In a circle with center O, where BOC is the diameter and ABCD is a cyclic quadrilateral, if \(\angle ADC = 110^\circ\), then find the value of \(\angle ACB\).

Q.In a circle with center O, where BOC is the diameter and ABCD is a cyclic quadrilateral, if \(\angle ADC = 110^\circ\), then find the value of \(\angle ACB\).

ABCD is a cyclic quadrilateral. \(\therefore \angle ADC + \angle ABC = 180^\circ\) \(\therefore \angle ABC = 180^\circ - 110^\circ = 70^\circ\) Again, \(\angle BAC\) is an angle subtended by the diameter, so it's a semicircular angle = \(90^\circ\)

\(\therefore \angle ACB\) in triangle ABC = \(180^\circ - (\angle ABC + \angle BAC)\) = \(180^\circ - (70^\circ + 90^\circ)\) = \(180^\circ - 160^\circ = 20^\circ\)