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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