Correct option is (B) 60°
Angles subtended by arc (AC) at centre of circle (O) is double than the angle subtended by that arc (AC) at point of circumference (B).
\(\therefore\) \(\angle AOC=2\angle ABC\)
But given that \(\angle AOC+\angle ABC\) = 2 right angle
\(\Rightarrow\) \(\angle AOC+\frac{\angle AOC}2\) \(=2\times90^\circ=180^\circ\)
\(\Rightarrow\) \(\frac{3}2\angle AOC\) \(=180^\circ\)
\(\Rightarrow\) \(\angle AOC\) \(=180^\circ\times\frac23=120^\circ\)
\(\because\) AB = BC
\(\Rightarrow\) arc AB = arc BC ___________(1)
\((\because\) Equal chords makes equal arcs)
Since, arc AB & arc AD forms a semi-circle and arc BC & arc CD forms another semi-circle.
\(\therefore\) arc AB + arc AD = arc BC + arc CD \(=\pi r\) ___________(2)
(Both will form semi-circle)
\(\Rightarrow\) arc AD = arc CD (From (1) & (2))
\(\Rightarrow\) \(\angle AOD=\angle COD\) (Equal arcs subtends equal angles at centre of circle)
But \(\angle AOD+\angle COD\) \(=\angle AOC=120^\circ\)
\(\Rightarrow\) \(2\angle AOD=120^\circ\)
\(\Rightarrow\) \(\angle AOD=60^\circ\)
