JAMB 1987 · UME · Q38✱✱

The sine, cosine and tangent of $210^\circ$ are respectively

Worked solution (try it first)

$210^\circ = 180^\circ + 30^\circ$ , so it lies in the third quadrant, with reference angle $30^\circ$ .
In the third quadrant sine and cosine are negative and tangent is positive.
So $\sin210^\circ = -\sin30^\circ = -\frac12$ and $\cos210^\circ = -\cos30^\circ$
$= -\frac{\sqrt3}{2}$ .
And $\tan210^\circ = \tan30^\circ$
$= \frac{1}{\sqrt3}$ , which is $\frac{\sqrt3}{3}$ .
That is option A.