on a horizontal plane there is …

Let us suppose that AB be the tower and BC be flagpole
Let us suppose that O be the point of observation.Then, OA = 9 m
According to question it is given that
{tex}\angle A O B = 30 ^ { \circ } \text { and } \angle A O C = 60 ^ { \circ }{/tex}​​

From right angled {tex}\triangle \mathrm { BOA }{/tex}
{tex}\frac { A B } { O A } = \tan 30 ^ { \circ }{/tex}
{tex}\Rightarrow \frac { A B } { 9 } = \frac { 1 } { \sqrt { 3 } } \Rightarrow A B = 3 \sqrt { 3 }{/tex}
From right angled {tex}\Delta O A C{/tex}
{tex}\frac { A C } { O A } = \tan 60 ^ { \circ }{/tex}
{tex}\frac { A C } { 9 } = \sqrt { 3 } \Rightarrow A C = 9 \sqrt { 3 } \mathrm { m }{/tex}
{tex}\therefore{/tex} BC = (AC - AB)   {tex}= 6 \sqrt { 3 } \mathrm { m }{/tex}
Thus {tex}A B = 3 \sqrt { 3 } m = 5.196 m{/tex} and {tex}B C = 6 \sqrt { 3 } m = 10.392 m{/tex}
Hence, height of the tower is 5.196m and the height of the flagpole is 10.392 m