Triangle Solving – Problem 1: In an acute , suppose that and

In an acute $\triangle ABC$, suppose that $\sqrt{3}\sin A\left(\frac{\cos A}{a}+\frac{\cos C}{c}\right)=\sin B\sin C$ and $\sqrt{3}\sin C+\cos C=2$. Which of the following values can $a+b$ take?

A. 5 B. 4 C. $2\sqrt{3}$ D. 3

Step 1. Since $\sqrt{3}\sin C+\cos C=2\sin\left(C+\frac{\pi}{6}\right)=2$, we have $\sin\left(C+\frac{\pi}{6}\right)=1$.

Step 2. Because $0<C<\frac{\pi}{2}$, it follows that $C+\frac{\pi}{6}\in\left(\frac{\pi}{6},\frac{2\pi}{3}\right)$, hence $C+\frac{\pi}{6}=\frac{\pi}{2}$ and $C=\frac{\pi}{3}$.

Step 3. Let the circumradius be $R$. By the Law of Sines, $a=2R\sin A,\ b=2R\sin B,\ c=2R\sin C$.

Step 4. Substitute into $\sqrt{3}\sin A\left(\frac{\cos A}{a}+\frac{\cos C}{c}\right)=\sin B\sin C$: $$\sqrt{3}\sin A\Bigl(\frac{\cos A}{2R\sin A}+\frac{\cos C}{2R\sin C}\Bigr)=\sin B\sin C.$$

Step 5. Simplifying gives $\sqrt{3}(\cos A\sin C+\sin A\cos C)=2R\sin B\sin^{2}C$.

Step 6. Note $\cos A\sin C+\sin A\cos C=\sin(A+C)=\sin B$. With $C=\frac{\pi}{3}$ we have $\sin^{2}C=\frac{3}{4}$, so $\sqrt{3}\sin B=2R\sin B\cdot\frac{3}{4}$, hence $2R=\frac{4\sqrt{3}}{3}$.

Step 7. Therefore $a+b=2R(\sin A+\sin B)=\frac{4\sqrt{3}}{3}\left[\sin A+\sin\left(\frac{2\pi}{3}-A\right)\right]$.

Step 8. Using sum-to-product: $$\sin A+\sin\left(\frac{2\pi}{3}-A\right)=2\sin\frac{\pi}{3}\cos\left(A-\frac{\pi}{3}\right)=\sqrt{3}\sin\left(A+\frac{\pi}{6}\right),$$ so $a+b=4\sin\left(A+\frac{\pi}{6}\right)$.

Step 9. Since $\triangle ABC$ is acute and $C=\frac{\pi}{3}$, we have $A+B=\frac{2\pi}{3}$ and $0<B<\frac{\pi}{2}$, hence $\frac{\pi}{6}<A<\frac{\pi}{2}$.

Step 10. Thus $A+\frac{\pi}{6}\in\left(\frac{\pi}{3},\frac{2\pi}{3}\right)$, so $\sin\left(A+\frac{\pi}{6}\right)\in\left(\frac{\sqrt{3}}{2},1\right]$.

Step 11. Hence $a+b\in(2\sqrt{3},4]$. Among the options, only 4 is attainable, so the correct choice is B.

B