The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one, then the sides of the triangle (in units) are

Let \(\mathrm{AC}=\mathrm{n}, \mathrm{AB}=\mathrm{n}+1, \mathrm{BC}=\mathrm{n}+2\)
\(\therefore\) Largest angle is \(\mathrm{A}\) and smallest angle is \(\mathrm{B}\).
\(\therefore \mathrm{A}=2 \mathrm{~B}\)
Since \(\mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ}\)
\(\therefore 3 \mathrm{~B}+\mathrm{C}=180^{\circ}\)
\(\Rightarrow \mathrm{C}=180^{\circ}-3 \mathrm{~B} \)
\(\Rightarrow \sin \mathrm{C}=\sin \left(180^{\circ}-3 \mathrm{~B}\right)=\sin 3 \mathrm{B}\)
By sine rule,
\(\frac{\sin A}{n+2}=\frac{\sin B}{n}=\frac{\sin C}{n+1} \)
\(\Rightarrow \frac{\sin 2 B}{n+2}=\frac{\sin B}{n}=\frac{\sin 3 B}{n+1} \)
\(\Rightarrow \frac{2 \sin B \cos B}{n+2}=\frac{\sin B}{n}=\frac{3 \sin B-4 \sin ^3 B}{n+1} \)
\(\Rightarrow \frac{2 \cos B}{n+2}=\frac{1}{n}=\frac{3-4 \sin ^2 B}{n+1} \)
\(\therefore \cos B=\frac{n+2}{2 n}, 3-4 \sin ^2 B=\frac{n+1}{n} \)
\(\therefore 3-4\left(1-\cos ^2 B\right)=\frac{n+1}{n} \)
\(\therefore -4+4\left(\frac{n+2}{2 n}\right)^2=\frac{n+1}{n} \)
\(\Rightarrow-1+\frac{n^2+4 n+4}{n^2}=\frac{n+1}{n} \)
\(\Rightarrow-n^2+n^2+4 n+4=n^2+n \)
\(\Rightarrow n^2-3 n-4=0 \)
\(\Rightarrow(n+1)(n-4)=0 \)
\(\Rightarrow n=-1 \text { or } n=4\)
But \(\mathrm{n}\) cannot be negative.
\(\therefore \mathrm{n}=4\)
\(\therefore \) The sides of the \(\Delta\) are \(4,5,6\).