If each side of length a of an equilateral triangle subtends an angle of 600 at the top of a tower h meter high situated at the center of the triangle, then

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,               SoOAa/2=sec300⇒OA=a2.23=a3           Now from right angle triangle POA          PA2=OP2+OA2   ⇒a2=h2+a23⇒2a23=h2       ⇒2a2=3h2