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

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(D) Let the sides be $b = 2a$ and $a$. The angles opposite to these sides are $B$ and $A$ respectively. Given $B - A = 60^{\circ}$.By the sine rule,$\

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right angled

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(D) Let the sides be $b = 2a$ and $a$. The angles opposite to these sides are $B$ and $A$ respectively. Given $B - A = 60^{\circ}$.By the sine rule,$\frac{b}{\sin B} = \frac{a}{\sin A}$ $\Rightarrow \frac{2a}{\sin B} = \frac{a}{\sin A}$ $\Rightarrow \sin B = 2 \sin A$.Substitute $B = A + 60^{\circ}$:$\sin(A + 60^{\circ}) = 2 \sin A$$\sin A \cos 60^{\circ} + \cos A \sin 60^{\circ} = 2 \sin A$$\frac{1}{2} \sin A + \frac{\sqrt{3}}{2} \cos A = 2 \sin A$$\frac{\sqrt{3}}{2} \cos A = \frac{3}{2} \sin A$$	an A = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$$A = 30^{\circ}$.Then $B = 30^{\circ} + 60^{\circ} = 90^{\circ}$.Since one angle is $90^{\circ}$,the triangle is a right-angled triangle.

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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