NECO 2022 · Paper 2 · Q5

1. $AD$ and $BC$ both pass through the centre $O$, so they are diameters.

(i)

1. $\angle AOB$ and $\angle COD$ are vertically opposite angles, so $\angle AOB = 85^\circ$.

(ii)

1. In triangle $OCD$, $OC = OD$ (radii), so the base angles are equal: $\angle OCD = \frac{180^\circ - 85^\circ}{2}$
   $= 47.5^\circ$.
2. $B$, $O$ and $C$ lie on one line, so $\angle BCD = \angle OCD = 47.5^\circ$.

(iii)

1. $\angle BAD$ and $\angle BCD$ both stand on the arc $BD$ (angles in the same segment), so $\angle BAD = 47.5^\circ$.

(iv)

1. $\angle BOD$ and $\angle COD$ lie on the straight line $BC$, so $\angle BOD = 180^\circ - 85^\circ$
   $= 95^\circ$.

(v)

1. In triangle $OBD$, $OB = OD$ (radii), so $\angle OBD = \frac{180^\circ - 95^\circ}{2}$
   $= 42.5^\circ$.