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Q.

A person walking along a straight road towards a hill observes at two points distance $\sqrt{3}$ km, the angle of elevation of the hill to be $30^{0}$ and $60^{0}$ . The height of the hill is

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a

\frac{3}{2}

km

b

\sqrt{\frac{21}{32}}

km

c

\frac{(\sqrt{3} + 1)}{2}

km

d

\sqrt{3}

km

answer is A.

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Detailed Solution

Given, distance between two points is

\sqrt{3}

km.
Angle of elevation of the hill to be

30^{0}

and

60^{0}

.

Question Image

We have to find the height of the hill.

\tan 60^{0} = \frac{h}{d}

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

[∵

\tan 60^{0} = \sqrt{3}

]

\Rightarrow h = \sqrt{3} d

……(1)

\tan 30^{0} = \frac{h}{\sqrt{3} + d}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{\sqrt{3} + d}

[∵

\tan 30^{0} = \frac{1}{\sqrt{3}}

]

\Rightarrow \sqrt{3} = \frac{\sqrt{3} + d}{h}

\Rightarrow \sqrt{3} = \frac{\sqrt{3} + d}{\sqrt{3 d}}

\Rightarrow 3 d = \sqrt{3} + d

\Rightarrow 3 d - d = \sqrt{3}

\Rightarrow 2 d = \sqrt{3}

\Rightarrow d = \frac{\sqrt{3}}{2}

…….(2)
Substituting (2) in (1),

\Rightarrow h = \sqrt{3} \times \frac{\sqrt{3}}{2}

\Rightarrow h = \frac{3}{2} km

Hence, the height of the hill is

\frac{3}{2} km

.
Therefore, the correct option is 1.

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