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WAEC Mathematics 2012 Paper

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Question : 15

Total: 49

The lengths of the minor and major arcs 54cm and 126cm respectively. Calculate the angle of the major sector

360^o
252^o
246^o
234^o

Solution:

Let 0 = angle of the minor sector

angle of the major sector = 360 -

heta

(angle at a point)

2

\pi r

= 54 + 126(i.e circumference of minor and major arc)

2

\pi r = 180^{\circ}

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^{\circ}

angle of the major sector = 360 - 108^o

= 252^o

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