JAMB 1998 · UME · Q35

In the diagram, $QTR$ is a straight line, $PQ = 15$ , $PR = 10$ , $PT = 8$ and $\angle PQT = 30^\circ$ . Find the sine of $\angle PTR$ .

Worked solution (try it first)

Work in triangle $PQT$ .
$PQ = 15$ faces $\angle PTQ$ , and $PT = 8$ faces the $30^\circ$ at $Q$ .
Sine rule: $\dfrac{\sin\angle PTQ}{15} = \dfrac{\sin30^\circ}{8}$ .
So $\sin\angle PTQ = \dfrac{15 \times \frac12}{8}$
$= \frac{15}{16}$ .
$QTR$ is a straight line, so $\angle PTR = 180^\circ - \angle PTQ$ , and $\sin(180^\circ - A) = \sin A$ .
So $\sin\angle PTR = \frac{15}{16}$ , option D.