Question

In the given figure AC is the diameter of the circle. If arc AXB is equal to half of arc BYC. \(\angle AOB\) = ___ degrees

\( \theta \)

Answer 1

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)

$\Rightarrow \angle AOB=\frac{1}{3}\times {180}^o$

$\Rightarrow \angle AOB={60}^o$