WAEC 2011 · Paper 2 · Q3

(a)
A sector of a circle with radius $21\text{ cm}$ has an area of $280\text{ cm}^2$ . Calculate, correct to 1 decimal place, the perimeter of the sector. $\left[\text{Take }\pi = \frac{22}{7}\right]$
(b)
If the sector is bent such that its straight edges coincide to form a cone, calculate, correct to the nearest degree, the vertical angle of the cone.

Worked solution (try it first)

Find the angle first: $\frac{\theta}{360} \times \frac{22}{7} \times 21^2 = 280$ , so $\theta = \frac{280 \times 360 \times 7}{22 \times 441}$
$\approx 72.73^\circ$ .
Arc $= \frac{72.73}{360} \times 2 \times \frac{22}{7} \times 21$
$\approx 26.67$ cm.
(Or use area $= \frac12 \times$ arc $\times\ r$ : arc $= \frac{2 \times 280}{21} \approx 26.67$ cm.)
Perimeter $= 26.67 + 21 + 21 \approx 68.7$ cm.

When the sector is bent into a cone, its radius (21 cm) becomes the slant height and its arc becomes the base circumference: $2\pi r = 26.67$ , so $r = \frac{26.67}{2 \times 22/7}$
$\approx 4.24$ cm.
The vertical angle is the angle at the tip of the cone.
Half of it is in a right-angled triangle with hypotenuse 21 (the slant height) and opposite side 4.24 (the radius): $\sin\frac{y}{2} = \frac{4.24}{21}$
$\approx 0.2020$ , so $\frac{y}{2} \approx 11.66^\circ$ and $y \approx 23^\circ$ .