A round balloon of radius r subtends an angle θ at the I of observer while the angle of elevation - Maths - Some Applications of Trigonometry

Question

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

Ajanta Trivedi · Accepted Answer

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 = $\frac{θ}{2}$

In right ΔACD,

$\sin \frac{θ}{2} = \frac{CD}{AC} ∴ \sin \frac{θ}{2} = \frac{r}{AC} \Rightarrow AC = \frac{r}{\sin \frac{θ}{2}} = r co \sec \frac{θ}{2} ... (1)$

In right ΔACB,

$\sin φ = \frac{BC}{A C} ∴ \sin φ = \frac{h}{r \cos e c \frac{θ}{2}} (Using (1)) \Rightarrow h = r \sin φ \cos e c \frac{θ}{2}$

Thus, the height of the centre of the balloon is $r \sin φ \cos e c \frac{θ}{2}$ .

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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