JAMB 1984 · UME · Q29

The sides of a triangle are $(x + 4)$ cm, $x$ cm and $(x - 4)$ cm. If the cosine of the largest angle is $\frac15$ , find the value of $x$ .

Worked solution (try it first)

The largest angle faces the longest side, $x + 4$ .
The cosine rule gives $(x + 4)^2 = x^2 + (x - 4)^2 - 2x(x - 4) \times \frac15$ .
Multiply through by 5 and expand: $5x^2 + 40x + 80 = 10x^2 - 40x + 80 - 2x^2 + 8x$ .
Collect terms: $3x^2 - 72x = 0$ , so $3x(x - 24) = 0$ and $x = 0$ or $x = 24$ .
A side of $x - 4$ must be positive, so $x = 24$ cm, option A.