WAEC 2008 · Paper 2 · Q11

(a)
In the diagram, $AB \parallel CD$ and $BC \parallel FE$ , $\angle CDE = 75^\circ$ and $\angle DEF = 26^\circ$ . Find the angles marked $x$ and $y$ .
Separate values with commas, e.g. 3, −2
(b)
The diagram shows a circle $ABCD$ with centre $O$ and radius $7\text{ cm}$ . The reflex angle $AOC = 190^\circ$ and $\angle DAO = 35^\circ$ . Find: (i) $\angle ABC$ ; (ii) $\angle ADC$ .
Separate values with commas, e.g. 3, −2
(c)
Using the diagram in (b), calculate, correct to 3 significant figures, the length of: (i) arc $ABC$ ; (ii) the chord $AD$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Extend $CD$ beyond $D$ to meet $EF$ at $G$ .
On the straight line $CDG$ , $\angle EDG = 180^\circ - 75^\circ$
$= 105^\circ$ .
In triangle $DEG$ the angles add up to $180^\circ$ : $\angle DGE = 180^\circ - 105^\circ - 26^\circ$
$= 49^\circ$ .
$BC \parallel FE$ , so the angle between $CD$ and $BC$ equals the angle between $CD$ and $FE$ (corresponding angles): $\angle BCD = 49^\circ$ .
$AB \parallel CD$ , so $x = \angle ABC = \angle BCD$ (alternate angles).
So $x = 49^\circ$ .
$y$ is the reflex angle at $C$ : $y = 360^\circ - 49^\circ = 311^\circ$ .

$B$ is on the arc opposite the reflex angle, and the angle at the centre is twice the angle at the circumference: $\angle ABC = \frac12 \times 190^\circ$
$= 95^\circ$ .

The other angle $AOC$ is $360^\circ - 190^\circ = 170^\circ$ , so $\angle ADC = \frac12 \times 170^\circ$
$= 85^\circ$ .
(Check: $95^\circ + 85^\circ = 180^\circ$ , as in any cyclic quadrilateral.)

Arc $ABC$ subtends the $170^\circ$ angle at $O$ .
Its length is $\frac{170}{360} \times 2 \times \frac{22}{7} \times 7 = 20.78$ , so $20.8\text{ cm}$ .

Triangle $AOD$ is isosceles ( $OA = OD = 7$ ), so $\angle ODA = 35^\circ$ and $\angle AOD = 180^\circ - 70^\circ$
$= 110^\circ$ .
Split it into two right-angled triangles: $AD = 2 \times 7 \sin 55^\circ$
$= 14 \times 0.8192$
$= 11.47$ , so $11.5\text{ cm}$ .