A round balloon of radius r subtends an angle α at the eye of the

observer while the angle of elevation of its centre is β. Prove that the
height of the centre of the balloon is r sin β cosecα/2.
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 Hi!

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 a .

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