Question

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

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

• A. This situation holds no signifiance
• B. OP need not be radius of circle
• C. OP is perpendicular to XY
• D. The angle between OP and XY is variable

Answer 1

The correct option is B

OP need not be radius of circle

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

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

$\Rightarrow$ arc APB : arc BQC : arc CRA = 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 ${360}^o$)

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