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 $\angle COA$.

• A. ${120}^o$
• B. ${160}^o$
• C. ${140}^o$
• D. ${150}^o$

Answer 1

The correct option is B ${160}^o$

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

$\therefore \angle COA=\frac{4}{9}\times {360}^o$ (angle made by a circle is ${360}^o$)

$\Rightarrow \angle COA={160}^o$