Triangle Solving – Problem 2: Determine angle

In $\triangle ABC$, let the sides opposite $A,B,C$ be $a,b,c$, respectively. Suppose $$\frac{c^{2}}{b^{2}+c^{2}-a^{2}}=\frac{\sin C}{\sin B}.$$ (1) Determine angle $A$. Let $D$ be the midpoint of $BC$. (2) If $a=\sqrt{7}$ and the area of $\triangle ABC$ is $\frac{3\sqrt{3}}{4}$, find $AD$.

Step 1. By the Law of Sines, $\frac{\sin C}{\sin B}=\frac{c}{b}$.

Step 2. The given relation becomes $\frac{c^{2}}{b^{2}+c^{2}-a^{2}}=\frac{c}{b}$, so $b^{2}+c^{2}-a^{2}=bc$.

Step 3. By the Law of Cosines, $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{1}{2}$, hence $A=\frac{\pi}{3}$.

Step 4. The area is $S_{\triangle ABC}=\frac12 bc\sin A=\frac{\sqrt{3}}{4}bc$. Since $S_{\triangle ABC}=\frac{3\sqrt{3}}{4}$, we get $bc=3$.

Step 5. Also $a^{2}=b^{2}+c^{2}-2bc\cos A=b^{2}+c^{2}-bc$. Substituting $a^{2}=7$ and $bc=3$ gives $b^{2}+c^{2}=10$.

Step 6. Let $D$ be the midpoint of $BC$. The median formula yields $AD^{2}=\frac{2b^{2}+2c^{2}-a^{2}}{4}=\frac{2\cdot10-7}{4}=\frac{13}{4}$.

Step 7. Therefore $AD=\frac{\sqrt{13}}{2}$.

(1) $A=\frac{\pi}{3}$. (2) $AD=\frac{\sqrt{13}}{2}$.