JAMB 1999 · UME · Q25

In triangle $XYZ$ , $XY = 2$ m, $YZ = 1$ m and $\angle XYZ = 120^\circ$ . Find the length $XZ$ .

Worked solution (try it first)

Two sides and the angle between them: use the cosine rule, $XZ^2 = 2^2 + 1^2 - 2(2)(1)\cos120^\circ$ .
$\cos120^\circ = -\frac12$ , so the last term becomes $+2$ and $XZ^2 = 5 + 2 = 7$ .
So $XZ = \sqrt7$ m, option A.