JAMB 1985 · UME · Q26

In $\triangle XYZ$ , $XY = 13$ cm, $YZ = 9$ cm, $XZ = 11$ cm and $\angle XYZ = \theta^\circ$ . Find $\cos\theta$ .

Worked solution (try it first)

$\theta$ is at $Y$ , between $XY = 13$ and $YZ = 9$ .
The side facing it is $XZ = 11$ .
Cosine rule: $\cos\theta = \dfrac{13^2 + 9^2 - 11^2}{2 \times 13 \times 9}$ , which is $\dfrac{169 + 81 - 121}{234} = \dfrac{129}{234}$ .
Divide top and bottom by 3: $\cos\theta = \frac{43}{78}$ , option E.