Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 50

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Question 1
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A vertical pole subtends an angle $$\tan^{-1}\dfrac{1}{2}$$ at point $$P$$ on the ground. The angle subtended by the upper half of the pole at the point $$P$$ is__

A
$$\tan^{-1}\dfrac{1}{4}$$

B
$$\tan^{-1}\dfrac{2}{9}$$

C
$$\tan^{-1}\dfrac{1}{8}$$

D
$$\tan^{-1}\dfrac{2}{3}$$

Solution

$$\angle ABP = 90^\circ $$ and $$\angle APB = ta{n^{ - 1}}\left( {\frac{1}{2}} \right)$$ (given)
Let height of the pole be $$h$$
In $$\Delta ABP$$
$$\dfrac{{AB}}{{BP}} = \tan \left( {\angle APB} \right)$$
$$ \Rightarrow \dfrac{h}{{BP}} = \tan \left( {{{\tan }^{ - 1}}\dfrac{1}{2}} \right)$$
$$ \Rightarrow \dfrac{h}{{BP}} = \dfrac{1}{2}$$
$$ \Rightarrow BP = 2h$$ ---(i)
in $$\Delta DBP$$
$$\dfrac{{DB}}{{BP}} = \tan \left( {\angle DPB} \right)$$
$$ \Rightarrow \dfrac{{\dfrac{h}{2}}}{{2h}} = \tan \left( {\angle DPB} \right)$$ (from (i))
$$ \Rightarrow \dfrac{h}{{4h}} = \tan \left( {\angle DPB} \right)$$
$$ \Rightarrow \tan \left( {\angle DPB} \right) = \dfrac{1}{4}$$
$$ \Rightarrow \angle DPB = {\tan ^{ - 1}}\left( {\dfrac{1}{4}} \right)$$
Angle subtended by upper half of the pole at point $$P$$ $$ = \angle APD = \angle APB - \angle DPB = {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) - {\tan ^{ - 1}}\left( {\dfrac{1}{4}} \right)$$
$$ = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{1}{2} - \dfrac{1}{4}}}{{1 + \dfrac{1}{2} \times \dfrac{1}{4}}}} \right)$$ $$[\because \tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\dfrac{x-y}{1+xy}\right)]$$
$$ = {\tan ^{ - 1}}\left( {\dfrac{2}{9}} \right)$$
Question 2
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The height of a tower is $$50$$ m. When angle of elevation changes from $$45^o $$ to $$ 30^o$$, the shadow of tower becomes $$x$$ metres more, find the value of $$x$$.

A
$$100(\sqrt{3}+1)$$

B
$$50(\sqrt{3}+1)$$

C
$$50(\sqrt{3}-1)$$

D
$$100(\sqrt{3}-1)$$
Question 3
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Find the angle of depression of a boat from the bridge at a horizontal distance of $$25$$m from the bridge, if the height of the bridge is $$25$$ m.

A
$$60^0$$

B
$$30^0$$

C
$$45^0$$

D
none

Solution

$$AB=BC=25m$$
$$\angle B=90^o$$
$$\angle BAC=\theta$$
Now, in $$\triangle ABC$$
$$\angle BAC=\angle DCA$$ (alternate angle)
$$\Rightarrow \tan \theta=\cfrac {AB}{BC}$$
$$\Rightarrow \tan\theta=1$$
$$\theta=45^o$$ ($$\because \tan 45^0=1$$)
Question 4
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From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a $$20\ meters$$ high building are $${45}^{o}$$ and $${60}^{o}$$ respectively. Then the height of the tower is

A
$$20(\sqrt{3}-1)$$

B
$$20(\sqrt{3}+1)$$

C
$$\dfrac{20}{\sqrt{3}-1}$$

D
$$\dfrac{20}{\sqrt{3}+1}$$

Solution
Question 5
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Two poles of equal heights are standing opposite each other on either side of the road which is $$80$$ m wide. From the points between them on the road, the  elevation of the top of the poles are $${60^ \circ }$$ and $${30^ \circ }$$ respectively. Find the height of the poles.

A
$$10\sqrt 3$$ m

B
$$20\sqrt 3$$ m

C
$$20$$ m

D
$$15 \sqrt 3$$ m

Solution

In $$\Delta DPC$$

$$\tan p = {{CD} \over {CP}}$$

$$\tan {30^0}$$

$${1 \over {\sqrt 3 }} = {{CD} \over {CP}}$$

$$CP = {{CP} \over {\sqrt 3 }}$$     (1)

IN $$\Delta APB$$

$$\tan p = {{AB} \over {PB}}$$

$$\tan {60^0} = {{AB} \over {BP}}$$

$$\sqrt 3 BP = AB$$

$$CD = \sqrt 3 BP$$ (2)

FROM (1)   & (2)

$${{CP} \over {\sqrt 3 }} = \sqrt 3 BP$$

$$CP = 3BP$$

NOW ,$$BC = BP + CP$$

$$80 = BP + 3BP$$

$$BP = 20m$$

$$CP = 60m$$

$$CD = \sqrt 3 BP$$

$$ = 20\sqrt 3 $$
Question 6
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The angle of elevation of jet fighter from $$A$$ point on the ground is $$60^0$$. After a fight of $$15$$ seconds angle of elevation changes to $$30^0$$. If the jet is flying at a speed of $$720 $$ km/hr. Find the constant height at which jet is flying. (Take $$\sqrt3= 1.732$$).

A
$$2.598km$$

B
$$2.598m$$

C
$$280m$$

D
$$2.97m$$

Solution

speed=$$720\ km/h$$
$$=720\times\cfrac{5}{18}=200m/s$$
distance travelled in $$16 sec$$
$$200\times15$$
now
$$\tan30^{\circ}=\cfrac{h}{x+3000}=\cfrac{1}{\sqrt{3}}$$
$$\tan60^{\circ}=\cfrac{h}{x}={\sqrt{3}}$$
$$=\cfrac { \cfrac { h }{ x } }{ \cfrac { h }{ x+3000 } } =\cfrac { \sqrt { 3 } }{ \cfrac { 1 }{ \sqrt { 3 } } } \\ =\cfrac { x+3000 }{ x } =3\\ =x+3000=3x=2x=3000\\ =x=1500$$
The constant height$$=1500\sqrt{3}$$
$$=1500\times1.732$$
$$=2598m=2.598km$$
Question 7
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The shadow of a vertical tower on a level ground increases by $$10$$ when the altitude of the sun changes from $$45^0$$ to $$30^0$$. Find the height of the tower, correct to two decimal places.

A
$$14$$ m

B
$$13.69$$ m

C
$$3.69$$ m

D
$$14.69$$ m

Solution

In $$\triangle ABC$$
$$\tan45^o=\cfrac {x}{y}$$
$$\Rightarrow x=y\longrightarrow (1)$$
In $$\triangle ABD$$
$$ \tan 30^o=\cfrac {x}{y+10}$$
$$\Rightarrow \cfrac {1}{\sqrt {3}}=\cfrac {x}{x+10}$$ {from equation (1)}
$$\Rightarrow x+10=\sqrt {3}x$$
$$\Rightarrow x=\cfrac {10}{\sqrt {3}-1}$$ $$\because \sqrt{3}=1.732$$
$$\Rightarrow x= 13.69m$$
Question 8
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The height of the tower is $$50$$ m when angle of elevation changes from $$45^0$$ to $$30^0$$, the shadow of tower becomes $$x$$ meters more, find the value of $$x$$.

A
$$50$$

B
$$50\sqrt{30}$$

C
$$50(\ 1-\sqrt3)$$

D
$$25$$

Solution

let $$BC=y$$

In $$\triangle ABC$$,
$$\tan { { 30 }^{ \circ } } $$ $$=\cfrac { 50 }{ y }$$

$$\cfrac { 1 }{ \sqrt { 3 } } =\cfrac { 50 }{ y } $$ $$\Rightarrow y=50\sqrt { 3 } $$

In $$\triangle ABD$$,
$$ \tan { { 45 }^{ \circ } } =\cfrac { 50 }{ y+x } $$

$$\Rightarrow 1=\cfrac { 50 }{ y+x } $$

$$y+x=50 \ \Rightarrow x=50-y$$

$$x=50-50\sqrt { 3 } $$

$$x=50(1-\sqrt { 3 } )$$
Question 9
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An aeroplane is moving one kilometer high from West to East horizontally. From a point on the ground the angle of elevation of the aeroplane is $${60}^{o}$$, and after $$10$$ seconds, the angle of elevation of the aeroplane is observed as $${30}^{o}$$, then the speed of the aeroplane in $$km/hour$$ is.

A
$$120\sqrt {3}$$

B
$$240\sqrt {3}$$

C
$$120$$

D
$$240$$
Question 10
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A tower $${T_1}$$ of height 60 m is located exactly opposite to a tower $${T_2}$$ of height 80 m on a straight road. From the top of $${T_1}$$, if the angle of depression of foot of $${T_2}$$ is twice the angle of elevation of the top of $${T_2}$$ , then the width (in m) of the road between the feet of the towers $${T_1}$$ and $${T_1}$$ is:

A
$$10\sqrt 2 $$

B
$$10\sqrt 3 $$

C
$$20\sqrt 3 $$

D
$$20\sqrt 2 $$

Solution

We know, $$tan(\Theta ) = \dfrac {Opposite}{Adjacent}$$

In $$\triangle ABC,\,\tan x = {{20} \over {AC}}$$

In $$ \triangle ACD,\,\tan 2x = {{60} \over {AC}}$$

$${\dfrac{2\tan x} {1 - {{\tan }^{2}x}}} = {\dfrac{60} {AC}}$$ ............. Using, $$tan2A=\dfrac { 2tanA }{ 1-{ tan }^{ 2 }A }$$

$${\dfrac {2 \times {\dfrac{20} {AC}}} {1 - {\dfrac{400} {A{C^2}}}}} = {\dfrac{60} {AC}}$$

$${\dfrac{40\,AC} {A{C^2} - 400}} = {\dfrac{60} {AC}}$$

$$40A{C^2} = 60A{C^2} - 24000$$

$$20A{C^2} = 24000$$

$$A{C^2} = 1200$$

$$AC = 20\sqrt 3 \,m$$