In triangle ABC,if a = 2b and |A - B| = frac{ | vedclass

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(B) Using the Law of Tangents: $\frac{a-b}{a+b} = \frac{	an(\frac{A-B}{2})}{	an(\frac{A+B}{2})}$.Given $a = 2b$,we have $\frac{2b-b}{2b+b} = \frac{b

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$\frac{\pi}{3}$

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(B) Using the Law of Tangents: $\frac{a-b}{a+b} = \frac{	an(\frac{A-B}{2})}{	an(\frac{A+B}{2})}$.Given $a = 2b$,we have $\frac{2b-b}{2b+b} = \frac{b}{3b} = \frac{1}{3}$.Also,$|A-B| = \frac{\pi}{3}$,so $\frac{A-B}{2} = \frac{\pi}{6}$.Thus,$\frac{1}{3} = \frac{	an(\pi/6)}{	an((A+B)/2)} = \frac{1/\sqrt{3}}{	an((A+B)/2)}$.This implies $	an(\frac{A+B}{2}) = \sqrt{3}$.Therefore,$\frac{A+B}{2} = \frac{\pi}{3}$,which means $A+B = \frac{2\pi}{3}$.Since $A+B+C = \pi$,we have $C = \pi - (A+B) = \pi - \frac{2\pi}{3} = \frac{\pi}{3}$.

In $\triangle ABC$,if $a = 2b$ and $|A - B| = \frac{\pi}{3}$,then $\angle C = $

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