WAEC 2017 · Paper 2 · Q8

(a)
Forces $F_1(18\text{ N}, 330^\circ)$ , $F_2(10\text{ N}, 090^\circ)$ and $F_3(25\text{ N}, 180^\circ)$ act on a body at rest. Find, correct to one decimal place, the magnitude and direction of the resultant force.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$F_1$ : $18\sin330^\circ = -9$ east and $18\cos330^\circ = 15.588$ north.
$F_2$ : $10$ east and $0$ north.
$F_3$ : $0$ east and $-25$ north.
Add: east $-9 + 10 = 1$ .
North $15.588 - 25 = -9.412$ .
$|\mathbf R| = \sqrt{1 + 88.59}$
$\approx 9.5\text{ N}$ .
It points slightly east of south: $\tan^{-1}\frac{1}{9.412} = 6.07^\circ$ east of south, a bearing of $180^\circ - 6.07^\circ \approx 173.9^\circ$ .