WAEC 2020 · Paper 2 · Q12

(a)
Given that $\cos(2y - 16)^\circ = \frac12$ , $0 \le y \le 90$ , find the value of $y$ .
(b)
(i) Using a ruler and a pair of compasses only, (α) construct a triangle $ABC$ in which $|AB| = 8\text{ cm}$ , $|BC| = 9\text{ cm}$ and $\angle ABC = 75^\circ$ ; (β) locate the point $P$ inside $ABC$ such that $|PA| = |PB|$ and $|PA| = 4.5\text{ cm}$ . (ii) Measure: (α) $|CP|$ ; (β) $\angle ACB$ .
Model answer
Draw $AB = 8$ cm, construct $75^\circ$ at $B$ ( $60^\circ$ plus half of the next $30^\circ$ ) and mark $C$ with $|BC| = 9$ cm. $P$ lies on the perpendicular bisector of $AB$ , where an arc of radius $4.5$ cm centred at $A$ cuts it inside the triangle. Measuring gives $|CP| \approx 6.8$ cm and $\angle ACB \approx 48^\circ$ .

The accurate construction: A(0, 0), B(8, 0), C(5.67, 8.69), P(4, 2.06).

Worked solution (try it first)

(α) Draw $|BC| = 9$ cm, construct an angle of $75^\circ$ at $B$ (a $60^\circ$ angle plus half of the $30^\circ$ next to it), and mark $A$ on its arm with $|AB| = 8$ cm.
Join $AC$ .

$|PA| = |PB|$ , so $P$ lies on the perpendicular bisector of $AB$ : construct it.
With centre $A$ and radius 4.5 cm, draw an arc to cut the bisector inside the triangle at $P$ .

Measure: $|CP| \approx 6.8$ cm and $\angle ACB \approx 48^\circ$ .
(By calculation, $|AC| \approx 10.38$ cm, $\angle ACB \approx 48.1^\circ$ and $|CP| \approx 6.84$ cm.)