WAEC 2019 · Paper 2 · Q10

(a)
The lengths of the sides of a triangle, in centimetres, are $(3 + \sqrt2)$ , $2\sqrt3$ and $3\sqrt2$ . Find the value of the largest angle of the triangle.
(b)
Express $\dfrac{2x - 1}{3x^2 + 4x + 1}$ in partial fractions.

Worked solution (try it first)

Compare the sides: $3 + \sqrt2 \approx 4.41$ , $2\sqrt3 \approx 3.46$ and $3\sqrt2 \approx 4.24$ .
The largest angle is opposite the longest side, $3 + \sqrt2$ .
By the cosine rule, $\cos\theta = \dfrac{(2\sqrt3)^2 + (3\sqrt2)^2 - (3 + \sqrt2)^2}{2(2\sqrt3)(3\sqrt2)}$ .
Work out each square: $(2\sqrt3)^2 = 12$ , $(3\sqrt2)^2 = 18$ and $(3 + \sqrt2)^2 = 11 + 6\sqrt2$ .
The top is $19 - 6\sqrt2 \approx 10.515$ .
The bottom is $12\sqrt6 \approx 29.394$ , so $\cos\theta \approx 0.3577$ .
So $\theta \approx 69.04^\circ$ .

Factorise the bottom: $3x^2 + 4x + 1 = (3x + 1)(x + 1)$ .
Write $\dfrac{2x - 1}{(3x + 1)(x + 1)} = \dfrac{P}{3x + 1} + \dfrac{Q}{x + 1}$ .
Multiply through: $2x - 1 = P(x + 1) + Q(3x + 1)$ .
Put $x = -1$ : $-3 = -2Q$ , so $Q = \frac32$ .
Put $x = -\frac13$ : $-\frac53 = \frac23P$ , so $P = -\frac52$ .
So $\dfrac{2x - 1}{3x^2 + 4x + 1} = -\dfrac{5}{2(3x + 1)} + \dfrac{3}{2(x + 1)}$ .