lf each side of length $a$ of an equilateral triangle subtends an angle of $60^0$ at the top of a tower of $h$ metre in height situated at the centre of the triangle, then

PA = PB = PC
Consider triangle PAB which is an isosceles triangle and since AB subtends an angle $60^0$ at P, this triangle becomes equilateral. Hence triangles PAB, PBC, PCA each are equilateral. Again with side, a $\Delta$ABC is equilateral with centre at O.
OA is bisector of angle A
(a/2)/OA = cos $30^0$ = $\sqrt{3}$/2
OA = a/$\sqrt{3}$
Now from right angled $\Delta$OPA, we have
$P{A}^2=O{P}^2+O{A}^2$ or  $a^2=h^2+a^2/3$
or $2a^2/3=h^2$ or $2a^2=3h^2$