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Question
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 centre of the balloon is a sin Φ cosec `θ/2`.
Solution
Let C be the centre of the balloon, O be the position of man's eye.
Let h be the height of the centre of the balloon.
Then, ∠AOB = θ
So, ∠BOC = ∠COA = `θ/2`
In ΔOAC,
sin θ/2 = `a/"OC"`
⇒ OC = a cosec `θ/2`
In ΔCOD,
`sin Φ = h/"OC"`
`⇒ h = OC sin Φ`
`⇒ h = a cosec θ/2 sin Φ`
`⇒ h = a sin Φ cosec θ/2`
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