Question on: WAEC Mathematics - 2012

Let 0 = angle of the minor sector

angle of the major sector = 360 - \(\theta\)(angle at a point)

2 \(\pi r\) = 54 + 126(i.e circumference of minor and major arc)

2\(\pi r = 180^o\)

r = \(\frac{180}{2\pi}\) = \(\frac{90}{\pi}\)

Lenght of ninor arc

= \(\frac{\theta}{360} \times 2 \pi r\)

54 = \(\frac{\theta}{360} \times 3 \pi r\)

\(\theta = \frac{360 \times 54}{2 \pi r}\)

but r = \(\frac{90}{\pi}\) substituting \(\frac{90}{\pi}\) for r

\(\theta = \frac{360 \times 54 \times \pi}{2 \times \pi \times 90}\)

\(\theta = 2 \times 54 = 108^o\)

angle of the major sector = 360 - 108^o

= 252^o