Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 17

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Question 1
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The angle of elevation of a cloud from a point $$h$$ metres above the lake water level is $$\theta$$ and the angle of depresion of its image in the lake is $$\phi$$. The height of the cloud is

A
$$\displaystyle \frac{h\sin(\phi+\theta)}{\sin(\phi-\theta)}$$

B
$$\displaystyle \frac{h\sin(\phi-\theta)}{\sin(\phi+\theta)}$$

C
$$\displaystyle \frac{h\sin(\theta+\phi)}{\sin(\theta-\phi)}$$

D
$$\displaystyle \frac{h\sin(\theta-\phi)}{\sin(\theta+\phi)}$$

Solution

$$\tan \theta = \dfrac{h_{1}-h}{d}$$

$$\tan \phi = \dfrac{h_{1}+h}{d}$$

$$\Rightarrow\dfrac{\tan \theta}{\tan \phi} = \dfrac{h_{1}-h}{h_{1}+h}$$

$$\Rightarrow \dfrac{\tan \theta +\tan \phi }{\tan \theta - \tan \phi}= \dfrac{h_{1}-h+h_{1}+h}{h_{1}-h-h_{1}-h}$$

Using C&D property, we get
$$\Rightarrow \dfrac{h_{1}}{h}= \dfrac{\tan \theta +\tan \phi }{\tan \theta - \tan \phi}$$

$$ \Rightarrow h_1= h \left ( \dfrac{\cos \theta \sin \phi + \sin \theta \cos \phi}{\sin \phi \cos \theta- \cos \phi \sin \theta} \right )$$

$$\Rightarrow h_1 = h\dfrac{\sin(\phi + \theta)}{\sin(\phi - \theta)}$$
Question 2
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$$A$$ and $$B$$ are two stations due north and south of a tower of height $$25m$$. The angles of depression of the stations from the top of the towe $$r$$ observed to be $$30^{0}$$ and $$45^{0}$$ respectively. The distance between the two stations is

A
$$25(\sqrt{3}+1)$$

B
$$25(\sqrt{3}-1)m$$

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

D
$$25(2+\sqrt{3})m$$

Solution

$$\tan 45 ^{0} = \dfrac{25}{d_{2}}$$
$$d_{2} = 25$$
$$\tan 30 ^{0} = \dfrac{25}{d_{1}}$$
$$d_{1} = 25 \sqrt{3}$$
$$d_{1} + d_{2} = 25(\sqrt{3}+1) m.$$
Question 3
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The angle of elevation of a cliff from a point $$A$$ on the ground and from the point $$B$$ $$100\ m$$ vertically above $$A$$ are $$\alpha$$ and $$\beta$$ respectively. The height of the cliff (in metres) is

A
$$\displaystyle \frac{100\tan\beta}{\cot\beta-\cot\alpha}$$

B
$$\displaystyle \frac{100\cot\beta}{\cot\beta-\cot\alpha}$$

C
$$\displaystyle \frac{100\tan\beta}{\cot\alpha+\cot\beta}$$

D
$$\displaystyle \frac{100\cot\beta}{\cot\alpha+\cot\beta}$$

Solution

$$\tan \alpha = \dfrac{h}{d}$$

$$\tan \beta = \dfrac{h-100}{d}$$

$$\Rightarrow \dfrac{\tan \alpha}{\tan \beta } = \dfrac{h}{h-100}$$

$$\Rightarrow \dfrac{\tan \alpha}{\tan \beta } = \dfrac{h}{h-100}$$

$$\Rightarrow \dfrac{\tan \beta}{\tan \alpha} = 1-\dfrac{100}{h}$$

$$\Rightarrow \dfrac{\tan \alpha - \tan \beta}{\tan \alpha} = \dfrac{100}{h}$$

$$\Rightarrow h = \dfrac{100\tan \alpha}{\tan \alpha-\tan \beta}$$
or
$$ h = \dfrac{100\cot \beta}{\cot \beta-\cot \alpha}$$
Question 4
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From the top of a cliff $$24\ m$$ 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
$$8\sqrt{3}\ m$$

B
$$8\sqrt{2}\ m$$

C
$$8\sqrt{5}\ m$$

D
$$8\ m$$
Question 5
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From the top of a tower, $$80\text{ m}$$ high, the angles of depression of two points $$P$$ and $$Q$$ in the same vertical plane with the tower are $$45^{^\circ}$$ and $$75^{^\circ}$$ respectively, find the value of $$PQ.$$
[Use $$\tan75^\circ=2+\sqrt3$$]

A
$$80 (\sqrt{3}+1)\text{ m}$$

B
$$80(\sqrt{3}-1)\text{ m}$$

C
$$160 (\sqrt{3}+1)\text{ m}$$

D
$$160 (\sqrt{3}-1)\text{ m}$$

Solution

Let the distance of point $$Q$$ from the base of the tower be $$y$$ and distance of point $$P$$ from $$Q$$ be $$x.$$

In $$\triangle QSR,$$
$$\tan 75^\circ = \dfrac{SR}{y}$$
$$\Rightarrow (2+\sqrt{3}) = \dfrac{80}{y}$$
$$\Rightarrow y= \dfrac{80}{(2+\sqrt{3})}$$

Now, in $$\triangle PSR,$$
$$\begin{aligned}{}\tan45^\circ&=\frac{SR}{x+y}\\=1& = \frac{{80}}{{x + y}}\\x + y& = 80\\x + \frac{{80}}{{2 + \sqrt 3 }} &= 80\\x &= 80 - \frac{{80}}{{2 + \sqrt 3 }}\\ &= \frac{{80\left( {\sqrt 3 + 1} \right)}}{{\sqrt 3 + 2}}\\& = 80\left( {\sqrt 3 - 1} \right)\text{ m}\end{aligned}$$

Hence, $$PQ$$ is equal to $$80(\sqrt3-1)\text{ m}.$$
Question 6
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In a prison wall there is a window of $$1$$ metre height, $$24$$ metres from the ground. An observer at a height of $$10 m$$ from ground, standing at a distance from the wall finds the angle of elevation of the top of the window and the top of the wall to be $$45^{ 0}$$ and $$60^{0}$$ respectively. The height of the wall above the window is

A
$$15\sqrt{3}$$

B
$$15\left(1-\displaystyle \frac{1}{\sqrt{3}}\right)$$

C
$$15(\sqrt{3}-1)$$

D
$$14(\sqrt{3}-1)$$

Solution

We can see that , $$CE=DF=10$$

Therefore, $$BC=BE-CE=24-10=14$$

in $$\triangle BCD$$
$$\tan45^o=\dfrac{14}{b}\Rightarrow b=14$$

Now, In $$\triangle ACD$$
$$\tan60^o=\dfrac{a+14}{14}\Rightarrow \sqrt3=\dfrac{a+14}{14}\Rightarrow a=14(\sqrt3-1)$$

Therefore, Answer is $$14(\sqrt3-1)$$
Question 7
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If from the top of a tower of $$60$$ metre high, the angles of depression of the top and floor of a house are $$\alpha$$ and $$\beta$$ respetivley and if the height of the house is $$\displaystyle \frac{60\sin(\beta-\alpha)}{x}$$, then $$x=$$

A
$$\sin\alpha\sin\beta$$

B
$$\cos\alpha\cos\beta$$

C
$$\sin\alpha\cos\beta$$

D
$$\cos\alpha\sin\beta$$

Solution

$$\tan \alpha=\dfrac{60-h}{d}$$
$$\tan \beta=\dfrac{60}{d}$$
$$\dfrac{\tan \alpha}{\tan \beta}=\dfrac{60-h}{60}$$

$$\dfrac{\tan \alpha}{\tan \beta}=1-\dfrac{h}{60}$$

$$\dfrac{h}{60}=\dfrac{\tan \beta-\tan \alpha}{\tan \beta}$$

$$\dfrac{h}{60}=\dfrac{\sin \beta \ \cos\alpha-\sin\alpha \ \cos\beta}{\cos\alpha \ \sin\beta}$$

$$\dfrac{h}{60}=\dfrac{\sin(\beta-\alpha)}{\sin\beta \ \cos\alpha}$$

$$h=\dfrac{60\ \sin(\beta-\alpha)}{\cos\alpha \sin\beta }$$

$$\therefore x=\cos\alpha \sin \beta $$
Question 8
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On the level ground the angle of elevation of the top of a tower is $$30^{0 }$$ On moving 20 metres nearer tower, the angle of elevation is found to be $$60^{0}$$ The height of the towerin metres is

A
10 $$\sqrt{3}$$

B
$$8\sqrt{3}$$

C
$$6\sqrt{3}$$

D
$$5\sqrt{3}$$

Solution

$$\tan\ 60=\dfrac{h}{d}$$
$$d=\dfrac{h}{\sqrt{3}}$$
$$\tan\ 30=\dfrac{h}{20+d}$$
$$\dfrac{1}{\sqrt{3}}=\dfrac{h}{20+\dfrac{h}{\sqrt{3}}}$$
$$20\sqrt{3}+h=3h$$
$$h=10\sqrt{3}$$
Question 9
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The angle of elevation of the top of a hill when observed from a certain point on the horizontal plane through its base is $$30^{0}$$. After walking 120 meters towards it on level ground the elevation is found to be $$60^{0}$$. Find the height of the hill(in meters).

A
120

B
$$60\sqrt{3}$$

C
$$120\sqrt{3}$$

D
60

Solution

From the triangle $$ABC$$, let
Height of the hill $$(BC) = h$$
$$AD = 120\,\mathrm{m}$$ and $$DB = d$$

In the $$\triangle DBC, \angle B = 90^\circ$$
$$\tan 60^\circ = \dfrac{BC}{DB}$$

$$\,\Rightarrow\sqrt{3} = \dfrac{h}{d}$$ ........(1)

In the $$\triangle ABC, \angle B = 90^\circ$$

$$\tan 30^{\circ}=\dfrac{BC}{AB}$$

$$\,\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{h}{d+120}$$......(2)

Dividing (1) by (2)

$$\dfrac{\dfrac{h}{d}}{\dfrac{h}{d+120}} = \dfrac{\sqrt 3}{\dfrac{1}{\sqrt3}}$$

$$\Rightarrow \dfrac{d+120}{d} =3$$
$$\Rightarrow 3d = d + 120$$
$$\Rightarrow d = 60$$
Substituting the value of $$d$$ in (1), we get
$$\sqrt{3} = \dfrac{h}{60}$$
$$\Rightarrow h = 60\sqrt{3}$$

$$\therefore$$ Height of the hill $$=60\sqrt3 \mathrm{m}$$
Question 10
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$$A$$ man observes a tower$$AB$$ of height $$h$$ from a point $$P$$ on the ground. He moves a distance $$d$$ towards the foot of the tower and finds that the angle of elevation is doubled. He further moves a distance $$\dfrac {3d}{4}$$ in the same direction and the angle of elevation is three times that at $$P$$. Then $$\displaystyle \frac{h^{2}}{d^{2}}=$$

A
$$\dfrac {35}{9}$$

B
$$\dfrac {35}{36}$$

C
$$\dfrac {36}{5}$$

D
$$\dfrac {36}{35}$$

Solution

In $$\triangle BQR$$ apply sine formula
$$\displaystyle\frac { d }{ \sin { \left( \pi -3\alpha \right) } } =\frac { \dfrac { 3 }{ 4 } d }{ \sin { \alpha } } =4\sin { \alpha } -3\sin { 3\alpha } $$
$$\Rightarrow 4\sin { \alpha } =3\left( 3\sin { \alpha } -4\sin ^{ 3 }{ \alpha } \right) $$
Since $$\displaystyle\sin { \alpha } \neq 0,4=9-12\sin ^{ 2 }{ \alpha } \Rightarrow \sin ^{ 2 }{ \alpha } =\frac { 5 }{ 12 } $$
From $$\triangle BQA$$
$$h=BQ\sin ^{ 2 }{ \alpha } =2d\sin { \alpha }\cos { \alpha } =2d\sin { d }\sqrt { 1-\sin ^{ 2 }{ \alpha } } $$
$$\displaystyle=2d\sqrt { \frac { 5 }{ 12 } } \sqrt { 1-\frac { 5 }{ 12 } } =\frac { 2d }{ 12 } \sqrt { 35 } $$
$$\Rightarrow 36{ h }^{ 2 }=35{ d }^{ 2 } $$
$$\Rightarrow \dfrac{h^2}{d^2}=\dfrac{35}{36}$$