Each side of an equilateral triangle subtends an angle of 60º at the top of a tower h meters high located at the centroid of the triangle. If the length of each side of the triangle is 1 unit , then  the value of $3h^2=$

Let O be the center of an equilateral triangle ABC and OP be the tower of height h. Then each of the triangle’s PAB, PBC and PCA equilateral. Thus, PA = PB = PC = 1. Therefore, from right-angle triangle POA, we have$P{A}^2=P{O}^2+O{A}^2\Rightarrow 1=h^2+{\left(\frac{1}{2}\sec {30}^{\circ }\right)}^2$$3h^2=$2