The lengths of the minor and major arcs 54cm an... - WAEC Mathematics 2012 Question

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