Question

From a point, 36 m above the surface of a lake, the angle of elevation of a bird is observed to be 30° and the angle of depression of its image in the water of the lake is observed to be 60°. Find the actual height of the bird above the surface of the lake.

Answer 1

Let A be a point 36 m above the surface of the lake and B be the position of the bird. Let B' be the image if the brid in the water. 

Here, AC = DE = 36 m, ∠BAE = 30° and ∠B’AE = 60°, 

Let BE = h m,

Then, B'D = BD = 36 + h   ...(∴ B’ is image of B about D)

∴ B’E = B’D + DE

= 36 + 36 + h

= 72 + h  ...(i)

In ΔABE, 

`(BE)/(AE) = tan 30^circ`

`=> AE = sqrt(3)h`  ...(ii)

In ΔAB’E, 

`(B^’E)/(AE) = tan 60^circ`

`=> (72 + h )/(AE) = sqrt(3)`  ...[From (i)]

`=> 72 + h = (sqrt(3)h)sqrt(3)`  ...[From (ii)]

`=>` 72 + h = 3h

∴ h = 36 m

Hence, the actual height of the bird above the surface of the lake = 36 + 36 = 72 m.