JAMB 1983 · UME · Q46

In a triangle $PQR$ , $QR = 3$ cm, $PR = 3$ cm, $PQ = 3\sqrt3$ cm and $\angle PQR = 30^\circ$ . Find the angles $P$ and $R$ .

Worked solution (try it first)

$PR = QR = 3$ cm, so the triangle is isosceles.
Equal sides face equal angles: $QR$ faces $\angle P$ and $PR$ faces $\angle Q$ , so $\angle P = \angle Q = 30^\circ$ .
The angles of a triangle add up to $180^\circ$ , so $\angle R = 180^\circ - 30^\circ - 30^\circ$
$= 120^\circ$ .
Check with the cosine rule: $PQ^2 = 3^2 + 3^2 - 2(3)(3)\cos120^\circ$ , which is $18 + 9 = 27$ , and $\sqrt{27} = 3\sqrt3$ ✓.
So $P = 30^\circ$ and $R = 120^\circ$ , option B.