JAMB 1998 · UME · Q29

A chord of a circle of radius $\sqrt3$ cm subtends an angle of $60^\circ$ at the circumference. Find the length of the chord.

Worked solution (try it first)

The angle at the centre is twice the angle at the circumference, so the chord makes $120^\circ$ at the centre.
The perpendicular from the centre halves the chord and the $120^\circ$ .
So half the chord is $\sqrt3\sin60^\circ = \sqrt3 \times \frac{\sqrt3}{2}$
$= \frac32$ .
So the chord is $2 \times \frac32 = 3$ cm, option D.