ABC is a triangular park with a right angle at B . A vertical tower...

Let the height of the tower at $B$ be $h$.

The angle of elevation of the top of the tower from $A$ is $\theta_A = \sin^{-1}\left(\dfrac{1}{\sqrt{17}}\right)$.
This gives $\sin \theta_A = \dfrac{1}{\sqrt{17}}$, which implies $\cot \theta_A = 4$.
The horizontal distance $AB = h \cot \theta_A = 4h$.

The angle of elevation from $C$ is $\theta_C = \operatorname{cosec}^{-1}(\sqrt{10})$.
This gives $\operatorname{cosec} \theta_C = \sqrt{10}$, which implies $\cot \theta_C = 3$.
The horizontal distance $BC = h \cot \theta_C = 3h$.

In the horizontal right-angled triangle $ABC$, applying the Pythagoras theorem:
$AB^2 + BC^2 = AC^2$
$(4h)^2 + (3h)^2 = 150^2$
$16h^2 + 9h^2 = 22500$
$25h^2 = 22500$
$h^2 = 900$
$h = 30$

Answer: $30$