Question

The angle of elevation of the cloud at a point 2500 m high from a lake is 15° and from the same point the angle of depression of its reflection in the lake is 45°. The height (in metres) of the cloud above the lake (given cot 15° = 2 + \(\sqrt3\) ) is

(a) 2500

(b) 2500\(\sqrt2\) 

(c) 2500\(\sqrt3\) 

(d) 5000

Answer 1

 (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.