Question

In the figure, $CD$ is the diameter of a semicircle $CBED$ with centre $O$ and $AB$ $=$ $OD$. If $\angle EOD={60}^{\circ }$, then $\angle BAC$ is

• A. ${15}^{\circ }$
• B. ${20}^{\circ }$
• C. ${30}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is B ${20}^{\circ }$

$OB=OD=AB$
$\therefore \angle BAC=\angle BOC$
As $OB=OE=radius$
$\angle EBO=\angle BEO=2\alpha$
(Exterior angle of traingle $=$ sum of opposite interior angle)
For $\Delta$ BOE :
$2\alpha +2\alpha +({180}^{\circ }-\alpha -{160}^{\circ })={180}^{\circ }$
$3\alpha ={60}^{\circ }\Rightarrow \alpha ={20}^{\circ }$