10}{2-\sqrt3}=37.3\ m$$

Therefore the distance between the two poles is $$37.3\ m$$
Question 2
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A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is $${60}^{o}$$, when he retires $$40m$$ from the bank he finds the angle to $${30}^{o}$$. The breadth of river is

A
$$40 m$$

B
$$60 m$$

C
$$20 m$$

D
$$30 m$$

Solution

Let x be the breadth of the river
In $$\triangle $$ ABC $$\tan { { 60 }^{ 0 } } =\cfrac { H }{ x } $$
$$\sqrt { 3 } =\cfrac { H }{ x } $$
$$H=\sqrt { 3 } x\longrightarrow 1$$
In $$\triangle $$ DBE $$\tan { { 30 }^{ 0 } } =\cfrac { H }{ BE } $$
$$\cfrac { 1 }{ \sqrt { 3 } } =\cfrac { H }{ x+40 } $$
$$\cfrac { 1 }{ \sqrt { 3 } } =\cfrac { \sqrt { 3 } x }{ x+40 } $$ (From equation 1)
$$3x=x+40$$
$$2x=40$$
$$x=20$$
$$\therefore$$ Breadth of river is $$20m$$
Question 3
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The angle of elevation of the top of a TV tower from three points $$A, B$$ and $$C$$ in a straight line through the foot of the tower are $$\alpha, 2\alpha$$ and $$2\alpha$$ respectively. If $$AB = a$$, the height of the tower is

A
$$a\tan \alpha$$

B
$$a\sin \alpha$$

C
$$a\sin 2\alpha$$

D
$$a\sin 3\alpha$$

Solution

Based on the given information, we can draw the figure shown above.
Here, $$AB=a$$
$$E$$ is the top of the TV tower
Let $$BD=x$$ and $$ED=h$$ be the height of the tower.

We know that, $$\tan \theta=\dfrac{\text{Opposite Side}}{\text{Adjacent Side}}$$

$$\therefore \ $$ In $$\Delta BDE,$$
$$\tan 2\alpha =\dfrac { h }{ x } \\$$
$$\Rightarrow x=\dfrac { h }{ \tan2\alpha } \\$$

Also, in $$\Delta ADE,$$
$$\tan\alpha =\dfrac { h }{ a+x } \\$$
$$\Rightarrow \tan\alpha =\dfrac { h }{ a +\dfrac { h }{ \tan2\alpha } }\\$$

We know that, $$\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$$

$$\therefore\ \tan\alpha =\dfrac { h }{ a +\dfrac { h(1-{ \tan }^{ 2 }\alpha ) }{ 2\tan\alpha } }$$

$$\Rightarrow \tan\alpha =\dfrac { 2h\tan\alpha }{ 2a \tan\alpha +h-h{ \tan }^{ 2 }\alpha }\\$$
$$\Rightarrow h(1+{ \tan }^{ 2 }\alpha )=2a \tan\alpha\\$$
$$\Rightarrow h=2a\sin\alpha \cos\alpha\\$$
$$\Rightarrow h=a \sin2\alpha $$ ..... $$[\because \ \sin 2\theta=2\sin \theta\cos \theta]$$

Hence, the height of the TV tower is $$a\sin2\alpha$$.
Question 4
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The base of a cliff is circular. From the extremities of a diameter of the base, angles of elevation of the top of the cliff are $${ 30 }^{ \circ }$$ and $${ 60 }^{ \circ }$$. If the height of the cliff be $$500 \text{ m}$$, then the diameter of the base of the cliff is

A
$$\dfrac { 2000 }{ \sqrt { 3 } } \text{m}$$

B
$$\dfrac { 1000 }{ \sqrt { 3 } } \text{m}$$

C
$$\dfrac { 2000 }{ \sqrt { 2 } } \text{m}$$

D
$$1000\sqrt { 3 } \text{m}$$

Solution

In $$\triangle AEC$$,
$$\tan { { 60 }^{ o } } =\dfrac { 500 }{ { d }_{ 1 } }$$
$$ \Rightarrow \sqrt { 3 } =\dfrac { 500 }{ { d }_{ 1 } } $$
$$\Rightarrow { d }_{ 1 }=\dfrac { 500 }{ \sqrt { 3 } } \text{m}$$
and in $$\triangle BEC$$,
$$\tan { { 30 }^{ o } } =\dfrac { 500 }{ { d }_{ 2 } }$$
$$ \Rightarrow { d }_{ 2 }=500\sqrt { 3 } \text{m}$$
$$\therefore$$ Required diameter $$ ={ d }_{ 1 }+{ d }_{ 2 }$$
$$=\dfrac { 500 }{ \sqrt { 3 } } +500\sqrt { 3 } =\dfrac { 2000 }{ \sqrt { 3 } } \text{m}$$
Question 5
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A kite is flying with the string inclined at $${75}^{o}$$ to the horizon. If the length of the string is $$25 m$$, then height of the kite is :

A
$$\left( \dfrac { 25 }{ 2 } \right) { \left( \sqrt { 3 } -1 \right) }^{ 2 }$$

B
$$\left( \dfrac { 25 }{ 2 } \right) \left( \dfrac{\sqrt { 3 } +1}{\sqrt { 2 }} \right) $$

C
$$\left( \dfrac { 25 }{ 2 } \right) { \left( \sqrt { 3 } +1 \right) }^{ 2 }$$

D
$$\left( \dfrac { 25 }{ 2 } \right) \left( \sqrt { 6 } +\sqrt { 2 } \right) $$

Solution

Let H be the height of kite $$\therefore \sin { 75 }^{ 0 }=\cfrac { H }{ 25 }
$$$$\sin { { 75 }^{ 0 } } =\sin { \left( { 45 }^{ 0 }+{ 30 }^{ ' } \right) } $$
$$\sin\ 75^o=\sin(30^o+45^o)\\=\sin { { 45 }^{ 0 } } \cos { { 30 }^{ 0 } } +\cos { { 45 }^{ 0 } } \sin { { 30 }^{ 0 }
\\=\cfrac { 1 }{ \sqrt { 2 } } \times { \sqrt { 3 } }{ 2 } +\cfrac { 1 }{ \sqrt { 2 } } \times { 1 }{ 2 } }$$
$$=\cfrac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } $$
Put value of $$\sin { { 75 }^{ 0 } } $$ in equation 1
$$H=25\left[ \cfrac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \right] $$
$$H=\cfrac { 25 }{ 2 } \left[ \cfrac { \sqrt { 3 } +1 }{ \sqrt { 2 } } \right] $$
Question 6
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From the top of a tower, the angle of depression of a point on the ground is $$60^o$$ . If the distance of this point from the tower is $$ \dfrac {1}{\sqrt{3}+1}$$ m, then the height of the tower is :

A
$$ \dfrac {4 \sqrt{3}}{2}$$ m

B
$$ \dfrac { \sqrt{3} + 3}{2}$$ m

C
$$ \dfrac {3 - \sqrt{3}}{2}$$ m

D
$$ \dfrac { \sqrt{3} }{2}$$ m

Solution

Let $$AC$$ be the height of the tower $$=h$$

And $$BC$$ be the distance of the point from the tower $$=\dfrac{1}{\sqrt3+1}\ m$$
Given $$\angle ABC=60^o$$

In $$\triangle ABC,$$

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

$$\Rightarrow\sqrt3 = \dfrac {h}{\dfrac {1}{\sqrt{3}+1}}$$

$$ \Rightarrow {\sqrt{3}} = \dfrac{h(\sqrt{3}+1)}{1} $$

$$ \Rightarrow \dfrac {\sqrt{3}}{\sqrt{3}+1} = h$$

$$ \Rightarrow h = \dfrac {\sqrt{3}}{\sqrt{3}+1} \times \dfrac {\sqrt{3}-1}{\sqrt3-1} $$

$$ \Rightarrow h = \dfrac {\sqrt{3}(\sqrt{3}-1)}{3-1} = \dfrac {3 - \sqrt{3}}{2} $$

So $$Op-C$$
Question 7
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The angle of elevation of the top of a tower at a point on the ground is $$30^o$$. If on walking 20 m toward the tower, the angle of elevation becomes $$60^o$$, then the height of the tower is:

A
10 m

B
$$\frac{10}{\sqrt{3}}m$$

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

D
None of these

Solution

Let say $$h$$ be the height of the tower, $$x$$ be the initial distance and $$x+20$$ be the final distance away from the foot of the tower of the given point.

Now using trigonometry for the above triangles,
$$ \tan 60^o = \dfrac{h}{x} $$

$${\sqrt{3}} = \dfrac{h}{x} $$

$$ x=\dfrac{h}{{\sqrt{3}}} $$ .............. (1)

Also, $$ \tan 30^o = \dfrac{h}{20+x} $$

$$ \dfrac{1}{{\sqrt{3}}} = \dfrac{h}{20+x} $$

$$ \dfrac{1}{{\sqrt{3}}}\times \left ( 20+x \right ) = h $$

$$ \dfrac{1}{{\sqrt{3}}}\times \left ( 20+\dfrac{h}{\sqrt{3}} \right ) = h $$ using eq. (1)

$$20\sqrt 3 +h = 3h$$

$$ h=10{\sqrt{3}} $$

Ans will be option C.
Question 8
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A vertical pole $$PS$$ has two marks at $$Q$$ and $$R$$ such that the portions $$PQ, PR$$ and $$PS$$ subtend angles $$\alpha, \beta$$ and $$\gamma$$ at a point on the ground distance $$x$$ from the bottom of pole. If $$PQ = a, PR = b, PS = c$$ and $$\alpha + \beta + \gamma = 180^{\circ}$$ then $$x^{2}$$ is equal to

A
$$\dfrac {a^{3}}{a + b + c}$$

B
$$\dfrac {b^{3}}{a + b + c}$$

C
$$\dfrac {c^{3}}{a + b + c}$$

D
$$\dfrac {abc}{a + b + c}$$

Solution

We have $$\tan \alpha = \dfrac {a}{x}, \tan \beta = \dfrac {b}{x}$$ and $$\tan \gamma + \dfrac {c}{x}$$
$$\therefore \alpha + \beta + \gamma = 180^{\circ}$$
So, $$\tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \cdot \tan \beta \cdot \tan \gamma$$
or $$\dfrac {a}{x} + \dfrac {b}{x} + \dfrac {c}{x} = \dfrac {a}{x}\cdot \dfrac {b}{x}\cdot \dfrac {c}{x}$$
or $$x^{2} = \dfrac {abc}{a + b + c}$$.
Question 9
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The length of the shadows of a vertical pole of height $$h$$, thrown by the sun's rays at three different moments are $$h$$, $$2h$$ and $$3h$$. The sum of the angles of elevation of the rays at these three moments is equal to

A
$$\displaystyle\frac{\pi}{2}$$

B
$$\displaystyle\frac{\pi}{3}$$

C
$$\displaystyle\frac{\pi}{4}$$

D
$$\displaystyle\frac{\pi}{6}$$

Solution

Let $$OA$$ be the vertical pole of height $$h$$ and $$OP_1, OP_2$$ and $$OP_3$$ be the lengths of shadow.
In $$\Delta$$ $$AOP_1$$, we have
$$\tan\theta_1=\displaystyle\frac{OA}{OP_1}=\frac{h}{h}=1\Rightarrow \theta_1=\frac{\pi}{4}$$
In $$\Delta AOP_2$$, we have
$$\tan\theta_2=\displaystyle\frac{OA}{OP_2}=\frac{h}{2h}=\frac{1}{2}$$
$$\theta_2=\tan^{-1}1/2$$
Similarly, in $$\Delta AOP_3$$, we have
$$\tan\theta_3=1/3$$
$$\theta_3=\tan^{-1}(1/3)$$
Therefore, sum of the angles of elevation of the eyes
$$=\theta_1+\theta_2+\theta_3$$
$$=\displaystyle\frac{\pi}{4}+\tan^{-1}\left(\displaystyle\frac{1}{2}\right)+\tan^{-1}\left(\displaystyle\frac{1}{3}\right)$$
$$=\displaystyle\frac{\pi}{4}+\tan^{-1}\left(\displaystyle\frac{\displaystyle\frac{1}{2}+\frac{1}{3}}{1-\displaystyle\frac{1}{3}\times \frac{1}{2}}\right)$$ $$\left[\because \tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\dfrac{x+y}{1-xy}\right)\right]$$
$$=\displaystyle\frac{\pi}{4}+\tan^{-1}\left(\displaystyle\frac{5/6}{5/6}\right)$$
$$=\displaystyle\frac{\pi}{4}+\tan^{-1}(1)=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}$$
Question 10
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The angle of elevation of the top of a tower from the top of a house is $${ 60 }^{ \circ }$$ and the angle of depression of its base is $${ 30 }^{ \circ }$$. If the horizontal distance between the house and the tower be $$12\ m$$, then the height of the tower is

A
$$48\sqrt { 3 }\ m$$

B
$$16\sqrt { 3 }\ m$$

C
$$24\sqrt { 3 }\ m$$

D
$$\dfrac { 16 }{ \sqrt { 3 } }\ m$$

Solution

Let $$AE$$ be the tower and $$BD$$ be the house with height $$h.$$

In $$\triangle BDE,$$
$$\tan { { 30 }^{ \circ } } =\dfrac { BD }{ DE }$$
$$ \Rightarrow \dfrac { 1 }{ \sqrt { 3 } }=\dfrac { h }{ 12 } $$
$$\Rightarrow h=\dfrac { 12 }{ \sqrt { 3 } }\ m $$

In $$\triangle ACB,$$
$$\tan { { 60 }^{ \circ } } =\dfrac { AC }{ BC }$$
$$\Rightarrow \sqrt3 =\dfrac { H }{ 12 } $$
$$\Rightarrow H=12\sqrt { 3 }\ m$$

$$\therefore $$ Height of tower $$=H+h$$
$$=12\sqrt { 3 } +\dfrac { 12 }{ \sqrt { 3 } } $$
$$=\dfrac{36+12}{\sqrt3}$$
$$=\dfrac{48}{\sqrt3}$$
$$=16\sqrt { 3 }\ m$$