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The height of the chimney when it is found that on walking towards it 50 m in the horizontal line through its base, the angle of elevation of its top changes from $30 °$ to $60 °$ is

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$25 \sqrt{3} m$

$25 m$

$25 \sqrt{4} m$

$\frac{25}{\sqrt{3}} m$

answer is A.

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Detailed Solution

Let $P Q$ be the chimney

$A and B$ are the two points $50 m$ a part

Let $P Q = h$

$\begin{matrix} From Δ B P Q & From Δ A P Q \\ \tan 60 ° = \frac{h}{B P} & \tan 30 ° = \frac{h}{A P} \\ \sqrt{3} = \frac{h}{B P} & \frac{1}{\sqrt{3}} = \frac{h}{A P} \\ B P = \frac{h}{\sqrt{3}} & A P = h \sqrt{3} \end{matrix}$

Now, $A B = A P - B P$

$50 = h \sqrt{3} - \frac{h}{\sqrt{3}}$

$= h (\sqrt{3} - \frac{1}{\sqrt{3}})$

$= \frac{h (3 - 1)}{\sqrt{3}}$

$50 = \frac{2 h}{\sqrt{3}}$

$h = 25 \sqrt{3} m$

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