A person walking along a straight road towards a hill observes at two points distance3km the angles of elevation of the hill to be 30° and 60°. The height of the hill is

In ΔABC,tan60°=hx⇒x=h3→(1)In ΔABD,tan30°=hx+3⇒13=hx+3→(2)From (1)&(2)3+h3=3h⇒h=32km∴The height of the hill (h) is =32km

A person walking along a straight road towards a hill observes at two points distance3km the angles of elevation of the hill to be 30° and 60°. The height of the hill is

In ΔABC,tan60°=hx⇒x=h3→(1)In ΔABD,tan30°=hx+3⇒13=hx+3→(2)From (1)&(2)3+h3=3h⇒h=32km∴The height of the hill (h) is =32km

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A person walking along a straight road towards a hill observes at two points distance $\sqrt{3} k m$ the angles of elevation of the hill to be $30 °$ and $60 °$ . The height of the hill is

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$\sqrt{\frac{2}{3}} k m$

$\sqrt{3} k m$

$(3 / 2) k m$

answer is A.

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Detailed Solution

In $Δ A B C, \tan 60 ° = \frac{h}{x}$

$\Rightarrow x = \frac{h}{\sqrt{3}} \to (1)$

In $Δ A B D, \tan 30 ° = \frac{h}{x + \sqrt{3}}$

$\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + \sqrt{3}} \to (2)$

From $(1) & (2)$

$\begin{array}{l} \sqrt{3} + \frac{h}{\sqrt{3}} = \sqrt{3} h \\ \Rightarrow h = \frac{3}{2} km \end{array}$

$∴$ The height of the hill (h) is $= \frac{3}{2} km$

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A person walking along a straight road towards a hill observes at two points distance3km the angles of elevation of the hill to be 30° and 60°. The height of the hill is

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A person walking along a straight road towards a hill observes at two points distance $\sqrt{3} k m$ the angles of elevation of the hill to be $30 °$ and $60 °$ . The height of the hill is