Question

In the figure, if $m\left(arcAPC\right)={60}^0$ and $\angle BAC={80}^0$, then m$\angle ABC$ in degrees is:

• A. ${150}^o$
• B. ${30}^o$
• C. ${60}^o$
• D. ${45}^o$

Answer 1

The correct option is B ${30}^o$

We join $AO,CO$ and $AC$

$m\left(arcAPC\right)={60}^o$ means the central angle subtended by $arcAPC={60}^o=\angle AOC$, subtended by the chord $AC$.

We know that the angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord at the cicumference of the circle.

$\therefore \angle ABC=\frac{1}{2}\times \angle ABC=\frac{1}{2}\times {60}^o={30}^o$