Question

In the given figure, chords $AB=CD,\angle COD={60}^{\circ }$ and $\angle AOD={170}^{\circ }$.

Find the value of $\angle BOC$.

Answer 1

We know that equal chords of a circle subtend equal angles at the centre.

$\Rightarrow \angle AOB=\angle COD={60}^{\circ }$

Since, the central angle of a circle is ${360}^{\circ }$

So, $\angle AOB+\angle BOC+\angle COD+\angle AOD={360}^{\circ }$

$\Rightarrow {60}^{\circ }+\angle BOC+{60}^{\circ }+{170}^{\circ }={360}^{\circ }$

$\Rightarrow {290}^{\circ }+\angle BOC={360}^{\circ }$

$\Rightarrow \angle BOC={360}^{\circ }-{290}^{\circ }={70}^{\circ }$