WAEC 2017 · Paper 2 · Q3

(a)
In the diagram, $PQR$ is a triangle with $|PR| = |PQ|$ , $\angle QPR = \left(2\frac14x + \frac12y\right)^\circ$ , $\angle PQR = 2y^\circ$ and $\angle PRQ = \left(2x - \frac25y\right)^\circ$ . Find the values of $x$ and $y$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$|PR| = |PQ|$ , so the triangle is isosceles and the base angles at $Q$ and $R$ are equal: $2y = 2x - \frac25y$ .
Multiply by 5: $10y = 10x - 2y$ , so $12y = 10x$ and $x = \frac65y$ (1).
The angles of a triangle add up to $180^\circ$ : $\left(2\frac14x + \frac12y\right) + 2y + \left(2x - \frac25y\right) = 180$ .
Collect terms: $\frac{17}{4}x + \frac{21}{10}y = 180$ .
Multiply by 20: $85x + 42y = 3600$ (2).
Substitute (1) into (2): $85 \times \frac65y + 42y = 3600$ , so $102y + 42y = 3600$ , $144y = 3600$ and $y = 25$ .
Then $x = \frac65 \times 25 = 30$ .
Check: the angles are $80^\circ$ , $50^\circ$ and $50^\circ$ , which add up to $180^\circ$ ✓.