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] |
A) \[h=r\]
B) \[h=\sqrt{2}r\]
C) \[h=\sqrt{3}r\]
D) \[h=2r\]
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 |
\[\frac{OB}{\sin 60{}^\circ }=\frac{OC}{\sin 30{}^\circ }\]\[\Rightarrow \]\[\frac{h}{\frac{\sqrt{3}}{2}}=\frac{r}{\frac{1}{2}}\] |
\[\therefore \] \[h=\sqrt{3}r\] |
You need to login to perform this action.
You will be redirected in
3 sec