In $\triangle \mathrm{ABC}$, by sine rule,
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
According to the given condition,
In $\triangle \mathrm{ABC}, \mathrm{a}=2 \mathrm{~b}$ and
$\begin{aligned}
& \mathrm{A}-\mathrm{B}=60^{\circ} \Rightarrow \mathrm{A}=60^{\circ}+\mathrm{B} \\
& \Rightarrow \frac{\sin \left(60^{\circ}+\mathrm{B}\right)}{2 \mathrm{~b}}=\frac{\sin \mathrm{B}}{\mathrm{b}} \\
& \Rightarrow \frac{\sin \mathrm{B}}{\sin \left(\mathrm{B}+60^{\circ}\right)}=\frac{1}{2} \\
& \Rightarrow 2 \sin \mathrm{B}=\sin \mathrm{B} \cos 60^{\circ}+\cos \mathrm{B} \sin 60^{\circ} \\
& \Rightarrow 2 \sin \mathrm{B}=\sin \mathrm{B}\left(\frac{1}{2}\right)+\cos \mathrm{B}\left(\frac{\sqrt{3}}{2}\right) \\
& \Rightarrow \frac{3}{2} \sin \mathrm{B}=\frac{\sqrt{3}}{2} \cos \mathrm{B}
\end{aligned}$
$\begin{aligned}
& \Rightarrow \tan \mathrm{B}=\frac{1}{\sqrt{3}} \Rightarrow \mathrm{B}=30^{\circ} \\
& \therefore \quad \mathrm{A}=30^{\circ}+60^{\circ}=90^{\circ}
\end{aligned}$
$\therefore \quad \triangle \mathrm{ABC}$ is right angled.