Question

In the given figure AC is the diameter of the circle. If arc AXB is equal to half of arc BYC. Find $\angle BOC$

• A. 30^o
• B. 60^o
• C. ^90o
• D. 120^o

Answer 1

The correct option is D

120^o

Given arc BYC is twice of arc AXB

$\therefore$ arc BYC : arc AXB = 2 : 1

$\Rightarrow \angle BOC:\angle AOB=2:1$

Angle made by arc AYC is ${180}^o$ ( since AC is the diameter)

$\Rightarrow \angle BOC=\frac{2}{3}\times {180}^o$

$\Rightarrow \angle BOC={120}^o$