WAEC 2019 · Paper 2 · Q13

(a)
In the diagram, $A$ , $B$ , $C$ , $D$ are points on the circle with centre $O$ . $\angle AOD = 130^\circ$ and $\angle BAO = 26^\circ$ . Find: (i) $\angle ODB$ ; (ii) $\angle BOD$ .
Separate values with commas, e.g. 3, −2
(b)
A number is selected at random from each of the sets $\{1, 2, 6\}$ and $\{3, 4, 5\}$ . Find the probability that the sum of the numbers selected is greater than seven.

Worked solution (try it first)

$OA = OD$ (radii), so triangle $OAD$ is isosceles: $\angle OAD = \angle ODA$
$= \frac{180^\circ - 130^\circ}{2}$
$= 25^\circ$ .
The angle at the centre is twice the angle at the circumference on the same arc, so $\angle ABD = \frac12 \times 130^\circ$
$= 65^\circ$ .
In triangle $ABD$ : $\angle BAD = 26^\circ + 25^\circ$
$= 51^\circ$ , so $\angle ADB = 180^\circ - 65^\circ - 51^\circ$
$= 64^\circ$ .
Then $\angle ODB = \angle ADB - \angle ODA$
$= 64^\circ - 25^\circ$
$= 39^\circ$ .

$OB = OD$ (radii), so triangle $OBD$ is isosceles with $\angle OBD = \angle ODB = 39^\circ$ : $\angle BOD = 180^\circ - 2 \times 39^\circ$
$= 102^\circ$ .
(Check: it is twice $\angle BAD = 51^\circ$ .)