Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 21

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Question 1
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Two vertical poles $$20m$$ and $$80m$$ high stand a part on a horizontal plane. The height of the point of intersection of the lines joining the top of each pole to the foot of the other is

A
$$13m$$

B
$$14m$$

C
$$15m$$

D
$$16m$$

Solution

$$\dfrac bh= \dfrac{a+b}{20}$$

$$\dfrac ah= \dfrac{a+b}{80}$$

$$\dfrac {80} h = 1+ \dfrac ba $$

$$\dfrac{80-h}{h}= \dfrac ba $$

$$\dfrac{20} h = \dfrac ab +1$$

$$\cfrac{20}h = \dfrac h {(80-h)}+1$$

$$\dfrac {20} h= \dfrac{80-h+h}{80-h}$$

$$100h= 1600$$
$$h= 16 m$$
Question 2
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On one side of a road of width $$d$$ metres there is a point of observation $$P$$ at a height $$h$$ metres from the ground. If a tree on the other side of the road, makes a right angle at $$P$$, height of the tree in metres is:

A
$$\displaystyle \frac{h^{2}-d^{2}}{h}$$

B
$$\displaystyle \frac{h^{2}+d^{2}}{h}$$

C
$$\displaystyle \frac{d^{2}-h^{2}}{h}$$

D
$$\displaystyle \frac{2d^{2}+h^{2}}{h}$$

Solution

In $$\triangle PBA,{ AP }^{ 2 }={ h }^{ 2 }+{ d }^{ 2 }$$
In $$\triangle PCD,{ PC }^{ 2 }={ (H-h) }^{ 2 }+{ d }^{ 2 }$$
In $$\triangle ADC,{ AC }^{ 2 }={ AP }^{ 2 }+{ PC }^{ 2 }$$
$${ H }^{ 2 }={ h }^{ 2 }+{ d }^{ 2 }+{ (H-h) }^{ 2 }+{ d }^{ 2 }$$
$${ H }^{ 2 }={ h }^{ 2 }+2{ d }^{ 2 }+{ H }^{ 2 }+{ h }^{ 2 }-2Hh$$
$$2Hh=2{ h }^{ 2 }+2{ d }^{ 2 }\Rightarrow H=\cfrac { { h }^{ 2 }+{ d }^{ 2 } }{ h } $$
Question 3
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A vertical tower of height $$50$$ meters high stands on a sloping ground. The bottom of the tower is at the same level as the middle point of a vertical flagpole. From the top of the tower, the angles of depression of the top and bottom of the flagpole are $$15^{0}$$ and $$45^{0}$$ respectively. The height of the flagpole is

A
$$\displaystyle \frac{50}{\sqrt{3}}m$$

B
$$50\sqrt{3}m$$

C
$$\displaystyle \frac{100}{\sqrt{3}}m$$

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

Solution

In, $$\triangle ABC,$$
$$\Rightarrow \tan { 15° } =\dfrac { 50-x }{ a }$$
$$\Rightarrow a=\dfrac { 50-x }{ \tan { 15° } } $$
In, $$\triangle ADE,$$
$$\Rightarrow \tan { 45° } =\frac { 50+x }{ a } $$
$$\Rightarrow a=50+x$$
$$\Rightarrow \dfrac { \left( 50-x \right) }{ \tan { 15° } } =50+x$$
$$\Rightarrow 50-x=\tan { 15° } \left( 50+x \right) $$
$$\Rightarrow 50-x=\dfrac { \left( \sqrt { 3 } -1 \right) }{ \left( \sqrt { 3 } +1 \right) } \left( 50+x \right) $$
$$\Rightarrow \left( 50-x \right) \left( \sqrt { 3 } +1 \right) =\left( \sqrt { 3 } -1 \right) \left( 50+x \right) $$
$$\Rightarrow 50\sqrt { 3 } -\sqrt { 3 } x+50-x=50\sqrt { 3 } -50+\sqrt { 3 } x-x$$
$$\Rightarrow 2\sqrt { 3 } x=100$$
$$\Rightarrow x=\dfrac { 50 }{ \sqrt { 3 } } $$
$$\therefore$$ Height of flagpole $$=2x=\dfrac { 100 }{ \sqrt { 3 } } m$$
Hence, the answer is $$\dfrac { 100 }{ \sqrt { 3 } } m.$$
Question 4
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The height of a hill is $$3,300$$ metres. From the point $$D$$ on the ground the angle of elevation of the top of the hill is $$60^{0}$$. A balloon is moving with constant speed vertically upwards from $$D.$$ After $$5$$ minutes of its movement a person sitting in it observes the angle of elevation of the top of the hill as $$30^{0}$$. The speed of the balloon is

A
$$2.64$$ km/hr

B
$$26.4$$ km/hr

C
$$22.4$$ km/hr

D
$$2.24$$ km/hr

Solution

We can see that, $$BCDE$$ is a Rectangle
So, $$BE=CD$$ and $$BC=ED$$

In $$\triangle AED$$
$$\tan60^o=\dfrac{AE}{ED}\Rightarrow \sqrt3=\dfrac{3300}{ED}\Rightarrow ED=\dfrac{3300}{\sqrt3}=1100\sqrt3$$

$$ED=BC=1100\sqrt3$$

in $$\triangle ABC$$
$$tan30^o=\dfrac{AB}{BC}\Rightarrow AB=BC\tan30^o\Rightarrow 1100$$

Distance Covered by Balloon is $$CD=BE=AE-AB=3300-1100=2200\ m=2.2 \ km$$

So, Speed will be $$=\dfrac{2.2\ km}{\dfrac{5}{60}\ hr}=26.4 \ km/hr$$
Question 5
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Two light posts are of equal height. A person standing mid-way between the line joining their feet, observes the elevation of the posts to be $$30^{0}$$. After walking $$12$$ metres towards one of them, he observes that the same post now subtends an angle of $$60^{0}$$. Find the distance between them.

A
$$18\ m$$

B
$$26\ m$$

C
$$36\ m$$

D
$$24\ m$$
Question 6
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The angle of elevation of a tower from a point $$A$$ due south of it, is $$x$$, from a point $$B$$ due east of $$A$$, is $$y$$. If $$AB=l,$$ then the height $$h$$ of the tower is given by

A
$$\displaystyle \frac { l }{ \sqrt { \cot ^{ 2 }{ y } -\cot ^{ 2 }{ x } } } $$

B
$$\displaystyle \frac { l }{ \sqrt { \tan ^{ 2 }{ y } -\tan ^{ 2 }{ x } } } $$

C
$$\displaystyle \frac { 2l }{ \sqrt { \cot ^{ 2 }{ y } -\cot ^{ 2 }{ x } } } $$

D
None of these

Solution

Let $$OP$$ be the tower of height $$h$$.
In right angles $$\triangle OAP$$
$$\displaystyle \angle OAP=x\Rightarrow \frac { OA }{ h } =\cot { x } \Rightarrow OA=h\cot { x } $$ ...(1)
In right angled $$\triangle OBP$$
$$\displaystyle \angle OBP=y\Rightarrow \frac { OB }{ h } =\cot { y } \Rightarrow OB=h\cot { y } $$ ...(2)
In right angles $$\triangle OAB$$
$${ AB }^{ 2 }+{ OA }^{ 2 }={ OB }^{ 2 }\Rightarrow { l }^{ 2 }+{ h }^{ 2 }\cot ^{ 2 }{ x } ={ h }^{ 2 }\cot ^{ 2 }{ y } $$
$$\displaystyle \Rightarrow { h }^{ 2 }\left( \cot ^{ 2 }{ y } -\cot ^{ 2 }{ x } \right) ={ l }^{ 2 }\Rightarrow h=\frac { l }{ \sqrt { \cot ^{ 2 }{ y } -\cot ^{ 2 }{ x } } } $$
Question 7
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A pole 6 m high casts a shadow $$2\sqrt{3}$$ m long on the ground, then the Sun's elevation is

A
$$60^{0}$$

B
$$45^{0}$$

C
$$30^{0}$$

D
$$90^{0}$$

Solution

The sun's elevation angle will be opposite to the side which depicts the height of the pole, and base will be the length of the shadow.
Base=$$2\sqrt{3}\ m$$
height=$$6\ m$$
$$\tan(\theta)=\dfrac{6}{2\sqrt3}$$$$=\sqrt{3}$$
$$\theta=\tan^{-1}(\sqrt{3})$$
$$=60^0$$
$$\theta$$ being the angle of elevation.
Question 8
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The angle of elevation a vertical tower standing inside a triangular at the vertices of the field are each equal to $$\theta$$. If the length of the sides of the field are $$30\ m,\ 50\ m$$ and $$70\ m$$, the height of the tower is:

A
$$70\sqrt{3}\tan\theta\ m$$

B
$$\displaystyle \frac{70}{\sqrt{3}}\tan\theta\ m$$

C
$$\displaystyle \frac{50}{\sqrt{3}}\tan\theta\ m$$

D
$$75\sqrt{3}\tan\theta\ m$$

Solution

$$tan \theta = \dfrac hR$$

$$R= \dfrac{abc}{40}$$

$$\Delta = \sqrt{\left ( \dfrac{3+5+7}{2} \right )\left ( \dfrac{3+5+7}{2} \right )\left ( \dfrac{7+3-5}{2} \right )\left ( \dfrac{7+5-3}{2} \right )\times 10^4} = \dfrac{15\times 1\times 5\times 9\times 10^4}{10}$$

$$\Delta = 25\times 5\times 3\sqrt{3}$$

$$R= \dfrac{30\times 50\times 70}{4\times 25\times 5\times 3\sqrt{3}}$$

$$R= \dfrac{70}{\sqrt{3}}$$

$$h= \dfrac {70}{\sqrt{3}} tan \theta $$
Question 9
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Two vertical poles $$20\ m$$ and $$80\ m$$ high stand apart $$50m$$ on a horizontal plane. The height of the point of intersection of the lines joining the top of each pole to the foot of the other is

A
$$15\ m$$

B
$$16\ m$$

C
$$18\ m$$

D
$$50\ m$$

Solution

Let A be the point of the top of the $$20m$$ pole.
Let B be the point of the foot of the$$ 20m$$ pole
Let C be the point of the top of the $$ 80m$$pole.
Let D be the point of the foot of the$$ 80m$$ pole.
Let E be the point of intersection of the line segments AD and BC.
From E drop a perpendicular to the ground.
Let F be the point on the ground where perpendicular from E meets the ground.
Let X be the distance from E to F.
Let P be the distance from B to F.
Let Q be the distance from F to D.
Triangles BEF and BCD are similar so we get Equation 1:
$$\dfrac { X }{ P } =\dfrac { 80 }{ \left( P+Q \right) } $$
Triangles DEF and DAB are similar so we get
$$\dfrac { X }{ Q } =\dfrac { 20}{ \left( P+Q \right) } $$
Multiplying both sides by 4 we get equation 2
$$\dfrac { 4X }{ Q } =\dfrac { 80 }{ \left( P+Q \right) } $$
Combining Equations 1 and 2 we get
$$\dfrac { X }{ P } =\dfrac { 4X }{ Q } $$
Simplifying this we get
$$\dfrac { 1 }{ P } =\dfrac { 4 }{ Q } $$
$$Q=4P$$
Substituting$$ Q=4P$$ into Equation 1 we get:
$$\dfrac { X }{ P } =\dfrac { 80 }{ \left( P+Q \right) } $$

$$\dfrac { X }{ P } =\dfrac { 80 }{ P+4P } $$

$$\dfrac { X }{ P } =\dfrac { 80 }{ 5P } $$

$$X=\dfrac { 80 }{ 5P } \times P$$

$$X=\dfrac { 8 }{ 5 } = 16 m$$
Question 10
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Two poles of height $$a$$ and $$b$$ stand at the centres of two circular plots which touch each other externally at a point and the two poles subtends angles of $$30^{o}$$ and $$60^{o}$$ respectively at this point. Then the distance between the centres of these plots is:

A
$$a+b$$

B
$$\displaystyle \frac{3a+b}{\sqrt{3}}$$

C
$$\displaystyle \frac{a+3b}{\sqrt{3}}$$

D
$${a}\sqrt{3}+b$$

Solution

In $$\triangle ABC,\tan { 30° } =\cfrac { a }{ BC } $$

$$BC=\sqrt { 3 } a.......(1)$$

In $$\triangle ECD,\tan { 60° } =\cfrac { b }{ CD } $$

$$CD=\cfrac { b }{ \sqrt { 3 } } .......(2)$$

Required distance is equal to $$BC+CD$$

$$BC+CD=a\sqrt { 3 } +\cfrac { b }{ \sqrt { 3 } } =\cfrac { 3a+b }{ \sqrt { 3 } } $$

Distance between poles$$=\cfrac { 3a+b }{ \sqrt { 3 } } $$