The sides of a triangle are three consecutive | vedclass

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(B) Let the sides be $n, n+1, n+2$. The smallest side is $n$ and the largest side is $n+2$. Let the angles opposite to these sides be $B$ and $A$ resp

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$4, 5, 6$

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(B) Let the sides be $n, n+1, n+2$. The smallest side is $n$ and the largest side is $n+2$. Let the angles opposite to these sides be $B$ and $A$ respectively. Thus,$A = 2B$.By the sine rule,$\frac{\sin A}{n+2} = \frac{\sin B}{n} = \frac{\sin C}{n+1}$.Since $A = 2B$,we have $\frac{\sin 2B}{n+2} = \frac{\sin B}{n}$ $\Rightarrow \frac{2 \sin B \cos B}{n+2} = \frac{\sin B}{n}$ $\Rightarrow \cos B = \frac{n+2}{2n}$.Using the cosine rule,$\cos B = \frac{(n+1)^2 + n^2 - (n+2)^2}{2n(n+1)} = \frac{n^2+2n+1+n^2-n^2-4n-4}{2n(n+1)} = \frac{n^2-2n-3}{2n(n+1)} = \frac{(n-3)(n+1)}{2n(n+1)} = \frac{n-3}{2n}$.Equating the two expressions for $\cos B$: $\frac{n+2}{2n} = \frac{n-3}{2n} \Rightarrow n+2 = n-3$,which is impossible. Re-evaluating: Let sides be $a=n, b=n+1, c=n+2$. Smallest angle is $A$ (opposite to $n$),largest is $C$ (opposite to $n+2$). Given $C = 2A$.By sine rule,$\frac{\sin A}{n} = \frac{\sin 2A}{n+2}$ $\Rightarrow \frac{\sin A}{n} = \frac{2 \sin A \cos A}{n+2}$ $\Rightarrow \cos A = \frac{n+2}{2n}$.By cosine rule,$\cos A = \frac{(n+1)^2 + (n+2)^2 - n^2}{2(n+1)(n+2)} = \frac{n^2+2n+1+n^2+4n+4-n^2}{2(n+1)(n+2)} = \frac{n^2+6n+5}{2(n+1)(n+2)} = \frac{(n+1)(n+5)}{2(n+1)(n+2)} = \frac{n+5}{2(n+2)}$.Equating: $\frac{n+2}{2n} = \frac{n+5}{2(n+2)}$ $\Rightarrow (n+2)^2 = n(n+5)$ $\Rightarrow n^2+4n+4 = n^2+5n$ $\Rightarrow n=4$.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 (in units) are

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