A round balloon of radius r subtends an angle θ at the I of observer while the angle of elevation of its centre is φ. Prove that the height of the centre of the balloon isr sin φ cosec θ/2

Let the height of centre of the balloon above the ground be *h* m.

Given, balloon subtends an θ angle at the observes eye.

∴ ∠ EAD = θ

In ΔACE and ΔACD,

AE = AD (Length of tangents drawn from an external point to the circle are equal)

AC = AC (Common)

CE = CD (Radius of the circle)

∴ ΔACE ΔACD (SSS congruence criterion)

⇒ ∠EAC = ∠DAC (CPCT)

∴ ∠EAC = ∠DAC =

In right ΔACD,

In right ΔACB,

Thus, the height of the centre of the balloon is .

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