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
B
23a
C
a3
D
a3

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)]        $$=\dfrac{a}{\sqrt{3}}$$Mathematics

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