If two vertical poles of height  $h_1\text{\hspace{0.17em}and\hspace{0.17em}}{h}_2$ standing at the middle point of the side  $BC$ and at the centroid of the triangle  $ABC$ respectively, make the same angle at the vertex  $A$, then  $\frac{h_1}{h_2}=$

Let $ABC$ be the given triangle with  $D$ as the midpoint of $BC$, $G$ be the centroid of the  $\Delta AB\text{}C$.

The poles of height  $h_1$ at  $D,h_2$ at  $G$ subtend the same angle at  $A$.

$\begin{array}{cc}\text{In\hspace{0.17em}}\Delta ADE& \text{In\hspace{0.17em}}\Delta AGF\\ \tan \alpha =\frac{DE}{AD}& \tan \alpha =\frac{GF}{AG}\\ \tan \alpha =\frac{h_1}{AD}& \tan \alpha =\frac{h_2}{AG}\\ AD=h_1\cot \alpha & AG=h_2\cot \alpha \end{array}$

Now, $AG=\frac{2}{3}AD$ ($G$ is centroid and $AD$ is median)

$h_2\cot \alpha =\frac{2}{3}{h}_1\cot \alpha$

$3h_2=2h_1$

$\frac{h_1}{h_2}=\frac{3}{2}$