If each side of length a of an equilateral
triangle subtends an angle of 60° at the top of a tower $h$ metre
high situated at the centre of the triangle, then

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$3 a^{2} = 2 h^{2}$

$2 a^{2} = 3 h^{2}$

$a^{2} = 3 h^{2}$

$3 a^{2} = h^{2}$

answer is B.

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Detailed Solution

Let $O$ be the centre of the equilateral triangle
$A B C$ and $O P$ be the tower of height $h$ . (Fig. 27.19)
Then each of the triangles P $A B, P B C$ and $P C A$ are equilateral
and thus

$P A = P B = A B = a .$

In triangle $A B C, O A$ is the bisector of the angle $A$ ,

So $\frac{O A}{a / 2} = \sec 30^{\circ}$

$\Rightarrow O A = \frac{a}{2} \cdot \frac{2}{\sqrt{3}} = \frac{a}{\sqrt{3}}$

Now from right angled triangle $P O A$

$\begin{array}{l} P A^{2} = O P^{2} + O A^{2} \\ \Rightarrow a^{2} = h^{2} + \frac{a^{2}}{3} \Rightarrow \frac{2 a^{2}}{3} = h^{2} \\ \Rightarrow 2 a^{2} = 3 h^{2} . \end{array}$