WAEC 2020 · Paper 2 · Q7

(a)
In the diagram, $\angle CAD = 35^\circ$ , $\angle ACB = 120^\circ$ , $|AC| = 12.7\text{ cm}$ and $CD$ is perpendicular to $AB$ . Calculate, correct to three significant figures, $|AB|$ .
(b)
Evaluate $\displaystyle\int_0^a (x^2 - 4)\,dx$ .
(c)
If $y = x^3 - 2x$ , find $\dfrac{dy}{dx}$ .

Worked solution (try it first)

In right-angled triangle $ADC$ (the right angle at $D$ ), $AC = 12.7$ cm is the hypotenuse: $|AD| = 12.7\cos 35^\circ \approx 10.403$ cm and $|CD| = 12.7\sin 35^\circ \approx 7.284$ cm.
The third angle of that triangle is $\angle ACD = 90^\circ - 35^\circ$
$= 55^\circ$ , so $\angle BCD = 120^\circ - 55^\circ$
$= 65^\circ$ .
In right-angled triangle $CDB$ : $|DB| = |CD|\tan 65^\circ$
$\approx 7.284 \times 2.1445$
$\approx 15.621$ cm.
So $|AB| = |AD| + |DB|$
$\approx 10.403 + 15.621$
$= 26.0$ cm.

Integrate term by term: $\int (x^2 - 4)\,dx = \frac{x^3}{3} - 4x$ .
Between the limits: $\left[\frac{x^3}{3} - 4x\right]_0^a = \frac{a^3}{3} - 4a - 0$
$= \frac{a^3}{3} - 4a$ .