Question

A round balloon of radius a subtends and equal angle theta at the eye of the observer while the angle of the elevation is its Centre is theta prove that the height of the centre of the balloon is sin theta cos theta by 2

Answer 1

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 is congruent to ΔACD  (SSS congruence criterion)

⇒ ∠EAC = ∠DAC  (CPCT)

∴ ∠EAC = ∠DAC = θ/2

In right ΔACD,

sin θ/2= CD/AC= r/AC
or AC= r/ sin θ/2 = r cos θ/2

In right ΔACB,

sin α = BC/AC= h/ r cosec θ/2
or h= r sin α cosec θ/2
Thus, the height of the centre of the balloon is r sin α cosec θ/2.

Answer 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 is congruent to ΔACD  (SSS congruence criterion)

⇒ ∠EAC = ∠DAC  (CPCT)

∴ ∠EAC = ∠DAC = θ/2

In right ΔACD,

sin θ/2= CD/AC= r/AC
or AC= r/ sin θ/2 = r cos θ/2

In right ΔACB,

sin α = BC/AC= h/ r cosec θ/2
or h= r sin α cosec θ/2
Thus, the height of the centre of the balloon is r sin α cosec θ/2.