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(D) Let the sides of the triangle be $a = n$,$b = n + 1$,and $c = n + 2$,where $n$ is a natural number.Since $c > b > a$,the angle $C$ is the largest

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$4, 5, 6$

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(D) Let the sides of the triangle be $a = n$,$b = n + 1$,and $c = n + 2$,where $n$ is a natural number.Since $c > b > a$,the angle $C$ is the largest and $A$ is the smallest. Given $C = 2A$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{c}{\sin C} = 2R$,so $\sin A = \frac{a}{2R}$ and $\sin C = \frac{c}{2R}$.Since $C = 2A$,$\sin C = \sin 2A = 2 \sin A \cos A$.Substituting the values,$\frac{c}{2R} = 2 \left( \frac{a}{2R} ight) \cos A$,which simplifies to $\cos A = \frac{c}{2a}$.Using the Cosine Rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Equating the two expressions for $\cos A$: $\frac{c}{2a} = \frac{b^2 + c^2 - a^2}{2bc} \implies c^2 b = a(b^2 + c^2 - a^2)$.Substituting $a = n, b = n + 1, c = n + 2$:$(n + 1)(n + 2)^2 = n((n + 1)^2 + (n + 2)^2 - n^2)$.$(n + 1)(n^2 + 4n + 4) = n(n^2 + 2n + 1 + n^2 + 4n + 4 - n^2)$.$(n + 1)(n^2 + 4n + 4) = n(n^2 + 6n + 5) = n(n + 1)(n + 5)$.Since $n$ is a natural number,$n+1 
eq 0$,so $(n + 2)^2 = n(n + 5)$.$n^2 + 4n + 4 = n^2 + 5n \implies n = 4$.The sides are $4, 5, 6$.

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 are

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