Triangle Solving – Problem 9: Determine angle

In $\triangle ABC$, let the sides opposite $A,B,C$ be $a,b,c$, respectively, and suppose $$a\sin A=\left(b-\frac{c}{2}\right)\sin B+\left(c-\frac{b}{2}\right)\sin C.$$ (1) Determine angle $A$. (2) If the area of $\triangle ABC$ is $\sqrt{3}$ and the perimeter is $8$, find $a$.

Step 1. Using the Law of Sines $\sin A=\frac{a}{2R}$, $\sin B=\frac{b}{2R}$, $\sin C=\frac{c}{2R}$, multiply the given equation by $2R$ to obtain $$a^{2}=\left(b-\frac{c}{2}\right)b+\left(c-\frac{b}{2}\right)c=b^{2}+c^{2}-bc.$$

Step 2. Hence $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{1}{2}$, so $A=\frac{\pi}{3}$.

Step 3. For (2), since the area is $\sqrt{3}$, $$\frac12 bc\sin A=\frac12 bc\sin\frac{\pi}{3}=\frac{\sqrt{3}}{4}bc=\sqrt{3},$$ so $bc=4$.

Step 4. From $a^{2}=b^{2}+c^{2}-bc$, we get $a^{2}=(b+c)^{2}-3bc=(b+c)^{2}-12$.

Step 5. Since $a+b+c=8$, we have $b+c=8-a$, so $$a^{2}=(8-a)^{2}-12.$$ Solving gives $a=\frac{13}{4}$.

(1) $A=\frac{\pi}{3}$. (2) $a=\frac{13}{4}$.