Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 32

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Question 1
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The angles of elevation of the top of a tower from two points at a distance, $$x$$ and $$y$$ meters from the base and in the same straight line with it are complementary. Find the height of the tower.

A
$$h=\sqrt{xy}$$

B
$$h=\dfrac{x}{y}$$

C
$$h=\sqrt{y}$$

D
$$h=\sqrt{x}$$

Solution

Let the height of the tower be $$h.$$

In right $$\triangle ABC$$
$$\tan \theta=\dfrac{h}{x}\quad \dots (i)$$

In right $$\triangle ABD$$
$$\tan(90−\theta)=\dfrac{h}{y}\quad \dots (ii)$$

From $$(i)$$ & $$(ii)$$ we get,
$$\tan \theta\times \tan(90-\theta)=\dfrac{h}{x}\times \dfrac{h}{y}$$
$$\Rightarrow \tan\theta\cdot \cot\theta=\dfrac{h^2}{xy}$$
$$\Rightarrow h^2=xy$$
$$\Rightarrow\displaystyle h=\sqrt {xy}$$

Hence, option $$A$$ is the correct answer.
Question 2
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Two villages are 2 km apart. If the angles of depression of these villages when observed from a plane are around to be $$ \displaystyle 45^{\circ}$$ and $$60^{\circ} $$ respectively , then height of the plane in km is:
(Plane is between two villages)

A
$$ \displaystyle (3+\sqrt{3}) $$

B
$$ \displaystyle (3-\sqrt{3}) $$

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

D
$$ \displaystyle 3\sqrt{3} $$

Solution

Let $$PQ= h$$ m be a plane

From right angled $$\displaystyle \bigtriangleup PAQ,$$

$$\displaystyle \tan 45^{\circ}=\dfrac{PQ}{AQ} $$

$$\displaystyle \Rightarrow 1=\dfrac{h}{x} $$

$$\Rightarrow x=h$$ $$...(i)$$

Again from right angled $$\displaystyle \bigtriangleup PBQ $$

$$\Rightarrow \tan 60^{\circ}=\dfrac{PQ}{QB} $$

$$\Rightarrow \sqrt{3}=\dfrac{h}{2-x} $$

$$\Rightarrow 2\sqrt{3}-\sqrt{3}x=h $$

From equation $$(i)$$

$$\Rightarrow 2\sqrt{3}-\sqrt{3}h=h $$

$$\Rightarrow h+\sqrt{3}h=2\sqrt{3} $$

$$\Rightarrow h(1+\sqrt{3})=2\sqrt{3} $$

$$\Rightarrow h=\dfrac{2\sqrt{3}}{1+\sqrt{3}}$$

$$\Rightarrow h=\dfrac{2\sqrt{3}}{1+\sqrt{3}}\times \dfrac{1-\sqrt{3}}{1-\sqrt{3}} $$

$$\Rightarrow h=\displaystyle \frac{2\sqrt{3}-6}{1-3}$$

$$\Rightarrow h=\dfrac{-2(3-\sqrt{3})}{-2} $$

$$\therefore\ h\displaystyle =(3-\sqrt{3})\ km $$
Question 3
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On the same side of a tower two objects are located. Observed from the top of the tower their angles of depression are $$\displaystyle 45^{0}$$ and $$\displaystyle 60^{0}$$. If the height of the tower is $$150\ m$$, the distance between the objects is

A
$$63.5$$ $$m$$

B
$$76.9$$ $$m$$

C
$$86.7$$ $$m$$

D
$$90$$ $$m$$

Solution

Let $$AB$$ be the tower and $$C$$ and $$D$$ be the point of object
Given $$AB =150\ m$$ and $$\angle ACB=45^{0}$$ and $$ \angle ADB=60^{0}$$

In $$\Delta BAD$$,
$$\tan 60^{0}=\cfrac{AB}{AD}$$
$$\Rightarrow \sqrt{3}= \cfrac{150}{AD}$$
$$\Rightarrow AD=\cfrac{150}{\sqrt{3}} \text{ m}$$

In triangle $$BAC$$,
$$\tan 45^{0}=\cfrac{AB}{AC}$$
$$\Rightarrow 1=\cfrac{150}{AC}$$
$$\Rightarrow AC=150 \text{ m}$$

$$\therefore CD=AC-AD$$
$$=150-\cfrac{150}{\sqrt{3}}$$
$$=\left [ \cfrac{150(\sqrt{3}-1)}{\sqrt{3}} \right ]$$
$$=50(3-\sqrt{3})$$
$$=50\times 1.27$$
$$=63.5$$ $$\text{m}$$
Question 4
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A man standing at a point C is watching the top of a tower which makes an angle of elevation of $$\displaystyle 30^{0}$$ with the man's eye. The man walks some distance towards the tower to watch its top, the angle of elevation becomes $$\displaystyle 60^{0}$$. What is the distance between the base of the tower and the point C?

A
$$\displaystyle 4\sqrt{3}$$ units

B
8 units

C
12 units

D
Data inadequate

E
None of these

Solution

One of AB, AD and CD must have been given.
So the data is inadequate.
Question 5
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$$AB$$ is a straight road leading to $$C$$, the foot of a tower. $$A$$ is at a distance $$125\text{ m}$$ from $$B$$ and $$B$$ at $$75\text{ m}$$ meters from $$C$$. If the angle of elevation of the tower at $$B$$ be double the angle of elevation at $$A$$, then the find the value of $$\cos2\alpha,$$

A
$$\dfrac{4}{5}$$

B
$$\dfrac{1}{5}$$

C
$$\dfrac{2}{5}$$

D
$$\dfrac{3}{5}$$

Solution

In $$\triangle APB,$$
$$ \angle CBP=\angle A+\angle APB $$
$$\Rightarrow2\alpha =\alpha +\angle APB $$
$$\Rightarrow\angle APB= \alpha $$

This implies that $$AB=BP=125 m$$ as sides opposite to equal angles in a triangle are equal.

In $$\triangle BCP,$$
$$ \displaystyle \cos 2\alpha =\frac{75}{125} $$
$$ \Rightarrow \cos 2\alpha =\dfrac{3}{5} $$
Question 6
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The angles of elevation of a tower from a point on the ground is $$ \displaystyle 30^{\circ} $$ . At a point on the horizontal line passing through the foot of the tower and $$100$$ meters closer to it than the previous point, if the angle of elevation is found to be $$ \displaystyle 60^{\circ} $$ , then height of the tower is

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

B
$$ \displaystyle \frac{50}{\sqrt{3}} $$ meters

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

D
$$ \displaystyle \frac{100}{\sqrt{3}} $$

Solution

Let $$AB=h$$ be the height of the tower, and let $$AD=x$$ m

From right angled $$ \displaystyle \bigtriangleup ABD $$

$$ \displaystyle \tan 60^{\circ}=\frac{AB}{AD} $$

or $$ \displaystyle \sqrt{3}=\frac{h}{x} $$ or $$ x= \dfrac{h}{\sqrt{3}} $$

From right angled $$ \displaystyle \bigtriangleup BCA $$

$$ \displaystyle \tan 30^{\circ}=\frac{AB}{CA} $$

or $$ \displaystyle \frac{1}{\sqrt{3}}=\frac{h}{100+x} $$

or $$ \displaystyle h=\frac{100+x}{\sqrt{3}}=\frac{100+\dfrac{h}{\sqrt{3}}}{\sqrt{3}} $$

or $$ \displaystyle \frac{2}{\sqrt{3}}h=100 $$

or $$ \displaystyle \therefore h=50\sqrt{3}m $$
Question 7
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From a point $$p$$ on a level ground the angle of elevation of the top of a tower is $$\displaystyle 30^{0}$$ If the tower is $$100\ m$$ high the distance of point $$p$$ from the foot of the tower is

A
$$149\ m$$

B
$$156\ m$$

C
$$173\ m$$

D
$$200\ m$$

Solution

Let $$AB$$ be the tower then $$\displaystyle \angle APB=30^{0}$$ and $$\displaystyle AB=100m$$

$$\displaystyle \frac{AB}{AP}=\tan 30^{0}\\\ \ \ \ \ \ \ \ =\dfrac{1}{\sqrt{3}}$$

$$\displaystyle \Rightarrow AP=\left ( AB\times \sqrt{3} \right )=100\sqrt{3}m$$

$$\displaystyle =\left ( 100\times 1.73 \right )m$$

$$\displaystyle = 173m$$
Question 8
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The angle of elevation of the sun when the length of the shadow of a tree is $$\displaystyle \sqrt{3}$$ times the height of the tree is:

A
$$\displaystyle 30^{0}$$

B
$$\displaystyle 45^{0}$$

C
$$\displaystyle 60^{0}$$

D
$$\displaystyle 90^{0}$$

Solution

Let $$AB$$ be the tree and $$AC$$ be its shadow.
Also, $$AC=\sqrt3AB$$.
Let the angle of elevation of the sun $$=\displaystyle \angle ACB=\theta $$
Then in $$\triangle ABC,$$

$$\displaystyle \tan \theta = \frac{AB}{AC}=\dfrac{1}{\sqrt{3}} $$ {$$\because AC=\sqrt3AB$$}

$$ \tan 30^o =\dfrac{1}{\sqrt{3}}$$

$$\displaystyle \Rightarrow \theta =30^{0}$$
Question 9
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Two ships are sailing in the sea on the two sides of a lighthouse The angles of elevation of the top of the lighthouse as observed from the two
ships are $$\displaystyle 30^{0}$$ and $$\displaystyle 45^{0}$$ respectively If the lighthouse is $$100$$ m high the distance between the two ships is

A
$$173$$m

B
$$200$$m

C
$$273$$m

D
$$300$$m

Solution

Let $$AB$$ be the height of lighthouse and $$C$$ and $$D$$ be the position of the ships.
Given $$AB=100$$ m and $$\angle ACB=30^{0}$$$$\angle ADB=45^{0}$$
In $$\Delta BCA$$
$$\dfrac{AB}{AC}=tan 30^{0}\\\ \ \ \ \ \ \ =\dfrac{1}{\sqrt{3}}$$
$$\Rightarrow AC=\sqrt{3}AB\\\ \ \ \ \ \ \ \ \ \ =100\sqrt{3}$$
In $$\Delta BAD$$
$$\dfrac{AB}{AD}=tan 45^{0}\\\ \ \ \ \ \ \ \ =1$$
$$\Rightarrow AD=AB=100 m$$
$$\therefore CD=AC+AD\\=100\sqrt{3}+100\\=100(\sqrt{3}+1)\\=100(1.73+1)\\=273 m$$
Question 10
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An observer 1.6m tall is $$\displaystyle 20\sqrt{3}$$m away from a tower The angle of elevation from his eye to the top of the tower is $$\displaystyle 30^{0}$$ The height of the tower is

A
21.6m

B
23.2m

C
24.72m

D
None of these

Solution

Let $$AB$$ be the observer and $$CD$$ tower
Draw $$BE$$ perpendicular to $$CD$$
Then $$CE=AB=1.6 \ m$$
And $$BE=AC=$$ $$20\sqrt{3}$$ m
Then right angle triangle DEB
$$\therefore tan 30^{o}=\frac{DE}{BE}$$
$$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{DE}{20\sqrt{3}}$$
$$\Rightarrow DE=20\sqrt{3}$$m
Then $$CD=CE+DE=1.6+20=21.6\ m$$