WAEC 2008 · Paper 2 · Q18

(a)
The position vectors of the points $A$ , $B$ and $C$ are $(4\mathbf i + 5\mathbf j)$ , $(2\mathbf i + 2\mathbf j)$ and $(6\mathbf i + \mathbf j)$ respectively. Calculate, correct to one decimal place, the acute angle between $\overrightarrow{AB}$ and $\overrightarrow{BC}$ .
(b)
A particle of mass $5\text{ kg}$ is suspended by two light inextensible strings making angles $30^\circ$ and $45^\circ$ respectively with the horizontal. Find the tensions in the strings. (Take $g = 10\text{ m s}^{-2}$ .)
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$\overrightarrow{AB} = \mathbf b - \mathbf a$
$= -2\mathbf i - 3\mathbf j$ and $\overrightarrow{BC} = \mathbf c - \mathbf b$
$= 4\mathbf i - \mathbf j$ .
Dot product: $(-2)(4) + (-3)(-1) = -8 + 3 = -5$ .
Magnitudes: $|\overrightarrow{AB}| = \sqrt{13}$ and $|\overrightarrow{BC}| = \sqrt{17}$ .
$\cos\theta = \dfrac{-5}{\sqrt{13}\sqrt{17}}$
$\approx -0.3363$ , so $\theta \approx 109.7^\circ$ .
The acute angle is $180^\circ - 109.7^\circ = 70.3^\circ$ .

The weight is $5 \times 10 = 50\text{ N}$ .
Let $T_1$ be the tension in the string at $30^\circ$ and $T_2$ the one at $45^\circ$ .
Horizontal components balance: $T_1\cos30^\circ = T_2\cos45^\circ$ , so $T_2 = \dfrac{T_1\cos30^\circ}{\cos45^\circ}$ .
Vertical components balance the weight: $T_1\sin30^\circ + T_2\sin45^\circ = 50$ .
Substitute, using $\cos45^\circ = \sin45^\circ$ : $T_1(\sin30^\circ + \cos30^\circ) = 50$ , so $T_1 = \dfrac{50}{0.5 + 0.8660}$
$\approx 36.6\text{ N}$ .
Then $T_2 = \dfrac{36.60 \times 0.8660}{0.7071}$
$\approx 44.8\text{ N}$ .