If one side of a triangle is double the other and the angles opposite to these sides differ by \(60^{\circ}\), then the triangle is

Given, \(A-B=60^{\circ}\)

By sine rule,
\(\frac{2 a}{\sin A}=\frac{a}{\sin B}\)
\(\Rightarrow \sin A-2 \sin B=0\)
\(\Rightarrow \sin \left(60^{\circ}+\mathrm{B}\right)-2 \sin \mathrm{B}=0\)
\(\Rightarrow \frac{\sqrt{3}}{2} \cos \mathrm{B}+\frac{1}{2} \sin \mathrm{B}-2 \sin \mathrm{B}=0 \)
\( \Rightarrow \frac{\sqrt{3}}{2} \cos \mathrm{B}-\frac{3}{2} \sin \mathrm{B}=0 \)
\( \Rightarrow \sqrt{3}\left(\frac{1}{2} \cos \mathrm{B}-\frac{\sqrt{3}}{2} \sin \mathrm{B}\right)=0 \)
\( \Rightarrow \sqrt{3}\left[\cos \left(60^{\circ}+\mathrm{B}\right)\right]=0 \)
\( \Rightarrow 60^{\circ}+\mathrm{B}=90^{\circ} \)
\( \Rightarrow \mathrm{B}=30^{\circ} \)
\( \Rightarrow \mathrm{A}=90^{\circ}\)
Hence, it is right angled triangle.