NECO 2022 · Paper 2 · Q6

(a)
Use logarithm tables to evaluate $\sqrt[3]{\left(\frac{565 \times 0.0536}{249.1}\right)^2}$ .
(b)
An arc $AB$ subtends an angle of $36^\circ$ at the centre $O$ of a circle of radius 7 cm. Calculate the area of the minor sector. $\left(\text{Take } \pi = \frac{22}{7}\right)$

Worked solution (try it first)

Read the logarithms from the tables: $\log 565 = 2.7520$ , $\log 0.0536 = \bar{2}.7292$ and $\log 249.1 = 2.3964$ .
Multiplying means adding logs: $2.7520 + \bar{2}.7292 = 1.4812$ .
Dividing means subtracting logs: $1.4812 - 2.3964 = \bar{1}.0848$ .
Squaring means doubling the log: $2 \times \bar{1}.0848 = \bar{2}.1696$ .
The cube root means dividing by 3.
Write $\bar{2}.1696$ as $\bar{3} + 1.1696$ so the negative part divides exactly: $\bar{1} + 0.3899 = \bar{1}.3899$ .
Take the antilog: $\text{antilog } \bar{1}.3899 \approx 0.2454$ .

The area of a sector is $\frac{\theta}{360} \times \pi r^2$ .
Substitute: $\frac{36}{360} \times \frac{22}{7} \times 7^2 = \frac{1}{10} \times 154$ .
The minor sector has area $15.4\text{ cm}^2$ .