Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 27

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Question 1
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Two poles of height 18 metres and 7 metres are erected on the ground. A wire of length $$22$$ metres tied to the top of the poles. Find the angle made by the wire with the horizontal.(in degree)

A
$$30$$

B
$$10$$

C
$$20$$

D
$$15$$

Solution

Angle of elevation from smaller pole to bigger = $$x$$
Height of smaller building $$= 7$$ m
Height of bigger pole $$= 18$$ m
Length of the rope $$= 22$$ m
Difference between the two buildings $$= 11$$ m

Now, $$\sin x = \cfrac{opposite\ side}{hypotaneous}$$
$$\sin x = \cfrac{11}{22} = \dfrac{1}{2}$$
$$\sin x = \sin 30$$
$$x = 30^{\circ}$$
Question 2
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Two flagstaffs stand on a horizontal plane. A and B are two points on the line joining their feet and between them. The angles of elevation of the tops of the flagstaff as seen from A are 30$$^o$$ and 60$$^o$$ and as seen from B are 60$$^o$$ and 45$$^o$$. If AB is 30 m, the distance between the flagstaffs in meters is:

A
16+ 15 $$\sqrt{3}$$

B
45+ 15 $$\sqrt{3}$$

C
60 - 15 $$\sqrt{3}$$

D
60 + 15 $$\sqrt{3}$$

Solution

Let x and y be the heights of the flagstaffs at P and Q respectively.

Then $$AP = x .cot 60^o = s\sqrt{3}, \ AQ = y .cot 30^o =y\sqrt{3}$$

$$BP = x cot 45^o = x, BQ = y cot 60^o = y/\sqrt{3}$$

$$\Rightarrow BP - AP = x - x/\sqrt{3} = AB$$

$$\Rightarrow 30\sqrt{3} = (\sqrt{3} - 1) x \Rightarrow x = 15 (3 + \sqrt{3})$$

Similarly $$30 = y (\sqrt{3} - 1 / \sqrt{3}) \Rightarrow y = 15 \sqrt{3}$$

so that $$PQ = BP + BQ = x + y / \sqrt{3} = 15 (3 + \sqrt{3}) + 15 $$

$$=(16 + 15 \sqrt{3})m$$
Question 3
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A kite is attached to a $$100\ m$$ long string. Find the greatest height reached by the kite when its string makes an angle of $$\displaystyle 60^{\circ}$$ with the level ground.

A
$$86.6$$

B
$$45.7$$

C
$$63.8$$

D
$$72.0$$

Solution

The highest height the kite reaches be P
The length of string is L = 100 m
The angle made by the string with the ground = $$60^{\circ}$$

now, $$\sin 60 = \dfrac{P}{L}$$
$$P = \dfrac{100 \sqrt{3}}{2}$$
$$P = 50 \sqrt{3}$$
$$P = 86.6$$ m
Question 4
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A ladder is placed against a vertical tower. If the ladder makes an angle of $$\displaystyle 30^{\circ}$$ with the ground and reaches up to a height of $$15\ m$$ of the tower; find length of the ladder in cm.

A
2311

B
3000

C
1688

D
1200

Solution

Consider the ladder and wall system as the right angled triangle, such that length of wall (W) till the ladder reaches be the perpendicular= 15 cm and angle (G) made by ladder (L) with the ground is $$30^{\circ}$$,
and length of ladder be the Hypotenuse.
Thus, $$\sin G = \dfrac{P}{H}$$
$$\sin 30 = \dfrac{W}{L}$$
$$\dfrac{1}{2} = \dfrac{15}{L}$$
$$L = 30\ m=3000\ cm$$
Question 5
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An observer $$1.5m$$ tall is $$29.5m$$ away from a chimney. The angle of elevation of the top of the chimney from her eyes is $${45}^{o}$$. The height of the chimney is:

A
$$31\ m$$

B
$$27\ m$$

C
$$29.5\ m$$

D
None of these

Solution

Consider the triangle.

$$\tan45^{0}=1$$

$$=\dfrac{h}{29.5}$$

Hence

$$h=29.5m$$

Hence the total height of the tower will be

$$=h+1.5m$$

$$(29.5+1.5)m$$

$$=31\ m$$
Question 6
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A flagstaff $$6$$ metres high throws a shadow $$2\sqrt 3$$ metres long on the ground. The angle of elevation is

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

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

C
$${90}^{o}$$

D
$${60}^{o}$$

Solution

Let $$\theta$$ be the angle of elevation.
Hence, $$\tan(\theta)$$ $$=\dfrac{Height\:of\:pole}{Length\:of\:shadow}$$

$$=\dfrac{6}{2\sqrt{3}}$$

$$=\sqrt{3}$$
Hence. $$\tan\theta=\sqrt{3}$$
$$\theta=60^{0}$$
Question 7
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An observer finds that the angular elevation of a tower is $$A$$. On advancing $$3$$ m towards the tower the elevation is $${ 45 }^{ \circ }$$ and on advancing $$2$$ m nearer, the elevation is $$({ 90 }^{ \circ }-A)$$. The height of the tower is

A
$$2$$ m

B
$$4$$ m

C
$$6$$ m

D
$$8$$ m

Solution

Let $$PQ$$ be the tower of height $$h$$
Let $$A$$ be the intial point of observing tower $$PQ$$, $$B$$ be 3m away from $$A$$ and $$C$$ be 2m away from $$B$$
Given,
$$\displaystyle\angle PAQ=A$$
$$\angle PBQ={ 45 }^{ o }$$
$$\angle PCQ={ 90 }^{ o }-A$$
$$\displaystyle AB=3,BC=2$$
$$\displaystyle\Rightarrow 3=AB=PA-PB=h\cot { A } -h$$
and $$\displaystyle 2=BC=PB-CP=h-h\tan { A }$$
$$\displaystyle \therefore \frac { h+3 }{ h } =\cot { A } ,\frac { h-2 }{ h } =\tan { A }$$
$$\displaystyle \therefore \frac { h+3 }{ h } .\frac { h-2 }{ h } =1\Rightarrow { h }^{ 2 }+h-6={ h }^{ 2 }\Rightarrow h=6$$ m
Question 8
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An observer $$\sqrt 3m $$ tall is $$3m$$ away from the pole $$2\sqrt 3m $$ high. The angle of elevation of the top from the pole is:

A
$${45}^{o}$$

B
$${30}^{o}$$

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

D
$${15}^{o}$$

Solution

Now consider the below figure
$$\tan \theta=\dfrac{(2\sqrt{3}-\sqrt{3})}{3}$$
$$=\dfrac{\sqrt{3}}{3}$$

$$=\dfrac{1}{\sqrt{3}}=\tan30^{0}$$
Hence
$$\tan\theta=\tan30^{0}$$
$$\theta=30^{0}$$
Question 9
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The top of a broken tree touches the ground at a distance of 15 m from its base. If the tree is broken at aheight of 8 m from the ground, then the actualheight of the tree is

A
20 m

B
25 m

C
30 m

D
17 m

Solution

$$\because\, \Delta\, ABC$$ is right angled
$$AC^2\,=\, AB^2\, +\, BC^2$$
$$AC^2\,=\, 8^2\,+\, 15^2$$
= 64 + 225
= 289
AC = $$\sqrt{289}\,=\, 17 m$$
$$\therefore$$ Actual length of tree = AB + AC
= 8 + 17
= 25 m
Question 10
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An aeroplane is observed at the same time by two anti-aircraft batteries distant $$6000m$$ apart to be at elevations of $${30}^{o}$$ and $${45}^{o}$$ respectively. Assuming that the aeroplane is travelling directly towards the two batteries, find its height and its horizontal distance from the nearer battery.

A
$$3000(\sqrt 3+1)m$$; $$3000(\sqrt 3+1)m$$

B
$$4000(\sqrt 2+1)m$$; $$3000(\sqrt 3+1)m$$

C
$$3000(\sqrt 3+1)m$$; $$4000(\sqrt 2+1)m$$

D
$$4000(\sqrt 2+1)m$$; $$4000(\sqrt 2+1)m$$