If one side of a triangle is double the other | vedclass

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(B) In $	riangle ABC$,by the sine rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$.Given $a = 2b$ and $A - B = 60^{\circ}$,so $A = 60^{\

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right angled

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(B) In $	riangle ABC$,by the sine rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$.Given $a = 2b$ and $A - B = 60^{\circ}$,so $A = 60^{\circ} + B$.Substituting these into the sine rule:$\frac{\sin(60^{\circ} + B)}{2b} = \frac{\sin B}{b}$$\sin(60^{\circ} + B) = 2 \sin B$$\sin 60^{\circ} \cos B + \cos 60^{\circ} \sin B = 2 \sin B$$\frac{\sqrt{3}}{2} \cos B + \frac{1}{2} \sin B = 2 \sin B$$\frac{\sqrt{3}}{2} \cos B = \frac{3}{2} \sin B$$	an B = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.Thus,$B = 30^{\circ}$.Then $A = 60^{\circ} + 30^{\circ} = 90^{\circ}$.Since one angle is $90^{\circ}$,the triangle is right angled.

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

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