WAEC 2023 · Paper 1 · Q39

Given that $\mathbf{r} = (10\ \text{N}, 200^\circ)$ and $\mathbf{n} = (16\ \text{N}, 020^\circ)$ , find $(3\mathbf{r} - 2\mathbf{n})$ .

Worked solution (try it first)

$3\mathbf{r} = (30\ \text{N}, 200^\circ)$ and $2\mathbf{n} = (32\ \text{N}, 020^\circ)$ .
$-2\mathbf{n}$ points the opposite way: $020^\circ + 180^\circ = 200^\circ$ , so $-2\mathbf{n} = (32\ \text{N}, 200^\circ)$ .
Both parts now act along $200^\circ$ , so add their sizes: $30 + 32 = 62$ N.
So $3\mathbf{r} - 2\mathbf{n} = (62\ \text{N}, 200^\circ)$ , option A.