Triangle Solving Solution – Problem 26: Find

Question

In $\triangle ABC$ , $3\sin A=2\sin C$ and $\cos B=\frac{1}{3}$ . Find $\sin A$ .

Step-by-step solution

Step 1. By the Law of Sines, $3\sin A=2\sin C\Rightarrow 3a=2c$ .

Step 2. By the Law of Cosines, $b^{2}=a^{2}+c^{2}-2ac\cos B$ . Using $c=\frac{3}{2}a$ and $\cos B=\frac13$ , we get $b^{2}=a^{2}+\left(\frac{3a}{2}\right)^{2}-2\cdot a\cdot \frac{3a}{2}\cdot\frac{1}{3}=a^{2}+\frac{9a^{2}}{4}-a^{2}=\frac{9a^{2}}{4}.$ So $b=\frac{3}{2}a$ , hence $b=c$ and therefore $B=C$ .

Step 3. Since $\cos B=\frac{1}{3}$ , $\sin B=\sqrt{1-\cos^{2}B}=\frac{2\sqrt{2}}{3}$ .

Step 4. From $3a=2c$ and $b=c$ , we have $3a=2b$ , so $\frac{a}{b}=\frac{2}{3}$ . By the Law of Sines, $\frac{\sin A}{\sin B}=\frac{a}{b}=\frac{2}{3}$ , hence $\sin A=\frac{2}{3}\sin B=\frac{2}{3}\cdot\frac{2\sqrt{2}}{3}=\frac{4\sqrt{2}}{9}.$

Final answer

$\frac{4\sqrt{2}}{9}$

Marking scheme

1. Checkpoints (max 7 pts total)

Chain A: Law of Sines approach

Set up side-angle relations [2 pts]: States and correctly advances the key derivation steps
Substitute and simplify [2 pts]: Substitutes correctly and simplifies accurately
Handle multiple cases / admissibility [1 pt]: Considers branches and rejects invalid cases
Conclusion and verification [1 pt]: States the conclusion and checks against constraints
Final answer [1 pt]: Gives the correct final result (for multiple-choice, include the option letter)

2. Zero-credit items

Copies formulas without concrete substitution or derivation
Guesses the answer / provides only a conclusion with no reasoning
Uses an approach incompatible with the problem conditions, leading to an invalid conclusion

3. Deductions

Computation error [-1]: Incorrect algebraic/trigonometric manipulation
Logical gap [-1]: Missing a key equivalence step or a necessary condition check
Nonstandard final statement [-1]: Missing units/range/option letter or wrong answer format

Triangle Solving – Problem 26: Find