Question

In the figure; AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If $\angle C={20}^0$, find $\angle AOD$

• A. ${90}^o$
• B. ${60}^o$
• C. ${50}^o$
• D. ${120}^o$

Answer 1

The correct option is B

${60}^o$

Join OB

In $\Delta$ OBC, BC= OD = OB (radii of the circle)

$\therefore \angle BOC=\angle BCO={20}^0$

and Exterior $\angle ABO=\angle BCO+\angle BOC$

= ${20}^0+{20}^0={40}^0$ ...(i)

In $\Delta$ OAB, OA =OB (radii of the circle)

$\therefore \angle OAB=\angle OBA={40}^0$ [from(i)]

$\angle AOB={180}^0-\angle OAB-\angle OBA$

= ${180}^0-{40}^0-{40}^0={100}^0$

$\because$ DOC is a line

$\therefore \angle AOD+\angle AOB+\angle BOC={180}^0$

$\Rightarrow \angle AOD+{100}^0+{20}^0={180}^0$

$\Rightarrow \angle AOD+{120}^0={180}^0$

$\therefore \angle AOD={180}^0-{120}^0={60}^0$