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Q.

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 $θ$ and $ϕ$ respectively then the distance between their tops when $tanθ = \frac{4}{3} and tanϕ = \frac{5}{2}$ is

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a

\frac{150}{\sqrt{3}} m

b

100 \sqrt{3} m

c

150 m

d

100 m

answer is D.

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Detailed Solution

It is given that the angle of depressions of the tap and the foot of a chimney which is seen from the top of a second chimney of 150 m height are

θ

and

ϕ

respectively.
Let the distance between their top is x meter and d is the distance between their bases.
Also assume that the second tower is h meter long than the first tower.
According to the given data the figure has been drawn below.
Question Image

Question Image

Now

tanθ = \frac{4}{3}

and

tanϕ = \frac{5}{2}

.
In ,

tan θ = P e r p e n d i c u l a r b a s e

tanϕ = \frac{EB}{AB}

= \frac{150}{d}

\Rightarrow d = 150 cotϕ

(since

tan∅ = \frac{1}{cot∅})

= 150 \times \frac{2}{5}

= 60

…(1)
In

ΔDEC,

tan θ = P e r p e n d i c u l a r b a s e

tan θ = \frac{CE}{DC}

\Rightarrow \frac{4}{3} = \frac{h}{DC}

\Rightarrow \frac{4}{3} = \frac{h}{AB}

(Since AB and DC are opposite sides of a rectangle)

\Rightarrow \frac{4}{3} = \frac{h}{d}

\Rightarrow h = \frac{4 d}{3}

= \frac{4 \times 60}{3}

( from the equation 1)

= 4 \times 20

= 80

….(2)
In

ΔDCE

,

D E^{2} = D C^{2} + C E^{2}

\Rightarrow x^{2} = 60^{2} + 80^{2}

(from equation 1 and 2)

= 10000

\Rightarrow x = 100

So the distance is 100 m.
So the correct option is 4.

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