WAEC 2018 · Paper 2 · Q7

(a)
A triangle $PQR$ has vertices $P(2, 2)$ , $Q(3, -1)$ and $R(4, 0)$ . Using the vector method, calculate angle $PQR$ .

Worked solution (try it first)

Angle $PQR$ is at $Q$ , so use vectors starting at $Q$ .
$\overrightarrow{QP} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$ and $\overrightarrow{QR} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ .
Their scalar product is $(-1)(1) + (3)(1) = 2$ .
$\cos\theta = \dfrac{2}{\sqrt{10}\sqrt2}$
$= \dfrac{1}{\sqrt5}$ , so $\theta \approx 63.43^\circ$ .