Triangle Solving – Problem 19: Find the area of

In $\triangle ABC$, the circumradius is $R$, and $$\frac{a^{2}-c^{2}}{2R}=(a-b)\sin B,\qquad \sin B=2\sin A,\qquad c=2.$$ Find the area of $\triangle ABC$.

A. $\frac{4}{3}$ B. $\frac{2\sqrt{3}}{3}$ C. $\frac{4\sqrt{3}}{3}$ D. $\frac{2}{3}$

Step 1. From $\frac{a^{2}-c^{2}}{2R}=(a-b)\sin B$ and $\frac{b}{\sin B}=2R$, we get $a^{2}-c^{2}=ab-b^{2}$.

Step 2. By the Law of Cosines, $a^{2}+b^{2}-c^{2}=2ab\cos C$. But $a^{2}+b^{2}-c^{2}=ab$, so $\cos C=\frac12$ and $C=\frac{\pi}{3}$.

Step 3. Since $\sin B=2\sin A$, the Law of Sines gives $b=2a$. Also $c=2$.

Step 4. Using the Law of Cosines with $C=\frac{\pi}{3}$: $$c^{2}=a^{2}+b^{2}-2ab\cos C=a^{2}+4a^{2}-2\cdot a\cdot 2a\cdot\frac12=3a^{2}.$$ So $4=3a^{2}\Rightarrow a^{2}=\frac{4}{3}$.

Step 5. The area is $S_{\triangle ABC}=\frac12 ab\sin C=\frac12\cdot a\cdot 2a\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}a^{2}=\frac{2\sqrt{3}}{3}$. Hence the correct choice is B.

B