Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 13

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Question 1
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A pole $$15$$ m long rests against a vertical wall at an angle of $$\displaystyle 30^{\circ}$$ with the ground How high up the wall does the pole reach?

A
$$5$$ m

B
$$7$$ m

C
$$7.5$$ m

D
$$8$$ m

Solution

As shown in the fig. let $$AB$$ be the length of the pole.

$$\displaystyle \sin 30^{\circ}=\frac{x}{15}$$

Or

$$x=15\displaystyle \sin 30^{\circ}\\=\frac{15}{2}\\=7.5$$m
Question 2
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In the given figure given, $$\beta$$ is _____

A
Angle of elevation

B
Angle of depression

C
Line of sight

D
Horizontal level

Solution

Here, $$A$$ is an object below the level of the eye and $$O$$ is eye.
Also, $$OB$$ is a horizontal line through $$O$$.
Thus $$\angle AOB = B$$ is called the angle of depression.
Question 3
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In the given figure, $$\alpha$$ is _____

A
Angle of elevation

B
Angle of depression

C
Line of sight

D
Horizontal level

Solution

The given object A is above the level of the eye O. Thus $$\alpha$$ is the angle of elevation.
So, option A is correct.
Question 4
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the shadow cast by a tower is 30 m long when the elevation of sum is $$\displaystyle 30^{\circ} $$ If the elevation of sum is $$\displaystyle 60^{\circ} $$ then the length of the shadow is

A
30 m

B
20 m

C
10 m

D
$$\displaystyle 10\sqrt{3} m $$

Solution

$$\displaystyle \tan 30^{\circ}=\frac{h}{30}$$ or $$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{30}$$
$$\displaystyle \therefore h=\frac{30}{\sqrt{3}}$$
$$\displaystyle \tan 60^{\circ}=\frac{30/\sqrt{3}}{x}$$
or $$\displaystyle \sqrt{3}=\frac{30}{x\sqrt{3}}$$
$$\displaystyle \therefore $$ x=10
Question 5
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The angle of elevation of the top of a tower at a horizontal distance equal to the height of the tower from the base of the tower is

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

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

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

D
any acute angle

Solution

Let the angle of elevation be $$\theta$$, from the same distance from pole as the length of the pole then
$$\tan \theta = \dfrac{distance~ from \ pole}{length\ of\ pole}$$
$$\tan \theta = 1$$
$$\tan \theta = \tan 45$$
$$\theta = 45^{\circ}$$
Question 6
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The length of the shadow of a pole is $$\displaystyle \sqrt{3} $$ times its height The elevation of the sum must be

A
$$\displaystyle 60^{\circ} $$

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

C
$$\displaystyle 30^{\circ} $$

D
$$\displaystyle 15^{\circ} $$

Solution

$$\displaystyle \tan \theta =\frac{h}{\sqrt{3}h}=\frac{1}{\sqrt{3}}$$
$$\displaystyle \therefore \theta =30^{\circ}$$
Question 7
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From the point B a perpendicular BD is drawn on AC. If $$\cos 30^o=0.8$$, find the length of AD.

A
$$45$$

B
$$30$$

C
$$50$$

D
$$80$$

Solution

In $$ \displaystyle \bigtriangleup ' S ABC and ABD , \angle A $$ is common and $$ \displaystyle \angle ABC = \angle ADB $$
Hence $$ \displaystyle \bigtriangleup ' ABC and \bigtriangleup ABD $$ are equiangular
$$ \displaystyle \therefore \frac{AD}{AB}= \frac{AB}{AC} $$ OR
$$ \displaystyle AD=\frac{AB^{2}}{AC}=\frac{AC^{2}-BC^{2}}{AC} $$
$$ \displaystyle =AC\left [ 1-\left ( \frac{BC}{AC} \right )^{2} \right ] $$
$$ \displaystyle =\frac{BC}{\cos 30^{\circ}}\left [ 1-\left ( \cos 30^{\circ} \right )^{2} \right ] $$
$$ \displaystyle =\frac{100}{0.8}\left [ 1-\left ( 0.8 \right )^{2} \right ] $$
$$ \displaystyle =\frac{100\times 10}{8}\left [ 1-\frac{64}{100} \right ] $$
$$ \displaystyle =\frac{100\times 10}{8}\times \frac{36}{100}=45 $$
Question 8
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In case of angle of elevation, the observer has to look _____ to view the object.

A
Straight

B
Anywhere

C
Down

D
Up

Solution

In case of angle of elevation. The observer has to look up to view the object.
Question 9
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From a point P on a level ground, the angle of elevation of the top tower is $$30^{\circ}$$. 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, $$\angle APB = 30^{\circ}$$ and $$AB = 100\ m$$.
$$\dfrac {AB}{AP} = \tan 30^{\circ} = \dfrac {1}{\sqrt {3}}$$
$$\Rightarrow AP = (AB \times \sqrt {3})m$$
$$= 100\sqrt {3}m$$
$$= (100\times 1.73)m$$
$$= 173\ m$$.
Question 10
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In angle of elevation the observer looks ____ to view the object whereas in angle of depression he looks ____ to view object.

A
Up, up

B
Down, down

C
Down, up

D
Up, down

Solution

In angle of elevation the observer looks up to view the object whereas in angle of depression he looks down to view the object.