Question

For the given figure, find $\angle ABC+\angle ACB,$ where $O$ is the centre of the circle.

• A. ${40}^o$
• B. ${120}^o$
• C. ${140}^o$
• D. ${180}^o$

Answer 1

The correct option is C ${140}^o$

The given figure is as follows:

Here, $OB=OC=$ radius of the circle
$\therefore △OBC$ is a isosceles triangle.
$\Rightarrow \angle OBC=\angle OCB={50}^o$

From interior angle sum property for $△OBC,$
$\angle BOC={180}^o-{50}^o-{50}^o={80}^o$

We know, the angle subtended by an arc at the centre of a cirlce is double the angle subtended by the arc at any point on the circle.
$\therefore \angle BOC=2\times \angle BAC$
$\Rightarrow \angle BAC=\frac{{80}^o}{2}={40}^o$
From interior angle sum property for $△ABC,$
$\angle ABC+\angle ACB={180}^o-\angle BAC$
$\Rightarrow \angle ABC+\angle ACB={180}^o-{40}^o={140}^o$