Question

In the following figure, O is centre of the circle, $\angle AOB={60}^{\circ }$ and $\angle BDC={100}^{\circ }$, Find $\angle OBC$

• A. $\angle OBC={50}^{\circ }$
• B. $\angle OBC={90}^{\circ }$
• C. $\angle OBC={10}^{\circ }$
• D. $\angle OBC={120}^{\circ }$

Answer 1

The correct option is A

$\angle OBC={50}^{\circ }$

Arc AB subtends $\angle AOB$ at the centre and $\angle ACB$ at the remaining part of circle,

$\therefore \angle AOB=2\angle ACB$
Or $\angle ACB=\frac{1}{2}\angle AOB=\frac{1}{2}\times {60}^{\circ }={30}^{\circ }$

Now in $\Delta DBC$

$\angle DBC+\angle ACB+\angle BDC={180}^{\circ }$

$\Rightarrow \angle DBC+{30}^{\circ }+{100}^{\circ }={180}^{\circ }$

$\therefore \angle DBC={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

Or $\angle OBC={50}^{\circ }$