Question

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

• A. ${40}^o$
• B. ${60}^o$
• C. ${160}^o$
• D. ${80}^o$

Answer 1

Answer: (C)

${160}^o$

We join $BO$ & $CO$

$m\left(arcBQC\right)$ means the central angle subtended by $arcBQC=\angle BOC$, subtended by the chord C at O.

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.

$\therefore \angle BOC=2\times \angle BAC=2\times 8^o={160}^o$