Question

The angle of elevation of a stationery cloud from a point 2500m above a lake is 30° and the angle of depression of its reflection in the lake is 45°. What is the height of the cloud above the lake level?

Answer 1

Answer:

Heyaa♡♡

Let EB = x m ,then

$\tan(15) = \frac{h}{x} \\ = > x = hcot(15)....(1) \\ and \: \tan(45) = \frac{h + 5000}{x} \\ = > 1 = \frac{h + 500}{x} \\ = > x = h + 5000....(2)$

From eq. (1) and (2):-

$h \ \cot(15) = h + 5000 \\ = > h = \frac{5000}{ \sqrt{3} + 1} \\ = > h = \frac{5000}{ \sqrt{3} + 1} \times \: \frac{ \sqrt{3} - 1}{ \sqrt{3} -1 } \\ = > h = \frac{5000( \sqrt{3} - 1)}{3 - 1} \\ = > h = \frac{5000( \sqrt{3} - 1) }{2} \\ = > h = 2500( \sqrt{3} - 1) \\ so \: the \: height \: of \: the \: cloud \: is \: 2500 + 2500( \sqrt{3} - 1) \\ = > 2500 \sqrt{3} m$

It's explanation is in photo.

Hope it helped you.