Question

The angle of elevation of a stationery cloud from a point 2500 m above a lake is 15° and the angle of depression of its reflection in the lake is 45°. What is the height of the cloud above the lake level? (Use tan 15° = 0.268)

Answer 1

Step-by-step explanation:

that is height of cloud above lake=height of reflection

Answer 2

Solution:

Let AB be the surface of the lake and P be the point of observation such as AP = 2500 m . Let C be the position of cloud and C¹ be the reflection in the lake. Then CB = C¹B

Let PQ be perpendicular from P on CB.

Let Pq =x m, CQ = h, QB = 2500 m , then CB = h + 2500 consequently,  C¹B = h + 2500 m and ∠CPQ -15°, ∠QPC¹ = 45°

To find the height of the cloud, we use trigonometry ratios,

InΔPCQ,

=> tan 15° = CQ / PQ

=> 2 - √3  = h / x

=> x = h / 2 - √3 ... (1)

Again in ΔPQC,

=> tan 45° = QB + BC¹ / PQ

=> 1 = ( 2500 + h + 2500 ) / x

=> x = 5000 + h

From (1) we get,

h / (2 -√3) = 5000 + h

h = 2500 ( √3 -1)

CB = 2500 + 2500 ( √3 -1)

CB = 2500 √3

CB = 2500 (1.732) = 4330

Hence, the height of the cloud above the lake level is 4330 m .