In the $\triangle ABC$, $b:c=2:1$ and $\sin(B-C)=\frac{3}{5}$. Then:
Step-by-Step Solution
Key Concept: Use the Weierstrass substitution formula and the law of sines combined with given side ratios to find angles and side ratios.
Given $\sin(B-C) = \frac{2}{5}$ and $\cos(B-C) = \frac{4}{5}$, let $t = \tan(\frac{B-C}{2})$. Using the Weierstrass substitution, $t^2 = \frac{1}{9}$, so $t = \frac{1}{3}$. Since $b = 2c$, we have $\cot(\frac{A}{2}) = \pm 1$, giving $\frac{A}{2} = \frac{\pi}{4}$ and $A = \frac{\pi}{2}$. Then $a : c = \sqrt{b^2 + c^2} : c = \sqrt{4c^2 + c^2} : c = \sqrt{5} : 1$.
Correct Answer: 1,4