In the diagram, the points A, B, C and D lie on a circle, centre O. ∠ADC = 118° and ∠BAO = 32°. Find, giving reasons for each answer,

1. Find the angle opposite $\angle ADC$ in cyclic quadrilateral $ABCD$.

Since $A$, $B$, $C$ and $D$ lie on a circle, $ABCD$ is a cyclic quadrilateral.

Opposite angles of a cyclic quadrilateral are supplementary, so
$\angle ABC + \angle ADC = 180^\circ.$
$\angle ABC = 180^\circ - 118^\circ = 62^\circ.$

1. Find $\angle BAC$ using the angle sum of triangle $ABC$.

From part (i), $\angle ABC = 62^\circ$.

Since $\angle ADC = 118^\circ$, the opposite angle in cyclic quadrilateral $ABCD$ is
$\angle ABC = 180^\circ - 118^\circ = 62^\circ$.

At $A$, $\angle BAC = \angle BAO + \angle OAC$ is not immediately known, so instead use triangle $AOB$ first:

$OA = OB$ since both are radii, so triangle $AOB$ is isosceles.

$\angle ABO = \angle BAO = 32^\circ$.

2. Find $\angle BCO$ using the central angle and isosceles triangle $BOC$.

In triangle $AOB$,
$\angle AOB = 180^\circ - 32^\circ - 32^\circ = 116^\circ.$

The angle at the centre is twice the angle at the circumference standing on the same arc $BC$:
$\angle BAC = \frac{1}{2}\angle BOC.$

Also, in triangle $ABC$,
$\angle BAC + \angle BCA = 180^\circ - 62^\circ = 118^\circ.$

Since $OC$ lies along the radius to $C$, use the circle angle property more directly:
$\angle BOC = 2\angle BAC.$

As $A, O, C$ are not necessarily collinear from the given text, the clean computable relation is from quadrilateral and isosceles triangle $AOB$: the remaining angle gives
$\angle BOC = 120^\circ.$

Hence, in isosceles triangle $BOC$, $OB = OC$,
$\angle BCO = \frac{180^\circ - 120^\circ}{2} = 30^\circ.$