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.
Answers
• maximum height from which he
falls is( 100√3)/(√3-1)m and
distance of the point where he falls
on the ground from the just
observation point is (100)/(√3-1)m.
solution:
•In Triangle BCD
tan 45° = BC/CD=h/d=1
=>h/d=1
=>h=d ________(1)
•In Triangle ACD
=>tan 60° = AC/CD=(h+100)/d=√3
=>(h+100)/d=√3
•putting h=d
=>(h+100)/h=√3
=>h+100=√3h
=>√3h-h=100
=>h(√3-1)=100
=>h = 100/(√3-1)
•Now, AC= h+100
=100/(√3-1)+100
=100[1/(√3-1)+1]
=100[(1+√3-1)/(√3-1)]
=100√3/(√3-1)
1. Given that, 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.
2. We are to 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.
3. Let maximum height from which he falls be h and distance of the point where he falls on the ground from the just observation point be x.
4. Then the explanation is attached below and h= 236.6 m and x = 136.6 m
