The angle of elevation of the …

Let {tex}\angle APB = \theta{/tex}
Then, {tex}\angle AQB = 90^o - \theta{/tex}
{tex}\because \angle A P B \text { and } \angle A Q B{/tex} are complementary
In right triangle ABP,
{tex}\tan \theta = \frac { A B } { P B }{/tex}
{tex}\Rightarrow \tan \theta = \frac { \mathrm { AB } } { 9 }{/tex}...........(1)
In right triangle ABQ,
{tex}\tan \left( 90 ^ { \circ } - \theta \right) = \frac { A B } { Q B }{/tex}
{tex}\Rightarrow \cot \theta = \frac { A B } { 4 }{/tex}.............(2)

Multiplying (1) and (2), we get
{tex}\frac { A B } { 9 } \cdot \frac { A B } { 4 } = \tan \theta , \cot \theta{/tex}
{tex}\Rightarrow \frac { A B ^ { 2 } } { 36 } = \tan \theta \cdot \frac { 1 } { \tan \theta } \Rightarrow \frac { A B ^ { 2 } } { 36 } = 1{/tex}
{tex}\Rightarrow A B ^ { 2 } = 36 \Rightarrow A B = \sqrt { 36 } = 6{/tex}
Hence, the height of the tower is 6 m.