Question

$△ABC$ is inscribed in a circle. Point $P$ lies between $A$ and $C$, whereas point $Q$ lies between $B$ and $C$. If $m\left(\text{arc}APC\right)={60}^{\circ }$ and $\angle BAC={80}^{\circ }$, find $m\left(\text{arc}BQC\right)$.

• A. ${180}^{\circ }$
• B. ${90}^{\circ }$
• C. ${160}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

Answer: (C)

${160}^{\circ }$

By inscribed angle theorem,

$\frac{1}{2}m\angle BAC=m\left(arcBQC\right)$

$m\left(arcBQC\right)=2\times \angle BAC$

$\therefore m\left(arcBQC\right)=2\times {80}^o={160}^o$