A parachutist is descending vertically and makes angles of elevation of 45° and 60° at two observing points 100 m apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of the point where he falls on the ground from the just observation point.

Let the height of the parachutist = h (m)


Let the distance of falling point from observation point = (m)


In ∆ABC,


tan 45° =


1 =


h = 100+ ----------(1)


In ∆ABD,


tan 60° =


√3 =


h = √3 -----------(2)


From eqn. (1) and eqn. (2) we get,


100+ = √3


100 = (√3-1)


=



136.6 m.



in eqn. (1)


100+


h = 100+136.6


h = 236.6 m.


Therefore height of parachutist is 236.6 m. and distance of point where he falls


is 136.6 m.


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