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Two posts are $x$ metres apart and the height of one is double that of the other. If from the mid-point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary, then the height (in meters) of the shorter post is

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x \sqrt{2}

\frac{x}{\sqrt{2}}

\frac{x}{2 \sqrt{2}}

\frac{x}{4}

answer is C.

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Detailed Solution

It is given that two posts are

x

metres apart and the height of one is double that of the other and from the mid-point of the line joining their feet, the angular elevations of their tops is complementary.
According to the given data the figure has been drawn below.
Question Image

Let the height of the shorter and the longer pole be

h

and

2 h

, respectively.
In

△ ABM,

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

tanθ = \frac{AB}{BM}

\Rightarrow \frac{h}{\frac{x}{2}}

\Rightarrow \frac{2 h}{x} \dots . (1)

△ CDM,

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

\tan (90 ° - θ) = \frac{CD}{MD}

\Rightarrow cotθ = \frac{2 h}{\frac{x}{2}}

(\tan (90 ° - θ) = cotθ)

= \frac{4 h}{x}

…………..(2)
On multiplying equations

(1)

and

(2)

, we get

tanθ \cdot cotθ = \frac{2 h}{x} \cdot \frac{4 h}{x}

\Rightarrow 1 = \frac{8 h^{2}}{x^{2}}

(tanθ = \frac{1}{cotθ})

\Rightarrow h^{2} = \frac{x^{2}}{8}

\Rightarrow h = \frac{x}{2 \sqrt{2}}

So the height (in meters) of the shorter post is

\frac{x}{2 \sqrt{2}}

m.
Hence, option

3

is the correct answer.

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