Question

In the given figure the perpendicular drawn from A to the base BC pass through the center O of the circle. Find the value of $\angle OCB$ if $\angle BAC={60}^{\circ }$.

• A. ${90}^{\circ }$
• B. ${45}^{\circ }$
• C. ${30}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is C

${30}^{\circ }$

Let the perpendicular drawn from A to BC meets BC at D.
$\angle BOC=2\angle BAC$ (Angle subtended by a chord at the centre is twice the angle subtended by it at the centre)
$\angle BOD=\angle COD$
Because, $\Delta BOD\cong \Delta COD$
So $\angle COD=\angle BAC$
$\angle COD={60}^{\circ }$
In $\Delta ODC$
$\angle OCD={180}^{\circ }-{90}^{\circ }-{60}^{\circ }$ (Sum of angles in a trinagle is ${180}^{\circ })$
$\angle OCD={30}^{\circ }\angle OCB={30}^{\circ }$