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\[\Delta AOC\], \[\Rightarrow \,\,\,\,\,\,\sin \left( \frac{{{60}^{{}^\circ }}}{2} \right)=\frac{OC}{AO}\] ? (ii) From Eqs. (i) and (ii), \[\frac{OB}{\sin {{60}^{{}^\circ }}}=\frac{OC}{\sin {{30}^{{}^\circ }}}\] \[\Rightarrow \,\,\,\,\,\frac{h}{\sqrt{3}}=\frac{r}{1}\Rightarrow \]You need to login to perform this action.
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