A round balloon of radius 'a' subtends an angle θ at the eye of the observer while the angle of elevation of its centre is Φ.Prove that the height of the center of the balloon is a sin θ cosec Φ /2.

Ans:    Let θ be the centre of the ballon of radius 'r' and 'p' the eye of the observer. Let PA, PB be tangents from P to ballong. Then ∠APB = θ .

∴∠APO = ∠BPO = θ/2

Let OL be perpendicular from O on the horizontal PX. We are given that the angle of the elevation of the centre of the ballon is φ i.e.,

∠OPL = φ

In ΔOAP, we have sin θ/2 = OA/OP

⇒ sin θ/2 =  a/OP

OP = a cosecθ/2

In ΔOPL, we have sinφ = OL/ OP

⇒ OL = OP sin φ = a cosec θ/2 sin θ.

Hence, the height of the center of the balloon is a sin θ cosec Φ /2.