From the top of a cliff, the angle of depression of a stationary...

Let the top of the cliff be $A(0, H)$ and the boat be $B(x, 0)$.

The angle of depression of the boat from $A$ is $60^\circ$, so $\tan 60^\circ = \dfrac{H}{x} \Rightarrow H = x\sqrt{3}$.

The observer moves $200$ m down a path inclined at $30^\circ$ to the horizontal, away from the sea.

The horizontal distance moved is $200 \cos 30^\circ = 100\sqrt{3}$ m (away from the boat).

The vertical distance moved is $200 \sin 30^\circ = 100$ m (downwards).

The new position $C$ has coordinates $(-100\sqrt{3}, H - 100)$.

From $C$, the angle of depression of the boat is $45^\circ$.

The horizontal distance from $C$ to the boat is $x - (-100\sqrt{3}) = x + 100\sqrt{3}$.

The vertical distance from $C$ to the boat is $H - 100$.

Thus, $\tan 45^\circ = \dfrac{H - 100}{x + 100\sqrt{3}} \Rightarrow 1 = \dfrac{x\sqrt{3} - 100}{x + 100\sqrt{3}}$.

$x + 100\sqrt{3} = x\sqrt{3} - 100 \Rightarrow x(\sqrt{3} - 1) = 100(\sqrt{3} + 1)$.

$x = 100 \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} = 100 \dfrac{(\sqrt{3} + 1)^2}{2} = 100(2 + \sqrt{3})$ m.

Answer: $100(2+\sqrt{3})$ m