WAEC 2011 · Paper 2 · Q8

(a)
Given that $\sin x = 0.6$ and $0^\circ \le x \le 90^\circ$ , evaluate $2\cos x + 3\sin x$ , leaving your answer in the form $\frac mn$ , where $m$ and $n$ are integers.
(b)
In the diagram, a semi-circle $WXYZ$ with centre $O$ is inscribed in an isosceles triangle $ABC$ . If $|AC| = |BC|$ , $|OC| = 30\text{ cm}$ and $A\hat CB = 130^\circ$ , calculate, correct to one decimal place, the: (i) radius of the circle; (ii) area of the shaded portion. $\left[\text{Take }\pi = \frac{22}{7}\right]$
Not to scale.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$OC$ is the line of symmetry of the isosceles triangle, so it cuts $\angle ACB$ in half: $\angle OCA = 65^\circ$ .
Let the semicircle touch $AC$ at $X$ .
A tangent is perpendicular to the radius, so $\angle OXC = 90^\circ$ and triangle $OXC$ is right-angled with hypotenuse $OC = 30$ .
The radius $OX$ is opposite the $65^\circ$ angle: $r = 30\sin 65^\circ$
$\approx 27.19$
$\approx 27.2\text{ cm}$ .

The shaded part is the triangle minus the semicircle.
$OC$ is perpendicular to $AB$ , so in triangle $OAC$ , $\angle AOC = 90^\circ$ and $|OA| = 30\tan 65^\circ \approx 64.34$ cm.
Area of triangle $ABC = \frac12 \times |AB| \times |OC|$
$= |OA| \times 30$
$\approx 1930.1\text{ cm}^2$ .
Area of the semicircle $= \frac12\pi r^2$
$= \frac12 \times \frac{22}{7} \times 27.19^2$
$\approx 1161.7\text{ cm}^2$ .
Shaded area $\approx 1930.1 - 1161.7$
$= 768.4\text{ cm}^2$ .