Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 22

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Question 1
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Each side of square subtends an angle of $$60^{o}$$ at the top of a tower of $$h$$ meter height standing in the centre of the square. If $$a$$ is the length of each side of the square then which of the following is/are correct?

A
$$2a^{2}=h^{2}$$

B
$$2h^{2}=a^{2}$$

C
$$3a^{2}=2h^{2}$$

D
$$2h^{2}=3a^{2}$$

Solution

As per question, the tower is standing on the center of square
Then the tower stands the point of intersection of two diagonal of square
Then the base of square of triangle be $$\dfrac{1}{2}a$$ and the diagonal of square $$=\dfrac{\sqrt{2}a}{2}=\dfrac{a}{\sqrt{2}}$$
And its perpendicular to be $$h$$ meters,
So in this triangle
$$\therefore \tan 60^o=\dfrac{h}{\frac{a}{\sqrt{2}}}$$

$$\sqrt{3}=\dfrac{\sqrt{2}h}{a}$$
$$\Rightarrow \sqrt{3}a=\sqrt{2}h$$
Squaring both side we get
$$3a^{2}=2h^{2}$$
Question 2
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The ratio of the length of a pole and its shadow is 1: $$\sqrt{3}$$. The angle of elevation of the sun is :

A
$$90^{\circ}$$

B
$$60^{\circ}$$

C
$$30^{\circ}$$

D
$$45^{\circ}$$

Solution

$$\tan { \theta } =\dfrac { h }{ x } $$
$$\Rightarrow \tan { \theta } =\dfrac { 1 }{ \sqrt { 3 } } $$
$$\Rightarrow \tan { \theta } =\tan { 30° } $$
$$\Rightarrow \theta =30°$$
Hence, the answer is $$ =30°.$$
Question 3
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If the ratio of height of a tower and the length of its shadow on the ground is $$\sqrt{3}:1 $$, then the angle of elevation of the sun is

A
$$60^{\circ}$$

B
$$45^{\circ}$$

C
$$30^{\circ}$$

D
$$90^{\circ}$$
Question 4
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The angle of elevation of the sun, when the length of the shadow of a pole is equal to its height, is

A
$$30^{\circ}$$

B
$$45^{\circ}$$

C
$$60^{\circ}$$

D
$$90^{\circ}$$

Solution

Let the height of the pole be $$AB=x\ m$$

Then the length of the shadow of the pole $$AB$$ will be $$OB=x\ m$$

Let the angle of elevation be $$\theta$$ i.e., $$\angle AOB=\theta$$

In the right-angled triangle $$OAB,$$ we have

$$\Rightarrow \tan \theta=\dfrac{AB}{OB}$$

$$=\dfrac{x}{x}$$

$$=1$$

$$\Rightarrow \theta=\tan^{-1} 45^\circ$$

Therefore angle of elevation is $$45^\circ$$
Question 5
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The angle of elevation of the top of a tower from a point on the ground is $$45^{\circ}$$. If the point is 42 m away from the foot of the tower, the height of the tower is.

A
$$63m$$

B
$$21m$$

C
$$84m$$

D
$$42m$$
Question 6
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A pole $$10\ m$$ high cast a shadow $$10\ m$$ long on the ground, then the sun's elevation is:

A
$$60^{\circ}$$

B
$$45^{\circ}$$

C
$$30^{\circ}$$

D
$$90^{\circ}$$

Solution

In $$\triangle BAC,$$
$$\Rightarrow \tan { \theta } =\dfrac { AB }{ BC } =\dfrac { 10 }{ 10 } =1$$
$$\Rightarrow \theta =45^{\circ}$$
Hence, the answer is $$ 45^{\circ}.$$
Question 7
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The measure of angle of elevation of top of tower $$75\sqrt{3}$$ m high from a point at a distance of 75 m from foot of tower in a horizontal plane is :

A
$$30^{\circ}$$

B
$$60^{\circ}$$

C
$$90^{\circ}$$

D
$$45^{\circ}$$

Solution

Let $$AB$$ be the tower and $$C$$ be the point from which angle of elevation is observed.
Let the angle of elevation be $$\theta.$$
In $$\triangle ACB,$$
$$\Rightarrow \tan { \theta } =\dfrac { AB }{ BC } =\dfrac { 75\sqrt { 3 } }{ 75 } $$
$$\Rightarrow \tan { \theta } =\sqrt { 3 } $$
$$\Rightarrow \theta =60^{\circ}$$
Hence, the answer is $$60^{\circ}.$$
Question 8
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The given figure, shows the observation of point $$C$$ from point $$A$$. The angle of depression from $$A$$ is:

A
$$60^{\circ}$$

B
$$30^{\circ}$$

C
$$45^{\circ}$$

D
$$75^{\circ}$$

Solution

Let angle of depression be $$\theta$$
Now, $$\angle ACB=\theta $$ {Alternate angles}
In $$\triangle ACB,$$
$$\Rightarrow \tan { \theta } =\dfrac { AB }{ BC } =\dfrac { 2 }{ 2\sqrt { 3 } } =\dfrac { 1 }{ \sqrt { 3 } } $$
$$\Rightarrow \tan { \theta } =\tan { 30^{\circ} }$$
$$\Rightarrow \theta =30^{\circ}$$
Hence, the answer is $$30^{\circ}.$$
Question 9
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If the angle of depression of an object from a $$75$$ m high tower is $$30^{\circ}$$, then the distance of the object from the base of the tower is:

A
$$125$$ m

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

C
$$75\sqrt{3}$$ m

D
$$150$$ m

Solution

Let the distance of the object from the base of tower be $$x$$ m
$$AB$$ is the height of the tower$$=75$$ m

In $$\triangle ACB,$$
$$\Rightarrow \tan { 30^{\circ} } =\dfrac { AB }{ BC } $$
$$\Rightarrow \dfrac { 1 }{ \sqrt { 3 } } =\dfrac { 75 }{ x } $$
$$\Rightarrow x= 75\sqrt { 3 } $$

Hence, the distance of the object is $$ 75\sqrt { 3 }m.$$
Question 10
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The angle formed by the line of sight with the horizontal, when the point being viewed is above the horizontal level is called:

A
Vertical Angle

B
Angle of Depression

C
Angle of Elevation

D
Obtuse Angle

Solution

The angle of Elevation is the angle formed by the line of sight with the horizontal, when the point being viewed is above the horizontal level.
Hence, the answer is angle of elevation.