Question

In the following figure, O is the centre of the circle,$\angle AOB={60}^0$ and $\angle BDC={100}^0.$Find $\angle OBC.$

Answer 1

Form the figure, we have
$\angle ACB=\frac{1}{2}\angle AOB=\frac{1}{2}\times {60}^{\circ }={30}^{\circ }$
[Angle at the centre is double the angle at the circumference subtended by the same chord]
Now, by applying angle sum property in $\Delta BDC$,
$\angle DBC={180}^{\circ }–{100}^{\circ }–{30}^{\circ }={50}^{\circ }$
Therefore,
$\angle OBC={50}^{\circ }$