WAEC Mathematics 2016 Paper

Concept:The perpendicular from the centre of a circle to a chord bisects the chord. This helps us use right-angled trigonometry to find half the chord length.Explanation:Given radius $OM = ON = 10\text{ cm}$ and central angle $\angle MON = 140^\circ$.Let $X$ be the midpoint of chord $MN$. Then $OX$ is perpendicular to $MN$ and bisects $\angle MON$.So, $\angle MOX = \frac{140^\circ}{2} = 70^\circ$.In right triangle $MOX$, we have:$\sin 70^\circ = \frac{MX}{OM}$.Let $MX = x$. Then $\sin 70^\circ = \frac{x}{10}$.So, $x = 10\sin 70^\circ = 10(0.93969) = 9.3969\text{ cm}$.Since $X$ is the midpoint, chord length $MN = 2x = 2(9.3969) = 18.7938\text{ cm}$.Correct to the nearest cm, $MN = 19\text{ cm}$.Answer:$19\text{ cm}$ (Option A).