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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Heights and Distances

NDA 2018 · Mathematics · Heights and Distances

A spherical balloon of radius r subtends an angle α at the eye of...

A spherical balloon of radius rr subtends an angle α\alpha at the eye of an observer, while the angle of elevation of its centre is β\beta. What is the height of the centre of the balloon (neglecting the height of the observer)?

Options

  1. A\frac{r \sin \beta}{\sin \left(\frac{\alpha}{2}\right)}
  2. B\frac{r \sin \beta}{\sin \left(\frac{\alpha}{4}\right)}
  3. Crsin⁡(β2)sin⁡α\frac{r \sin \left(\frac{\beta}{2}\right)}{\sin \alpha}
  4. D\frac{r \sin \alpha}{\sin \left(\frac{\beta}{2}\right)}

Correct answer

A. \frac{r \sin \beta}{\sin \left(\frac{\alpha}{2}\right)}

Step-by-step solution

Let ' AA ' be the position of eye. Let ' OO ' be the centre of spherical balloon. Let ' hh ' be the height of centre of balloon.
From figure, in △OAD,sin⁡α2=ODOA=rOA\triangle \mathrm{OAD}, \sin \frac{\alpha}{2}=\frac{\mathrm{OD}}{\mathrm{OA}}=\frac{\mathrm{r}}{\mathrm{OA}}⇒OA=rsin⁡α2...(1) In △OAB,sin⁡β=OBOA=hOA⇒h=OA⋅sin⁡β=r⋅sin⁡βsin⁡α2...from(1)\begin{aligned} & \Rightarrow \mathrm{OA}=\frac{\mathrm{r}}{\sin \frac{\alpha}{2}}...(1) \\ & \text{ In } \triangle \mathrm{OAB}, \sin \beta=\frac{\mathrm{OB}}{\mathrm{OA}}=\frac{\mathrm{h}}{\mathrm{OA}} \\ & \Rightarrow \mathrm{ h}=\mathrm{OA} \cdot \sin \beta \\ & =\frac{\mathrm{r} \cdot \sin \beta}{\sin \frac{\alpha}{2}}...from (1) \end{aligned}$$
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