JAMB 2002 · UME · Q17

In the diagram, $\angle PQR = 50^\circ$ and the exterior angle at $R$ is $128^\circ$ . The triangle $PQR$ is

Worked solution (try it first)

Angles on a straight line: $\angle QRP = 180^\circ - 128^\circ$
$= 52^\circ$ .
The exterior angle equals the sum of the two opposite interior angles: $\angle QPR = 128^\circ - 50^\circ$
$= 78^\circ$ .
The angles are $50^\circ$ , $52^\circ$ and $78^\circ$ : all different and all acute.
So the sides are all different, and the triangle is scalene, option A.