WAEC 2017 · Paper 2 · Q1

(a)
Simplify: $\dfrac{2\frac12 + 1\frac34 \div 1\frac25}{2\frac14 - 1\frac12}$ .
(b)
In the diagram, $\angle SRP = \angle RPQ = 90^\circ$ , $\angle PSR = 60^\circ$ and $\angle PQR = 45^\circ$ . If $|SR| = 3\sqrt2\text{ cm}$ and $|QR| = x$ , find the value of $x$ , leaving your answer in surd form (radicals).

Worked solution (try it first)

Top: divide first, $1\frac34 \div 1\frac25 = \frac74 \times \frac57$
$= \frac54$ .
Then $2\frac12 + \frac54 = \frac{10}{4} + \frac54$
$= \frac{15}{4}$ .
Bottom: $2\frac14 - 1\frac12 = \frac94 - \frac64$
$= \frac34$ .
So the value is $\frac{15}{4} \div \frac34 = \frac{15}{4} \times \frac43$
$= 5$ .

In $\triangle PSR$ , right-angled at $R$ : $PR$ is opposite the $60^\circ$ angle at $S$ and $SR$ is adjacent, so $\tan 60^\circ = \frac{|PR|}{3\sqrt2}$ .
Then $|PR| = 3\sqrt2 \times \sqrt3 = 3\sqrt6$ cm.
In $\triangle PRQ$ , right-angled at $P$ : $PR$ is opposite the $45^\circ$ angle at $Q$ and $QR = x$ is the hypotenuse, so $\sin 45^\circ = \frac{3\sqrt6}{x}$ .
Then $x = \frac{3\sqrt6}{\frac{1}{\sqrt2}}$
$= 3\sqrt6 \times \sqrt2$
$= 3\sqrt{12}$
$= 6\sqrt3\text{ cm}$ .