Question

A round balloon of radius ‘r’ subtends an angle θ at the eye of the observer, while angle of elevation of its center is ϕ. Find the height of the center of the balloon.

Answer 1

Let center of balloon of radius r of O and A is the position of observer. 

Let AP and AQ are too tangents to the balloon.

OB = h m

Balloon of radius r, subtends an angle θ and angle of  elevation of its center is ϕ. Then

∠PAQ = θ

and ∠PAO = ∠QAO = θ/2

OP = r and ∠PAB = ϕ

∠APO = ∠AQO = 90°

[∵ Radius and tangents are perpendicular to each other]

From right angled ∆OAB,

⇒ sin ϕ = OB/AO

⇒ sin ϕ = h/AO

⇒ h = AO sin ϕ ….(i)

From right angled ∆AOP,

sin θ/2 = OP/AO

⇒ sin θ/2 = r/AO

AO = r cosec θ/2 ……(ii)

Put the value of AO in equation (i),

h = r cosec θ/2 sin ϕ

Hence, height of the center of the balloon (h) = r cosec θ/2 sin ϕ