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(B) Let the sides of the triangle be $a, a+1, a+2$ where $a \in \mathbb{N}$. Let the angles opposite to these sides be $A, B, C$ respectively,such tha

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$4, 5, 6$

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(B) Let the sides of the triangle be $a, a+1, a+2$ where $a \in \mathbb{N}$. Let the angles opposite to these sides be $A, B, C$ respectively,such that $A Using the Sine Rule: $\frac{\sin A}{a} = \frac{\sin C}{a+2} = k$.Thus,$\sin C = k(a+2)$ and $\sin A = ka$.Since $C = 2A$,$\sin C = 2 \sin A \cos A$,which implies $k(a+2) = 2(ka) \cos A$,so $\cos A = \frac{a+2}{2a}$.Using the Cosine Rule: $\cos A = \frac{(a+1)^2 + (a+2)^2 - a^2}{2(a+1)(a+2)} = \frac{a^2+6a+5}{2(a^2+3a+2)}$.Equating the two expressions for $\cos A$: $\frac{a+2}{2a} = \frac{(a+1)(a+5)}{2(a+1)(a+2)} = \frac{a+5}{2(a+2)}$.$(a+2)^2 = a(a+5) \Rightarrow a^2+4a+4 = a^2+5a$.$a = 4$.The sides are $4, 5, 6$.

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

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