JAMB 1988 · UME · Q42

In triangle $PQR$ , $PQ = 1$ cm, $QR = 2$ cm and $\angle PQR = 120^\circ$ . Find the longest side of the triangle.

Worked solution (try it first)

The longest side faces the largest angle, the $120^\circ$ at $Q$ .
So the longest side is $PR$ .
Cosine rule: $PR^2 = 1^2 + 2^2 - 2(1)(2)\cos120^\circ$ .
$\cos120^\circ = -\frac12$ , so $PR^2 = 5 + 2 = 7$ and $PR = \sqrt7$ cm, option D.