Triangle Solving – Problem 20: In , let the sides opposite be

In $\triangle ABC$, let the sides opposite $A,B,C$ be $a,b,c$. Suppose $B=2C$ and $b=\sqrt{2}\,a$. Which statement is correct?

A. $\triangle ABC$ is right-angled B. $\triangle ABC$ is acute C. $\triangle ABC$ is obtuse D. The type of $\triangle ABC$ cannot be determined

Step 1. By the Law of Sines, $\frac{b}{a}=\frac{\sin B}{\sin A}$. Since $b=\sqrt{2}\,a$, we get $\sin B=\sqrt{2}\sin A$.

Step 2. Because $B=2C$ and $A=\pi-B-C=\pi-3C$, we have $\sin B=\sin 2C$ and $\sin A=\sin 3C$. Thus $\sin 2C=\sqrt{2}\sin 3C$.

Step 3. Using $\sin 2C=2\sin C\cos C$ and $\sin 3C=3\sin C-4\sin^{3}C$, and dividing by $\sin C>0$, we obtain $$2\cos C=\sqrt{2}(3-4\sin^{2}C).$$

Step 4. Let $x=\cos C$. Since $\sin^{2}C=1-x^{2}$, this becomes $2x=\sqrt{2}(4x^{2}-1)$, i.e. $$4\sqrt{2}x^{2}-2x-\sqrt{2}=0.$$

Step 5. Solving gives $x=\frac{\sqrt{2}}{2}$ (discard the negative root), so $C=\frac{\pi}{4}$ and $B=\frac{\pi}{2}$.

Step 6. Therefore $\triangle ABC$ is right-angled, and the correct choice is A.

A