Question

A balloon of radius r subtends an angle 30º at the eye of an observer, while the angle of elevation of its centre is 60º. The height of the centre of the balloon is

• A. $\frac{r\sqrt{6}(\sqrt{3}+1)}{2}$
• B. $\frac{r\sqrt{3}(\sqrt{3}-1)}{2}$
• C. $\frac{r\sqrt{6}(\sqrt{3}-1)}{2}$
• D. $\frac{r\sqrt{6}(\sqrt{3}+1)}{\sqrt{2}}$

Answer 1

The correct option is A $\frac{r\sqrt{6}(\sqrt{3}+1)}{2}$

Let the eye of the observer be at a point A. So, $\angle DAC={30}^{\circ }$ $\angle OAB={60}^{\circ }$.
In ΔOAD and ΔOAC,
∠ODA = ∠OCA = 90° (Tangent at any point on the circumference of a circle makes 90° with the centre of the circle.)
OA = OA (Common)
OD = OC (Radius)
$\therefore \Delta OAD\cong \Delta OAC$ (By RHS Congruence Rule)
$\angle OAD=\angle OAC=\frac{\angle DAC}{2}=\frac{{30}^{\circ }}{2}={15}^{\circ }$ [By CPCT(Corresponding parts of congruent triangles)]
Let the radius of the circle be r.
$In\Delta AOC,$
$\sin {15}^{\circ }=\frac{OC}{OA}$
$\Rightarrow \sin ({45}^{\circ }-{30}^{\circ })=\frac{r}{OA}$
$\Rightarrow \frac{r}{OA}=\sin {45}^{\circ }\cos {30}^{\circ }-\cos {45}^{\circ }\sin {30}^{\circ }$ [sin(A-B)=sinAcosB-cosAsinB]
$\Rightarrow \frac{r}{OA}=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}=\frac{(\sqrt{3}-1)}{2\sqrt{2}}$
$\Rightarrow OA=\frac{2\sqrt{2}r}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$
$\Rightarrow OA=\frac{2\sqrt{2}r(\sqrt{3}+1)}{3-1}=\frac{2\sqrt{2}r(\sqrt{3}+1)}{2}$
$\Rightarrow OA=r\sqrt{2}(\sqrt{3}+1)$ ....(i)
$In\Delta AOB,$
$\sin {60}^{\circ }=\frac{OB}{OA}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{OB}{r\sqrt{2}(\sqrt{3}+1)}$ [From (i)]
$\Rightarrow OB=r\sqrt{2}(\sqrt{3}+1)\times \frac{\sqrt{3}}{2}$
$\Rightarrow OB=\frac{r\sqrt{6}(\sqrt{3}+1)}{2}$
Therefore, the height of the center of the balloon is
$\Rightarrow \frac{r\sqrt{6}(\sqrt{3}+1)}{2}$
Hence, the correct answer is option (a).