Q. 294.8( 21 Votes )

# The angle of elev

Answer :

The figure is given below: Let AB be the surface of water.

Let P be the point that is 60 m above the surface of water.

AP = 60 m

Let C be the position of cloud. Let C’ be the reflection of cloud in the water.

Join PC’ and draw PMCB.

Let CM = h m

PM = AB

CB = (60+h) m

Given: CPM = 30°

C’PM = 60°

Now, in ΔCPM,

tan 30° = CM/PM

1/√3 = h/PM

PM = √3h ……… (1)

In ΔPMC,

tan 60° = C’M/PM

√3 = (h+60+60)/PM

√3 = (h+120)/PM

PM = (h+120)/√3 ……… (2)

From equation (1) and (2), we get:

(h+120)/√3 = √3h

h + 120 = 3h

3h – h = 120

2h = 120

h = 120/2 = 60 m

Height of the cloud from the surface = CB = CM + MB

= (h + 60) m

= (60 + 60) m

= 120 m

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