If each side of length a of an equilateral triangle subtends an angle of $60^{0}$ at the top of a tower h meter high situated at the center of the triangle, then

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

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

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

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

answer is B.

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Detailed Solution

Let O be the center of the equilateral triangle ABC and OP be the tower of height h. Then each of the triangle PAB, PBC and PAC are equilateral and thus PA=PB=AB=a .

In triangle ABC,OA is the bisector of the angle A,

$S o \frac{O A}{a / 2} = \sec 30^{0} \Rightarrow O A = \frac{a}{2} . \frac{2}{\sqrt{3}} = \frac{a}{\sqrt{3}}$

Now from right angle triangle POA

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