A spherical balloon of radius 10 m subtends an angle of 60^ at the...

Let $d$ be the distance from the observer's eye to the center of the balloon. The balloon subtends an angle of $60^\circ$, so the half-angle is $30^\circ$.
$\sin 30^\circ = \dfrac{\text{radius}}{d} \implies \dfrac{1}{2} = \dfrac{10}{d} \implies d = 20$ m.

Let $y$ be the vertical height of the balloon's center above the observer's eye level, and $x$ be the horizontal distance to the center.
Since the angle of elevation of the center is $30^\circ$:
$y = d \sin 30^\circ = 20 \times \dfrac{1}{2} = 10$ m.
$x = d \cos 30^\circ = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3}$ m.

The observer is at a height $h$ above the lake. Therefore, the height of the balloon's center above the lake surface is $H = h + y = h + 10$.
The reflection of the balloon's center in the lake will be at a depth $H = h + 10$ below the lake surface.

The vertical distance from the observer's eye down to the reflection is:
$D = h + H = h + (h + 10) = 2h + 10$.

The angle of depression of this reflection from the observer's eye is $60^\circ$. Using the horizontal distance $x$:
$\tan 60^\circ = \dfrac{D}{x}$
$\sqrt{3} = \dfrac{2h + 10}{10\sqrt{3}}$
$\sqrt{3} \times 10\sqrt{3} = 2h + 10$
$30 = 2h + 10$
$2h = 20 \implies h = 10$ m.

Answer: $10$