Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 45

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Question 1
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The shadow of tower standing on a level ground is $$x\ m$$ long when the Sun's altitude is $$30^{\circ}$$, while it is $$y\ m$$ long when the altitude is $$60^{\circ}$$. If the height of the tower is $$45\cdot \dfrac {\sqrt {3}}{2}m$$, then $$x - y$$ is

A
$$45\ m$$

B
$$45\sqrt {3}m$$

C
$$\dfrac {45}{\sqrt {3}}m$$

D
$$45\dfrac {\sqrt {3}}{2}m$$

Solution

In $$\triangle CAD, \tan 30^{\circ} = \dfrac {45\sqrt {3}/ 2}{x}$$

$$\Rightarrow x = \dfrac {45\sqrt {3}}{2\times \dfrac {1}{\sqrt {3}}} = \dfrac {135}{2} m$$

And in $$\triangle BAD$$,

$$\tan 60^{\circ} = \dfrac {45\sqrt {3}/2}{y}$$

$$\Rightarrow y = \dfrac {45}{2}$$

$$\therefore x - y = \dfrac {135}{2} - \dfrac {45}{2} = \dfrac {90}{2} = 45\ m$$.
Question 2
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A man on the top of a cliff 100 m high, observe the angles of depression of two points on the opposite sides of the cliff as 30$$^o$$ and 60$$^o$$ respectively. Then, the distance between the two points is equal to

A
$$400 \sqrt 3 m$$

B
$$\dfrac{400}{\sqrt 3} m$$

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

D
$$200 \sqrt 3 m$$

Solution

As shown n the figure, let $$h$$ be the height of the cliff.
$$\tan 60^o = \dfrac{100}{x} $$
$$\sqrt{3}= \dfrac{100}{x} $$
$$ x= \dfrac{100}{\sqrt{3}}$$

Similarly ,
$$\tan 30^o = \dfrac{100}{y} $$
$$\dfrac{1}{\sqrt{3}}= \dfrac{100}{y} $$
$$ y= 100\sqrt{3}$$

Therefore,
$$ x+y= \dfrac{100}{\sqrt{3}}+100\sqrt{3}$$
$$ x+y = \dfrac{400}{\sqrt{3}} m$$

Hence option (B) is correct.
Question 3
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Find the height of the chimney when it is found that on walking towards it $$50$$ m in the horizontal line through its base, the angle of elevation of its top changes from $$30^{\circ}$$ and to $$60^{\circ}$$

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

B
$$25$$ m

C
$$25\sqrt {4} $$ m

D
$$\dfrac {25}{\sqrt {3}} $$ m

Solution

Let $$PQ = h$$ be the height of chimney.
$$A$$ and $$B$$ are the two points $$50$$ m apart.
In $$\triangle APQ$$, we have
$$\tan 30^{\circ} = \dfrac {h}{AP}$$
$$\Rightarrow AP = h\cot 30^{\circ} = \dfrac {h}{AP}$$
$$\Rightarrow AP = h\cot 30^{\circ} $$ .... (i)
and in $$\triangle QBP$$, we have
$$\tan 60^{\circ} = \dfrac {h}{BP}$$
$$\Rightarrow BP = h\cot 60^{\circ}$$ .... (ii)
Since $$ AP - BP = 50$$
Thus $$ h(\cot 30^{\circ} - \cot 60^{\circ}) = 50$$
$$\Rightarrow h = \dfrac {50}{\left (\sqrt {3} - \dfrac {1}{\sqrt {3}}\right )} = \dfrac {50\sqrt {3}}{3 - 1}$$
$$= \dfrac {50\sqrt {3}}{2}$$
$$ = 25\sqrt {3} $$ m , which is the required height
Question 4
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A house subtends a right angle at the window of the opposite house and the angle of elevation of the window from the bottom of the first house is $$60^o$$. If the distance between the two houses be 6 m, then the height of the first house is?

A
$$6\sqrt{3} \text{ m}$$

B
$$8\sqrt{3} \text{ m}$$

C
$$4\sqrt{3} \text{ m}$$

D
None of these

Solution

In $$\triangle ADC$$
$$ \dfrac { AD }{ AB } =\tan { { 60 }^{ \circ } } \\ \dfrac { 6\sqrt { 3 } }{ AB } =\sqrt { 3 } \\ \Rightarrow AB=6\\ \Rightarrow DC=6$$

In $$\triangle DEC$$

$$\dfrac { DC }{ EC } =\tan { { 60 }^{ \circ } } \\ \dfrac { 6 }{ EC } =\sqrt { 3 } \\ \Rightarrow EC=2\sqrt { 3 } \\ h=EC+CB\\ h=2\sqrt { 3 } +6\sqrt { 3 } \\ h=8\sqrt { 3 } $$

So option $$B$$ is correct.
Question 5
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Two posts are $$x$$ metres apart and the height of one is double that of the other. If from the mid-point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary, then the height (in meters) of the shorter post is

A
$$x\sqrt{2}$$

B
$$\dfrac {x}{\sqrt{2}}$$

C
$$\dfrac {x}{2\sqrt{2}}$$

D
$$\dfrac {x}{4}$$

Solution

Let the height of the shorter and the longer pole be $$h$$ and $$2h$$, respectively.
In $$\triangle ABM,$$
$$\begin{aligned}{}\tan \theta & = \dfrac{{AB}}{{BM}}\\& = \dfrac{h}{{x/2}}\\ &= \dfrac{{2h}}{x}\quad \dots (i)\end{aligned}$$

In $$\triangle CDM,$$
$$\begin{aligned}{}\tan ({90^\circ} - \theta )& = \dfrac{{CD}}{{MD}}\\\cot \theta & = \dfrac{{2h}}{{x/2}}\\ &= \dfrac{{4h}}{x}\quad \dots (ii)\end{aligned}$$

On multiplying equations $$(i)$$ and $$(ii)$$, we get
$$\tan \theta \cdot \cot \theta=\dfrac{2h}{x}\cdot \dfrac{4h}{x}$$
$$\Rightarrow 1=\dfrac{8h^2}{x^2}$$

$$\Rightarrow h^2=\dfrac{x^2}{8}$$

$$\Rightarrow h=\dfrac{x}{2\sqrt2}$$

Hence, option $$C$$ is the correct answer.
Question 6
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At a distance $$2h$$ $$m$$ from the foot of a tower of height $$h$$ $$m$$, the top of the tower and a pole at the top of the tower subtend equal angles. Height of the pole should be

A
$$\dfrac { 5h }{ 3 } m$$

B
$$\dfrac { 4h }{ 3 } m$$

C
$$\dfrac { 7h }{ 5 } m$$

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

Solution

In $$\Delta ABD$$, $$\tan { \alpha } =\dfrac { h }{ 2h } $$
$$\Rightarrow \tan { \alpha } =\dfrac { 1 }{ 2 }$$
In $$ \Delta ABC$$,
$$\tan { 2\alpha } =\dfrac { h+p }{ 2h } $$
$$\Rightarrow \dfrac { 2\tan { \alpha } }{ 1-\tan ^{ 2 }{ \alpha } } =\dfrac { h+p }{ 2h } $$
$$\Rightarrow \dfrac { 2\left( \dfrac { 1 }{ 2 } \right) }{ 1-{ \left( \dfrac { 1 }{ 2 } \right) }^{ 2 } } =\dfrac { h+p }{ 2h } $$
$$\Rightarrow \dfrac { 4 }{ 3 } =\dfrac { h+p }{ 2h } $$
$$\Rightarrow 8h=3h+3p$$
$$\Rightarrow 5h=3p\Rightarrow p=\dfrac { 5h }{ 3 } m$$
Question 7
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From a point $$\mathrm{a}$$ metre above the lake, the angle of elevation of a cloud is $$\alpha$$ and the angle of depression of its reflection is $$\beta$$, then height of the cloud is .......

A
$$\cfrac { a\sin { \left( \alpha +\beta \right) } }{ \sin { \left( \alpha -\beta \right) } } m$$

B
$$\cfrac { a\sin { \left( \alpha +\beta \right) } }{ \sin { \left( \beta -\alpha \right) } } m$$

C
$$\cfrac { a\sin { \left( \beta -\alpha \right) } }{ \sin { \left( \alpha +\beta \right) } } m$$

D
None of these

Solution

Let $$OQ$$ be the surface of lake ,$$P$$ be the point $$a$$ metre above the lake ,$$C$$ is the point where cloud is present at height $$H$$ and $$C'$$ is reflection of cloud under the lake.
In $$\quad \triangle PAC,\quad \tan { \alpha } =\cfrac { H-a }{ x } $$
$$\Rightarrow x=(H-a)\cot { \alpha }$$ ....(i)

In $$\triangle PAC',\tan { \beta } =\cfrac { H+a }{ x } \quad $$
$$\Rightarrow x=(H+a)\cot { \beta }$$ ....(ii)

From equations $$(i)$$ and $$(ii)$$, we get
$$(H+a)\cot { \beta } =(H-a)\cot { \alpha } $$
$$H \cot {\beta} + a \cot {\beta} = H \cot {\alpha} -a\cot {\alpha} $$
$$H \cot {\alpha} -H\cot {\beta} = a\cot {\alpha} +a\cot {\beta} $$
$$\Rightarrow H=\cfrac { a(\cot { \alpha } +\cot { \beta } ) }{ \cot { \alpha } -\cot { \beta } }$$
$$\dfrac{\cot {\alpha}+\cot {\beta}}{\cot {\alpha}-\cot {\beta}}=\dfrac{\dfrac{\cos{\alpha}}{\sin{\alpha}}+\dfrac{\cos{\beta}}{\sin{\beta}}}{\dfrac{\cos{\alpha}}{\sin{\alpha}}-\dfrac{\cos{\beta}}{\sin{\beta}}}$$
$$=\dfrac{\cos{\alpha}\sin{\beta}+\cos{\beta}\sin{\alpha}}{\cos{\alpha}\sin{\beta}-\cos{\beta}\sin{\alpha}}=\cfrac { \sin { \left( \alpha +\beta \right) } }{ \sin { \left( \beta -\alpha \right) } }$$
$$\therefore H=\cfrac { a\sin { \left( \alpha +\beta \right) } }{ \sin { \left( \beta -\alpha \right) } } $$
Question 8
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At $$4:00$$ p.m. on a sunny day, a stick $$2$$ feet tall casts a shadow $$5$$ feet long. At the same time, a tree nearby casts a shadow $$55$$ feet long. What is the height in feet, of the tree ?

A
$$137.5$$

B
$$27.5$$

C
$$22$$

D
$$10$$

Solution

At the same time, the position of the sun will be fixed for both, hence the angle subtended by shadow on the tip of the tree will also be the same.
Let $$x$$ be the height of the tree in feet.

For stick,
$$\tan \theta = \dfrac{2}{5}$$

For tree,
$$\begin{aligned}{}\tan \theta & = \frac{x}{{55}}\\\frac{2}{5}& = \frac{x}{{55}}\\x& = \frac{{2 \times 55}}{5}\\& = 22\end{aligned}$$
Question 9
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From an airplane flying vertically above a horizontal road, the angle of depression of two consecutive kilometer stones on the same side of the airplane are observed to be $${ 30 }^{ \circ }$$ and $${ 60 }^{ \circ}$$, respectively. The height at which the airplane is flying (in km) is:

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

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

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

D
$$2$$

Solution

Let the distance of two consecutive stones are $$x$$, $$x+1$$.
In $$\Delta BCD$$, we have
$$\tan { { 60 }^{ o } } =\dfrac { h }{ x } $$
$$\Rightarrow x=\dfrac { h }{ \sqrt { 3 } } $$ .....(i)
In $$\Delta ABC$$, we have
$$\tan { { 30 }^{ o } } =\dfrac { h }{ x+1 } $$
$$\Rightarrow \dfrac { 1 }{ \sqrt { 3 } } =\dfrac { h }{ x+1 } $$
$$\Rightarrow \dfrac { h }{ \sqrt { 3 } } +1=\sqrt { 3 } h$$ ......[from equation (i)]
$$\Rightarrow \dfrac { 2h }{ \sqrt { 3 } } =1$$
$$\Rightarrow h=\dfrac { \sqrt { 3 } }{ 2 }\ \mathrm{km} $$
Question 10
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The angle of elevation of the sum when the length of the shadow of a tower is equal to its height is:

A
$${ 30 }^{ o }$$

B
$${ 45 }^{ o }$$

C
$${ 60 }^{ o }$$

D
$${ 90 }^{ o }$$

Solution

$$\overline {AB}$$ is a building and $$\overline {BC}$$ is its shadow.
Given that the height of a building and the length of its shadow are equal.
$$\therefore AB = BC$$
Let the angle of elevation of the sun is $$\theta$$.
In $$\triangle ABC, \tan \theta = \dfrac {AB}{BC}$$
$$\therefore \tan \theta = 1 (\because AB = BC)$$
$$\therefore \theta = 45^{\circ}$$
So, the angle of elevation of the sun is $$45^{\circ}$$.