WAEC 2024 · Paper 2 · Q12

(a)
In the diagram, $PR$ is a tangent to the circle centre $O$ at $Q$ , $\angle POQ = 56^\circ$ , and $PO$ meets the chord $SQ$ at $V$ such that $\angle SVP = 109^\circ$ . Calculate: (i) $\angle TQP$ ; (ii) $\angle QTS$ .
Separate values with commas, e.g. 3, −2
(b)
Simplify $\dfrac{2n^2 - 3n - 2}{2n^2 + 3n + 1} \times \dfrac{n^2 - 1}{n^2 - 4}$ .

Worked solution (try it first)

$T$ lies on $PO$ , so $\angle TOQ = 56^\circ$ .
$OT = OQ$ (radii), so $\angle OQT = \frac{180^\circ - 56^\circ}{2}$
$= 62^\circ$ .
A tangent is perpendicular to the radius at the point of contact, so $\angle OQP = 90^\circ$ .
Then $\angle TQP = 90^\circ - 62^\circ$
$= 28^\circ$ .

The angle at the circumference is half the angle at the centre on the same arc $TQ$ : $\angle TSQ = \frac12 \times 56^\circ$
$= 28^\circ$ .
In triangle $STV$ : $\angle SVT = 109^\circ$ and $\angle TSV = 28^\circ$ , so $\angle STV = 180^\circ - 109^\circ - 28^\circ$
$= 43^\circ$ .
$V$ lies on $TO$ , so $\angle VTQ = \angle OTQ$ , and $\angle OTQ = \angle OQT = 62^\circ$ (isosceles triangle $OTQ$ ).
So $\angle QTS = \angle STV + \angle VTQ$
$= 43^\circ + 62^\circ$
$= 105^\circ$ .

Factorise each part: $2n^2 - 3n - 2 = (2n + 1)(n - 2)$ , $2n^2 + 3n + 1 = (2n + 1)(n + 1)$ , $n^2 - 1 = (n - 1)(n + 1)$ and $n^2 - 4 = (n - 2)(n + 2)$ .
Cancel the common factors $(2n + 1)$ , $(n + 1)$ and $(n - 2)$ .
What's left is $\dfrac{n - 1}{n + 2}$ .