Question

A spherical balloon of radius 4 cm subtends an angle 60º at the eye of an observer. If the angle of elevation of its centre is 45º, then the height of the centre of the balloon is

• A. $\sqrt{2}cm$
• B. $\sqrt{3}cm$
• C. $4\sqrt{2}cm$
• D. $2\sqrt{3}cm$

Answer 1

The correct option is C $4\sqrt{2}cm$

Let the height of the centre of the balloon be h and radius of the balloon be r.
Given that r = 4 cm.
Let d be the distance between observer’s eye and the centre of balloon.
$\therefore OB=h$ and $OA=d$
In ΔOAQ and ΔOAP,
∠OQA = ∠OPA = 90° (Tangent at any point on the circumference of a circle makes 90° with the centre of the circle.)
OA = OA (Common)
OQ = OP (Radius)
$\therefore \Delta OAQ\cong \Delta OAP$ (By RHS Congruence Rule)
$\Rightarrow \angle OAQ=\angle OAP=\frac{\angle QAP}{2}=\frac{{60}^{\circ }}{2}={30}^{\circ }$ (By CPCT)
$In\Delta OAP,$
$\sin {30}^{\circ }=\frac{OP}{OA}$
$\Rightarrow \frac{1}{2}=\frac{r}{d}$
$\Rightarrow d=2r$ .....(i)
$In\Delta OAB,$
$\sin {45}^{\circ }=\frac{OB}{OA}$
$\Rightarrow \sin {45}^{\circ }=\frac{h}{d}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{h}{d}$
$\Rightarrow d=h\sqrt{2}$
From (i), $h\sqrt{2}=2r$
$\Rightarrow h=r\sqrt{2}$
$\Rightarrow h=4\sqrt{2}cm$ $(\because r=4cm)$
Hence, the correct answer is option (c).