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$A$ person walking $50$ metres towards a chimney in a horizontal line. The angle of elevation of its top changes from $30^{\circ}$ to $45^{\circ}$ . Height of the chimney (in metres) is

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25 (3 + \sqrt{3}) m

50 (\sqrt{3} + 1) m

25 (\sqrt{3} + 1) m

25 (\sqrt{3} - 1) m

answer is C.

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Detailed Solution

Given that when a person walks 50 m distance towards chimney, then angle of elevation changes from

30^{\circ}

45^{\circ}

Let h be the height of the chimney and x be the distance from the foot of the chimney to the new position of person.
Using trigonometric ratios,

\tan 30^{\circ} = \frac{h}{x + 50}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 50}

\Rightarrow \frac{x + 50}{\sqrt{3}} = h

... (1)

\tan 45^{\circ} = \frac{h}{x}

\Rightarrow 1 = \frac{h}{x}

\Rightarrow x = h

...(ii)
From (i) and (ii), we get

\frac{x + 50}{\sqrt{3}} = x

\Rightarrow

x + 50 = \sqrt{3} x

\Rightarrow 50 = \sqrt{3} x - x

\Rightarrow

50 = (\sqrt{3} - 1) x

\Rightarrow

\frac{50}{\sqrt{3} - 1} = x

\Rightarrow

x = \frac{50}{\sqrt{3} - 1}

Rationalising the denominator, we get,

\Rightarrow x = \frac{50}{\sqrt{3} - 1} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)}

\Rightarrow x = \frac{50 \times (\sqrt{3} + 1)}{{\sqrt{3}}^{2} - 1^{2}}

\Rightarrow x = \frac{50 \times (\sqrt{3} + 1)}{3 - 1}

\Rightarrow x = \frac{50 \times (\sqrt{3} + 1)}{2}

\Rightarrow x = 25 (\sqrt{3} + 1) m

∴ h = x = 25 (\sqrt{3} + 1) m

So, the height of chimney is

25 (\sqrt{3} + 1) m

.
Hence, the correct option is 3.

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