Question

$\text{In the given figure, O is the centre of a circle. If}\angle AOB={40}^{\circ }and\angle BDC={100}^{\circ }thenfind\angle OBC.$

• A. ${60}^{\circ }$
• B. ${40}^{\circ }$
• C. ${80}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is A

${60}^{\circ }$

$\angle AOB=2\times \angle ACB$
$\Rightarrow \angle ACB=\frac{1}{2}\times \angle AOB$
$=\frac{1}{2}\times {40}^{\circ }={20}^{\circ }$



In $\Delta BDC,\angle DBC+\angle BDC+\angle DCB={180}^{\circ }$
$\Rightarrow \angle DBC+{100}^{\circ }+{20}^{\circ }={180}^{\circ }$
$\therefore \angle DBC={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

$Hence\angle OBC={60}^{\circ }$