Quantitative Aptitude Ques 1585
Question: A spherical balloon of radius r subtends angle $ 60{}^\circ $ at the eye of an observer. If the angle of elevation of its centre is $ 60{}^\circ $ and h is the height of the centre of the balloon, then which one of the following is correct? [CDS 2013]
Options:
A) $ h=r $
B) $ h=\sqrt{2}r $
C) $ h=\sqrt{3}r $
D) $ h=2r $
Show Answer
Answer:
Correct Answer: C
Solution:
- In $ \Delta ABO, $ $ \sin 60{}^\circ =\frac{OB}{AO} $
$ \Rightarrow $ $ AO=\frac{OB}{\sin 60{}^\circ } $ (i) Now, in $ \Delta AOC, $ $ \sin \frac{60{}^\circ }{2}=\frac{OC}{AO} $
$ \Rightarrow $ $ AO=\frac{OC}{\sin 30{}^\circ } $
(ii)
From Eqs. (i) and (ii), we get
$ \frac{OB}{\sin 60{}^\circ }=\frac{OC}{\sin 30{}^\circ } $
$ \Rightarrow $ $ \frac{h}{\frac{\sqrt{3}}{2}}=\frac{r}{\frac{1}{2}} $
$ \therefore $ $ h=\sqrt{3}r $
BETA
