A, B, C and D are points on the circle with centre O and radius 5 cm. AE is a tangent to the circle at A and OE is a straight line that passes through C. Angle OEA = 28° and CD = 5 cm.

1. Find the angle between $OA$ and $OC$.

Since $AE$ is a tangent at $A$, $OA \perp AE$.

So $\angle OAE = 90^\circ$.

In $\triangle OAE$,
$\angle AOE = 180^\circ - 90^\circ - 28^\circ = 62^\circ.$
Since $O, C, E$ are collinear,
$\angle AOC = 62^\circ.$

2. Find the central angle $\angle COD$.

$OC=OD=5$ cm because they are radii of the circle, and $CD=5$ cm.

Hence $\triangle OCD$ is equilateral.

Therefore
$\angle COD = 60^\circ.$

3. Find $\angle OAD$ using isosceles triangle $AOD$.

From the diagram, $\angle AOD = \angle AOC + \angle COD$.

$\angle AOD = 62^\circ + 60^\circ = 122^\circ.$

Since $OA=OD$ are radii, $\triangle AOD$ is isosceles.

Thus
$\angle OAD = \angle ODA = \frac{180^\circ - 122^\circ}{2} = 29^\circ.$

1. Find $\angle ADC$ using the angle at $D$ split by $DO$.

From part (a)(i), $\angle OAD = 29^\circ$.

Since $OA=OD$, triangle $AOD$ is isosceles, so

$\angle ODA = 29^\circ$.

Also, $OC=OD=CD=5\text{ cm}$, so triangle $OCD$ is equilateral and

$\angle ODC = 60^\circ$.

Hence

$\angle ADC = \angle ADO + \angle ODC = 29^\circ + 60^\circ = 89^\circ$.

2. Use opposite angles in a cyclic quadrilateral to find $\angle ABC$.

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

Opposite angles in a cyclic quadrilateral are supplementary, so

$\angle ABC + \angle ADC = 180^\circ$.

$\angle ABC = 180^\circ - 89^\circ = 91^\circ$.