WAEC 2016 · Paper 2 · Q10

(a)
Using a ruler and a pair of compasses only, construct: (i) $\triangle XYZ$ such that $|XY| = 10\text{ cm}$ , $\angle XYZ = 30^\circ$ and $\angle YXZ = 45^\circ$ ; (ii) the locus $l_1$ of points equidistant from $Y$ and $Z$ ; (iii) the locus $l_2$ of points on the line through $Z$ parallel to $XY$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $XY = 10$ cm. At $Y$ , construct $60^\circ$ and bisect it to get $30^\circ$ . At $X$ , construct $90^\circ$ and bisect it to get $45^\circ$ . The two arms meet at $Z$ . Then (ii) bisect $YZ$ perpendicularly to get $l_1$ , and (iii) draw the line through $Z$ parallel to $XY$ (copy an angle, or use two equal perpendiculars) to get $l_2$ .
(b)
Locate the point $M$ , the point of intersection of $l_1$ and $l_2$ .
Model answer
$M$ is where $l_1$ crosses $l_2$ . Because $MZ = MY$ and $\angle MZY = 30^\circ$ (alternate to $\angle ZYX$ ), triangle $ZMY$ is isosceles with base angles $30^\circ$ , so $\angle ZMY$ measures $120^\circ$ .
(c)
Measure $\angle ZMY$ .

Worked solution (try it first)

$l_1$ , equidistant from $Y$ and $Z$ : construct the perpendicular bisector of $YZ$ .

Measure $\angle ZMY = 120^\circ$ .
Check: $M$ is on $l_1$ , so $MZ = MY$ .
$l_2 \parallel XY$ , so $\angle MZY = \angle ZYX = 30^\circ$ (alternate angles).
Triangle $ZMY$ is isosceles with base angles $30^\circ$ , so $\angle ZMY = 180^\circ - 60^\circ$
$= 120^\circ$ .