Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 41

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Question 1
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The angles of elevation of the top from two points on either side of a tree are $$30^{\circ}$$ and $$60^{\circ}$$. Find the height of the tree if the two points are $$52\ m$$ apart.

A
$$39\ m$$

B
$$22.5\ m$$

C
$$65\ m$$

D
$$225\ m$$

Solution

Let $$B$$ and $$C$$ are two points, such that distance between them is $$52\ m$$ and $$'h'$$ be the height of tree.
Now, in $$\triangle ABD,$$
$$\Rightarrow \tan { 30° } =\dfrac { AD }{ BD } =\dfrac { h }{ x } $$
$$\Rightarrow \dfrac { 1 }{ \sqrt { 3 } } =\dfrac { h }{ x } $$
$$\Rightarrow x=\sqrt { 3 } h\longrightarrow \left( 1 \right) $$
In $$\triangle ACD,$$
$$\Rightarrow \tan { 60° } =\dfrac { AD }{ DC } =\dfrac { h }{ 52-x } $$
$$\Rightarrow \sqrt { 3 } =\dfrac { h }{ 52-x } $$
$$\Rightarrow 52-x=\dfrac { h }{ \sqrt { 3 } } $$
Put $$x=\sqrt { 3 } h$$ from equation $$(1),$$ we get
$$\Rightarrow 52-\sqrt { 3 } h=\dfrac { h }{ \sqrt { 3 } } $$
$$\Rightarrow 52\sqrt { 3 } -3h=h$$
$$\Rightarrow 4h=52\sqrt { 3 } $$
$$\Rightarrow 4h=13\sqrt { 3 } $$
$$\Rightarrow h=22.5$$
$$\therefore$$ the height of the tree $$22.5\ m.$$
Hence, the answer is $$22.5\ m.$$
Question 2
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The height of a tower is 'h' and $$\alpha$$ is the angle of elevation of the top of the tower. On moving a distance $$\dfrac {h}{2}$$ towards the tower, the angle of elevation becomes $$\beta$$. What is the value of $$\cot \alpha - \cot \beta$$?

A
$$\dfrac {1}{2}$$

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

C
$$1$$

D
$$2$$

Solution

In, $$\triangle ACB,$$
$$\cot { \beta } =\dfrac { BC }{ AB } =\dfrac { x }{ h } $$
$$\Rightarrow \cot { \beta } =\dfrac { x }{ h } $$
$$x=h\cot { \beta } \longrightarrow \left( 1 \right) $$
In $$\triangle ADB,$$
$$\cot { \alpha } =\dfrac { BD }{ AB } =\dfrac { x+\dfrac { h }{ 2 } }{ h } $$
$$h\cot { \alpha } =x+\dfrac { h }{ 2 } $$
$$x=h\cot { \alpha } -\dfrac { h }{ 2 } \longrightarrow \left( 2 \right) $$
Subtracting $$(2)\;\&\;(1),$$ we get
$$h\cot { \alpha } -\dfrac { h }{ 2 } -h\cot { \beta } =0$$
$$\left( \cot { \alpha } -\cot { \beta } \right) =\dfrac { h }{ 2 } $$
$$ \cot { \alpha } -\cot { \beta } =\dfrac { 1 }{ 2 } $$
Hence, the answer is $$\dfrac { 1 }{ 2 }.$$
Question 3
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A pole is $$20$$ feet high. A taut wire that is $$46$$ feet extends from the top of the pole to the ground. What is the angle of depression, to the nearest degree, from the top of the pole to the bottom of the wire?

A
$$23$$

B
$$26$$

C
$$43$$

D
$$64$$

E
$$67$$

Solution

The given setup will form a right angled triangle with hypotenuse length $$46$$ feet and height of pole is $$20$$ fee
Let the angle of depression be $$\alpha$$
$$\because \sin \alpha=\dfrac{opposite}{hypotenuse}$$
$$\therefore$$ Angle of depression $$=\alpha =\sin ^{ -1 }{ \left( \dfrac { 20 }{ 46 } \right) } =26°$$
Hence, the angle of depression is $$26^{o}$$
Question 4
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The angle of elevation of a ladder leaning against a wall is $$60^{\circ}$$ and the foot of the ladder is $$4.6\ m$$ away from the wall. The length of the ladder is:

A
$$2.3\ m$$

B
$$4.6\ m$$

C
$$7.8\ m$$

D
$$9.2\ m$$

Solution

Let $$AB$$ be the wall and $$BC$$ be the ladder.
Then, $$\angle ACB = 60^{\circ}$$ and $$AC = 4.6\ m$$.
$$\dfrac {AC}{BC} = \cos 60^{\circ} = \dfrac {1}{2}$$
$$\Rightarrow BC = 2\times AC$$
$$= (2\times 4.6)m$$
$$= 9.2\ m$$
Question 5
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A man standing at a point P is watching the top of a tower, which makes an angle of elevation of $$30^{\circ}$$ with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes $$60^{\circ}$$. What is the distance between the base of the tower and the point P?

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

B
$$8\ units$$

C
$$12\ units$$

D
Data inadequate

Solution

From the data, we conclude that one of $$AB, AD$$ and $$CD$$ length are not provided.
So, the data is inadequate.
Question 6
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The angle of depressions of the tap and the foot of a chimney as seen from the top of a second chimney, which is 150 m high and standing on the same level as the first are $$\theta$$ and $$\phi$$ respectively then the distance between their tops when $$tan\, \theta =\frac{4}{3}\,\,and\,\, tan\, \phi =\frac{5}{2}$$ is.

A
$$\dfrac{150}{\sqrt{3}} m$$

B
$$100 \sqrt {3} m$$

C
150 m

D
100 m

Solution

Given: $$tan \theta =\frac{4}{3}$$ and then $$tan \phi =\frac{5}{2}$$
In $$\Delta ABE$$
$$tan\phi =\frac{150}{d}$$$$\Rightarrow d=150 cot \phi $$
$$=150 \times \frac{2}{5}=60 m$$
In $$\Delta DCE$$,
$$DE^2=DC^2+CE^2$$
$$\Rightarrow x^2=60^2+80^2=10000$$
$$\Rightarrow x=100 m$$
Question 7
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An observer from the top of the light house, the angle of depression of two ships $$P$$ and $$Q$$ anchored in the sea to the same side are found to have measure $$35$$ and $$50$$ respectively. Then from the light house ...............

A
The distance of $$P$$ is more than $$Q$$.

B
The distance of $$Q$$ is more than $$P$$.

C
$$P$$ and $$Q$$ are at equal distance.

D
The relation about the distance of $$P$$ and $$Q$$ cannot be determined.

Solution

We know that the less the angle of depression the more it is far from the base
Hence distance of $$P$$ is more than $$Q$$.
From the figure we can see that,
The distance of $$P$$ from the lighthouse is more than the distance of $$Q$$ from the lighthouse.
Question 8
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From the top of a building $$h$$ metre high, the angle of depression of an object on the ground has a measure $$\theta$$. The distance of the object from the building is

A
$$h\cos { \theta } $$ metre

B
$$h\sin { \theta } $$ metre

C
$$\tan { \theta } $$ metre

D
$$h\cot { \theta } $$

Solution

Height of the building is AB $$=h$$ metre and the object is located at point $$C$$

Now, $$\tan{\theta}=\cfrac {\text{opposite side}}{\text{adjacent side}}$$

$$\tan{\theta}=\cfrac {h}{\text{adjacent side}}$$

We need to find the adjacent side,

$$\implies \text{adjacent side}=\cfrac { h }{ \tan { \theta } } =h\cot { \theta } $$

Hence, the answer is $$h\cot \theta$$.
Question 9
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On walking ............... metres on a slope at an angle of measure 30 with the ground, one can reach the height '$$a$$' metres from the ground.

A
$$\cfrac { 2a }{ \sqrt { 3 } } $$

B
$$\cfrac { \sqrt { 3 } }{ 2 } a$$

C
$$2a$$

D
$$\cfrac{a}{2}$$

Solution

Assume that on walking '$$y$$' metres, one can reach the height of '$$a$$' metres from the ground.
$$\sin{30^{o}} =\cfrac{a}{y} $$
$$\implies \cfrac { 1 }{ 2 } =\cfrac { a }{ y } $$
$$\implies y=2a$$
Hence, the answer is $$2a$$
Question 10
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When observed from top of a tower, the angle of depression of two houses $$A$$ and $$B$$ in the eastern and western directions is $${30}^{o}$$ and $${60}^{o}$$ respectively, then

A
House $$A$$ is nearer to the tower than house $$B$$.

B
House $$B$$ is nearer to the tower than house $$A$$.

C
House $$A$$ and $$B$$ are equidistant from the tower.

D
None of the above.

Solution

Let $$MN$$ be the tower of height $$h$$.
Houses $$A$$ and $$B$$ are at distances $$x$$ and $$y$$ from tower respectively.
$$\tan{30^{o}} =\dfrac {h}{ x } $$
$$\implies x\tan{30^o} =h$$
$$\implies \dfrac { x }{ \sqrt { 3 } } =h...(1)$$
$$\tan {60^{o}} =\dfrac{h}{y} $$
$$\implies \sqrt { 3 } y=h....(2)$$
From (1) and (2), we write
$$\cfrac { x }{ \sqrt { 3 } } =\sqrt { 3 } y$$
$$\implies x=3y$$
Thus, distance of $$A$$ from tower is $$3$$ times the distance of $$B$$ from the tower.
Hence, house $$B$$ is nearer to tower than house $$A$$.