JAMB 2001 · UME · Q32

Find the value of $\theta$ in the diagram (an isosceles triangle with sides $t$ , $t$ and base $\sqrt3t$ ).

Worked solution (try it first)

$\theta$ is between the two sides $t$ , and the base $\sqrt3t$ faces it.
Cosine rule: $\cos\theta = \dfrac{t^2 + t^2 - (\sqrt3t)^2}{2 \times t \times t}$ .
$(\sqrt3t)^2 = 3t^2$ , so $\cos\theta = \dfrac{-t^2}{2t^2}$
$= -\frac12$ .
The cosine is negative, so $\theta$ is obtuse: $\theta = 180^\circ - 60^\circ = 120^\circ$ , option D.