the sides of a triangleare three consecutive natural numbers and its largest angle is twice the smallest one . determine the sides of the triangle ?
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$$\begin{gathered} \mathrm{Let} \mathrm{us} \mathrm{consider} \mathrm{first} \mathrm{and} \mathrm{second} \mathrm{terms} \\ \frac{\mathrm{x}}{\mathrm{sin\theta }}=\frac{\mathrm{x}+2}{\sin \left(2\mathrm{\theta }\right)} \\ \Rightarrow \frac{\mathrm{x}}{\mathrm{sin\theta }}=\frac{\mathrm{x}+2}{2\mathrm{sin\theta cos\theta }} \\ \Rightarrow \mathrm{cos\theta }=\frac{\mathrm{x}+2}{2\mathrm{x}}....\left(\mathrm{i}\right) \\ \mathrm{Let} \mathrm{us} \mathrm{consider} \mathrm{first} \mathrm{and} \mathrm{third} \mathrm{terms} \\ \frac{\mathrm{x}}{\mathrm{sin\theta }}=\frac{\mathrm{x}+1}{\sin \left(3\mathrm{\theta }\right)} \\ \Rightarrow \frac{\mathrm{x}}{\mathrm{sin\theta }}=\frac{\mathrm{x}+1}{3\mathrm{sin\theta }-4{\sin }^3\mathrm{\theta }} \\ \Rightarrow \frac{\mathrm{x}}{\mathrm{sin\theta }}=\frac{\mathrm{x}+1}{\mathrm{sin\theta }\left(3-4{\sin }^2\mathrm{\theta }\right)} \\ \Rightarrow 3-4{\sin }^2\mathrm{\theta }=\frac{\mathrm{x}+1}{\mathrm{x}} \\ \Rightarrow 3-4\left(1-{\cos }^2\mathrm{\theta }\right)=\frac{\mathrm{x}+1}{\mathrm{x}} \\ \Rightarrow 3-4+4{\cos }^2\mathrm{\theta }=\frac{\mathrm{x}+1}{\mathrm{x}} \\ \Rightarrow 4{\cos }^2\mathrm{\theta }=\frac{\mathrm{x}+1}{\mathrm{x}}+1 \\ \Rightarrow 4{\cos }^2\mathrm{\theta }=\frac{\mathrm{x}+1+\mathrm{x}}{\mathrm{x}} \\ \Rightarrow 4{\cos }^2\mathrm{\theta }=\frac{2\mathrm{x}+1}{\mathrm{x}} \\ \mathrm{Using} \left(\mathrm{i}\right) \\ \Rightarrow 4{\left(\frac{\mathrm{x}+2}{2\mathrm{x}}\right)}^2=\frac{2\mathrm{x}+1}{\mathrm{x}} \\ \Rightarrow 4\frac{{\left(\mathrm{x}+2\right)}^2}{4{\mathrm{x}}^2}=\frac{2\mathrm{x}+1}{\mathrm{x}} \\ \Rightarrow 4\frac{{\left(\mathrm{x}+2\right)}^2}{\mathrm{x}}=2\mathrm{x}+1 \\ \Rightarrow \left({\mathrm{x}}^2+4+4\mathrm{x}\right)=2{\mathrm{x}}^2+\mathrm{x} \\ \Rightarrow 2{\mathrm{x}}^2+\mathrm{x}-{\mathrm{x}}^2-4-4\mathrm{x}=0 \\ \Rightarrow {\mathrm{x}}^2-3\mathrm{x}-4=0 \\ \Rightarrow {\mathrm{x}}^2-4\mathrm{x}+\mathrm{x}-4=0 \\ \Rightarrow \mathrm{x}\left(\mathrm{x}-4\right)+1\left(\mathrm{x}-4\right)=0 \\ \Rightarrow \left(\mathrm{x}+1\right)\left(\mathrm{x}-4\right)=0 \\ \mathrm{x}=-1 \mathrm{is} \mathrm{not} \mathrm{possible}. \\ \Rightarrow \mathrm{x}=4 \\ \Rightarrow \mathrm{x}+1=5 \\ \Rightarrow \mathrm{x}+2=6 \\ \mathrm{Hence} \mathrm{sides} \mathrm{of} \mathrm{triangle} \mathrm{are} 4,5,6. \end{gathered}$$

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