(C) 2500\(\sqrt3\) m.
Let FE be the level of the lake.
Let A be a point 2500 m above the level of the lake from where the angle of elevation of cloud at B is 15°.
Let D be the reflection of the cloud in the lake.
Given,
BE = ED = H metres (say)
Draw AC || FE meeting BE in C.
∠BAC = 15°, ∠CAD = 45°,
AF = CE = h = 2500 m.
Now, in rt. Δ ABC,
tan 15° = \(\frac{BC}{AC}\)
⇒ AC = BC cot 15°
⇒ AC = (H – h) cot 15° ...(i)
In rt. Δ ACD,
tan 45° = \(\frac{CD}{AC}\)
⇒ AC = CD cot 45°
⇒ AC = (H + h) cot 45° ...(ii)
From eqns (i) and (ii),
(H – h) cot 15° = (H + h) cot 45°
⇒ H (cot 15° – cot 45°) = h(cot 45° + cot 15°)
⇒ H = \(\frac{h\,(1\,+cot\,15^°)}{cot\,15^°\,-\,1}\)
= \(\frac{2500\,(2\,+\sqrt3\,+\,1)}{2\,+\,\sqrt3\,-\,1}\)
= \(\frac{2500\,(3\,+\sqrt3)}{\sqrt3\,+\,1}\times\frac{(\sqrt3\,-\,1)}{(\sqrt3\,-\,1)}\)
= 2500\(\sqrt3\) m.