Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 48

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Question 1
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The angle of elevation of the top of a tower as observed from a point on the horizontal ground is $$x$$. If we move a distance $$d$$ towards the flux of the tower, the angle of elevation increases to $$y$$, then the height of the tower is

A
$$\cfrac { d\tan { x } \tan { y } }{ \tan { y- } \tan { x } } $$

B
$$d\left( \tan { y } +\tan { x } \right) $$

C
$$d\left( \tan { y } -\tan { x } \right) $$

D
$$\cfrac { d\tan { x } \tan { y } }{ \tan { y } +\tan { x } } $$

Solution

In $$\triangle ABC,$$

$$\Rightarrow \tan { y }=\dfrac hz$$

$$\Rightarrow h=z\tan { y } \quad ...(1)$$

In $$\triangle ADC,$$

$$\tan { x }=\dfrac{h}{d+z}$$

$$\Rightarrow \tan x=\dfrac { z\tan { y } }{ d+z } $$ [from (1)]

$$\Rightarrow z\left( \tan { y- } \tan { x } \right) =d\tan { x } $$

$$\therefore\ h=\cfrac { d\tan { x } \tan { y } }{ \tan { y- } \tan { x } } $$
Question 2
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The angle of elevation of the top of a tower at a distance $$500$$ meters from the foot is $$30^{\circ}$$. The height of the tower is

A
$$250\sqrt 3$$ meters

B
$$\dfrac{500}{\sqrt3} $$ meters

C
$$500\sqrt 3 $$ meters

D
$$250$$ meters

Solution

In $$\Delta ABC$$,
$$\dfrac { h }{ 500 } =\tan{ 30 }^{ \circ }=\dfrac { 1 }{ \sqrt { 3 } } \Rightarrow h=\dfrac { 500 }{ \sqrt { 3 } } $$ meters
Question 3
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If the angles of depression and elevation of the top of a tower of height $$h$$ from the top and bottom of a second tower and $$x$$ and $$y$$ respectively, then the height of the second tower is

A
$$h(\cot y + \cot x)$$

B
$$h(\tan x + \tan y)$$

C
$$h(1 + \tan x \cot y)$$

D
$$h(\tan y\cot x + 1)$$
Question 4
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The angle of elevation of a cloud from a point $$h$$ meters above the surface of a lake is $${30}^{o}$$ and the angle of depression of its reflection is $${60}^{o}$$. Then the height of the cloud above the surface of the lake is

A
$$h\sqrt { 3 } $$

B
$$h$$

C
$$\sqrt { 2 } h$$

D
$$2h$$

Solution

$$ED=E'D=x$$
$$\cfrac { EB }{ AB } =\tan { { 30 }^{ o } } =\cfrac { 1 }{ \sqrt { 3 } } \Rightarrow (x-h)=\cfrac { AB }{ \sqrt { 3 } } ...(1)$$
Similarly $$(x+h)=AB\sqrt { 3 } $$
dividing (1) by (2)
$$\cfrac { x-h }{ x+h } =\cfrac { AB }{ \sqrt { 3 } .\sqrt { 3 } .AB } =\cfrac { 1 }{ 3 } $$
$$\Rightarrow x=2h$$
$$\therefore$$ Height of cloud $$=2h$$
Question 5
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The shadow of a stick of height $$1$$ metre, when the angle of elevation of the sun is $${ 60 }^{ o }$$, will be

A
$$\cfrac { 1 }{ \sqrt { 3 } }$$ m

B
$$\cfrac { 1 }{ 3 } $$ m

C
$$\sqrt { 3 }$$ m

D
$$3$$ m

Solution

$$\dfrac { AB }{ BC } ={ \tan60 }^{ 0 }\quad ;\quad BC=\dfrac { AB }{ { \tan60 }^{ 0 } } =\dfrac { 1 }{ \sqrt { 3 } } m$$
Question 6
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The angle of elevation of a tower from a point on the ground is $${30}^{o}$$. At a point on the horizontal line passing through the foot of the tower and $$100$$ meters nearer to it, the angle of elevation is found to be $${60}^{o}$$, then the height of the tower is

A
$$50\sqrt { 3 } $$ meters

B
$$\dfrac{50}{\sqrt { 3 }} $$ meters

C
$$\dfrac{100}{\sqrt { 3 } }$$ meters

D
$$100\sqrt { 3 } $$ meters

Solution

Let $$CD=x\ m$$
$$BC=100m$$

From $$\triangle ACD,$$
$$tan60^o=\dfrac{AD}{CD}$$
$$\Rightarrow \sqrt{3}=\dfrac{AD}{x}$$
$$\Rightarrow AD=x\sqrt { 3 } ..........(i)$$

From $$\triangle ABD$$
$$tan30^o=\dfrac{AD}{BD}$$
$$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{AD}{x+100}$$
$$\Rightarrow \sqrt{3}AD=x+100$$
$$\Rightarrow 3x=x+100$$ $$[By\ (i)]$$
$$\Rightarrow x=50$$

Hence the height of the tower AD$$=50\times\sqrt{3}\ m=50\sqrt { 3 }\ m $$
Question 7
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If the angle of elevation of the sun at $$8:00$$ O'clock is $$\alpha$$ and at $$10:00$$ O'clock is $$\beta$$ then ______ holds.

A
$$\alpha > \beta$$

B
$$\alpha < \beta$$

C
$$\alpha \geq \beta$$

D
$$\alpha = \beta$$

Solution
Question 8
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The angles of depression of two boats as observed from the mast head of a ship $$50$$ metres high are $${45}^{o}$$ and $${30}^{o}$$. The distance between the boats, if they are on the same side of mast head in line with it, is

A
$$50\sqrt { 3 } $$ metres

B
$$0(\sqrt { 3 } +1)$$ metres

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

D
$$50(1-\sqrt { 3 } )$$ metres

Solution

From $$\triangle ADC,\quad \dfrac { 50 }{ x } =Tan{ 45 }^{ 0 }\Rightarrow x=50m$$
From $$\triangle ABD,\quad \dfrac { 50 }{ h+50 } =Tan{ 30 }^{ 0 }$$
$$\Rightarrow \quad h=50\left( \sqrt { 3 } -1 \right) m$$
Question 9
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The angle of elevation of a stationary cloud from a point $$25\ m$$ above a lake is $$15^o$$ and the angle of depression of its image in the lake is $$45^o$$. The height of the cloud above the lake level is

A
$$ 25\ m$$

B
$$25\sqrt { 3 } \ m$$

C
$$50\ m$$

D
$$50\sqrt { 3 }\ m$$
Question 10
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The angle of elevation of the top of the tower from the point $$a$$ meter away from the base of the tower is $$60^{\circ}$$. The height of the tower is _______ meter.

A
$$\dfrac {1}{\sqrt {3}}a$$

B
$$2a$$

C
$$\sqrt {3}a$$

D
$$a$$

Solution

Let $$AB$$ be the height of the tower.
The angle of elevation of the top of the tower from the point $$a$$ meter away from the base of the tower is $$60^{\circ}$$.
$$\therefore BC =a$$ meter and $$m\angle ACB = 60^{\circ}$$
In $$\triangle ABC, \tan 60^{\circ} = \dfrac {AB}{BC}$$
$$\therefore \sqrt {3} = \dfrac {AB}{a}\\\Rightarrow AB = \sqrt {3}a$$
$$\therefore$$ The height of the tower is $$\sqrt {3}a$$ meter.