The angles of elevation of the top of a tower at the top and the foot of a pole of height  $10m$ are  $30°\text{\hspace{0.17em}and\hspace{0.17em}}60°$ respectively.  The height of the tower is

Let  $AB\text{\hspace{0.17em}and\hspace{0.17em}}CD$ be the pale and tower respectively,

$\angle DAC=60°\text{\hspace{0.17em}and\hspace{0.17em}}\angle DBE=30°$

$\begin{array}{cc}\text{In\hspace{0.17em}}\Delta ACD& \text{In }\Delta DBE\\ \tan 60°=\frac{CD}{AC}& \tan 30°=\frac{DE}{BE}\\ \sqrt{3}=\frac{CD}{AC}& \frac{1}{\sqrt{3}}=\frac{DE}{AC}\\ AC=\frac{CD}{\sqrt{3}}& AC=DE\sqrt{3}\\ AC=\frac{h}{\sqrt{3}}\to \left(1\right)& =(h-10)\sqrt{3}\to \left(2\right)\end{array}$

From (1) and (2),

$\frac{h}{\sqrt{3}}=(h-10)\sqrt{3}$

$\frac{h}{3}=h-10$

$10=h(1-\frac{1}{3})$

$10=\frac{2h}{3}$

$h=\frac{30}{2}=15m$