The lengths of sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Then the length of the sides of the triangle (in units) are

Let \(a, a+1, a+2\) be the sides of the triangle and \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) be the angles opposite to them respectively.
According to the given condition,
\(\mathrm{C}=2 \mathrm{~A}\)
\(\therefore \sin C =\sin 2 A \)
\(\sin C =2 \sin A \cos A\)
Note that \(\frac{\sin A}{a}=\frac{\sin C}{a+2}=k\)
\(\Rightarrow \sin A=k a \text { and } \sin C=k(a+2) \)
\(\text {Also, } \cos A=\frac{(a+1)^2+(a+2)^2-a^2}{2(a+1)(a+2)} \)
\(=\frac{a^2+2 a+1+a^2+4 a+4-a^2}{2\left(a^2+3 a+2\right)} \)
\(=\frac{a^2+6 a+5}{2\left(a^2+3 a+2\right)} \)
\(\therefore (i) \Rightarrow k(a+2)=2 \times k a \times \frac{a^2+6 a+5}{2\left(a^2+3 a+2\right)} \)
\(\therefore a+2=\frac{a\left(a^2+6 a+5\right)}{\left(a^2+3 a+2\right)} \)
\( \therefore (a+2)\left(a^2+3 a+2\right)=a^3+6 a^2+5 a\)
\(\therefore a^3+5 a^2+8 a+4=a^3+6 a^2+5 a \)
\(\therefore a^2-3 a-4=0 \)
\(\therefore (a-4)(a+1)=0 \)
\(\Rightarrow a=4 \text { or }-1\)
But \(\mathrm{a}=-1\) is not possible.
\(\therefore 4,5,6\) are the lengths of the sides of the triangle.