from a point 100m above a lake the angle of elevation of a stationary helicopter is 30 and the angle of depression of reflection of the helicopter in the lake is 60 ..find the height of the helicopter.
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BD= 100m=CE
let AC be h
BC=DE= x
in ∆ ABC,
tan 30°= Ac /BC
1/√3= h/x
x= h√ 3→[1]
in ∆ BCE,
tan 60°=CF/ BC
√3= h+100+100/x
x=h+200/√3→[2]
from 1 and 2
h√3= h+200/√3
h√3*√3 =h+200
3h=h+200
3h-h=200
2h=200
h=200/2
h=100
height of helicopter= AE=h+100=100+100= 200m
let AC be h
BC=DE= x
in ∆ ABC,
tan 30°= Ac /BC
1/√3= h/x
x= h√ 3→[1]
in ∆ BCE,
tan 60°=CF/ BC
√3= h+100+100/x
x=h+200/√3→[2]
from 1 and 2
h√3= h+200/√3
h√3*√3 =h+200
3h=h+200
3h-h=200
2h=200
h=200/2
h=100
height of helicopter= AE=h+100=100+100= 200m
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The height of helicopter is 200 metres.
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