Question

The sides of a triangle are $3$ consecutive natural numbers and its largest angle is twice the smallest one. Determine the sides of the triangle.

Answer 1

Let the sides be $n,n+1$ and $n-1$

$\angle C$ is the longest angle while $\angle A$ is the smallest.

$\therefore 2\angle A=\angle C$

By sine rule,

$\therefore \frac{\sin A}{n-1}=\frac{\sin 2A}{n+1}$

$\frac{\sin A}{n-1}=\frac{2\sin A\cos A}{n+1}$

$\frac{1}{2}\left(\frac{n+1}{n-1}\right)=\frac{n^2+{(n+1)}^2-{(n-1)}^2}{2n(n+1)}$

$\frac{1}{2}\left(\frac{n+1}{n-1}\right)=\frac{n(n+4)}{2n(n+1)}$

${(n+1)}^2=(n+4)(n-1)$

$n^2+2n+1=n^2+3n-4$

$n=5$

$\therefore$The sides of the triangle will be $4,5$ and $6$