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Question

A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that AB (= a) subtends an angle of $$60^{\circ}$$ at the foot of the tower, and the angle of elevation of the top of the tower from A or B is $$30^{\circ}$$. The height of the tower is


A
2a3
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B
23a
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C
a3
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D
a3
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Solution

The correct option is C $$\dfrac{a}{\sqrt{3}}$$

Let us consider the height of the tower as $$h$$
We know that, $$h = AC \tan 30^\circ = BC  \tan 30^\circ$$      ...(1)


We know that $$AC = BC$$       [$$\because$$ radius of circle]

Hence, we can say that $$\triangle ABC$$ is an isosceles triangle with $$AC=BC$$.
So, $$\angle ABC=\angle BAC$$    ...(2)

But $$\angle ACB=60^\circ$$     [given]
$$\Rightarrow \angle ABC+\angle BAC+\angle ACB=180^\circ$$
$$\Rightarrow \angle ABC+\angle ABC+60^\circ \,\,\,\,\,\,\,\,=180^\circ$$            [from (2)]
$$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\times \angle ABC=120^\circ$$
$$\therefore\, \angle ABC=\angle BAC=60^\circ$$

Hence $$\triangle ABC$$ is an equilateral triangle.

$$\therefore\, AB = BC = CA = a$$    ...(3)


$$\therefore h = a \tan 30^\circ$$        [form equation (1)]
        $$=\dfrac{a}{\sqrt{3}}$$

1313031_40828_ans_4b75b9a1ba7547f29af3f270334a9bdf.png

Mathematics

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