A trigonometry question...pls help....?

Question

If the angle of elevation of a cloud from the point h metres above a lake is α(alpha) and the angle of depression of its reflection in the lake is β(Beta),prove that the distance of the cloud from the point of observation is [2h sec α]/[tan β-tan α].
Pls explain step wise.....its urgent....

Answer 1

Let H be the height of the cloud above the lake. Therefore H-h is the height of the cloud above the observation point. The reflection of the cloud is therefore H+h below the observation point. Let x be the horizontal distance from the observation point to the cloud.

Let d be the actual distance from the observation point to the cloud.

For simplicity alpha = a and beta = b

From above tan a = (H - h) / x yielding H - h = x tan a

Using pythagoras

d^2 = x^2 + (H - h)^2 = x^2(1 + tan^2 a) =x^2 sec^2 a

Thus d = x sec a

Also tan b = (H + h) / x. Equating x leads to

(H - h) / tan a = (H + h) / tan b

Solving for H leads to

H = h (tan a + tan b) / (tan b - tan a)

And thus H - h = 2 h tan a / (tan b - tan a)

From above x = (H - h) / tan a

x = 2 h tan a /(tan b - tan a)/tan a = 2h/(tan b - tan a)

From above d = x seca = 2h sec a /(tan b - tan a)

q.e.d.

Answer 2

Draw the diagram,
There are two right triangles,
i'll use a for alpha, b for beta.
Let H = height of cloud over the lake.
for one of them tan a = (H-h) / x, where x is the side common to the upright and inverted triangles.
tan b = H/x
Now, this gives (eliminating x between the two eqns)
H = htanb/ (tanb - tana)
=> x = Htanb = htan^2b/(tanb - tana)
Now to find the distance of the cloud,
it will be sqrt( (H-h)^2 + x^2) and this will give u the desired result on simplification.