Question

The circumference of a circle with centre O is divided into three arcs APB, BQC and CRA such that

$\frac{arcAPB}{2}=\frac{arcBQC}{3}=\frac{arcCRA}{4}$. Find $\angle BOA.$

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${80}^{\circ }$

Answer 1

The correct option is D

${80}^{\circ }$

Let $\frac{arcAPB}{2}=\frac{arcBQC}{3}=\frac{arcCRA}{4}=k$.

$\Rightarrow arcAPB=2k,arcBQC=3k,CRA=4k$.

$\Rightarrow arcAPB:arcBQC:arcCRA=2:3:4$

$\Rightarrow \angle AOB:\angle BOC:\angle COA=2:3:4$

as $\angle AOB=\angle BOA$,

$\therefore \angle BOA=\frac{2}{9}\times {360}^o$ (angle made by a circle is ${360}^o$)

$\Rightarrow \angle BOA={80}^o$