Triangle Solving Solution – Problem 28: compute

Question

In $\triangle ABC$ , if $AC^{2}+BC^{2}=5AB^{2}$ , compute $\frac{\tan C}{\tan A}+\frac{\tan C}{\tan B}.$

Step-by-step solution

Step 1. $\frac{\tan C}{\tan A}+\frac{\tan C}{\tan B}=\tan C\left(\cot A+\cot B\right)=\frac{\sin C}{\cos C}\cdot\frac{\sin A\cos B+\cos A\sin B}{\sin A\sin B}=\frac{\sin^{2}C}{\sin A\sin B\cos C}.$

Step 2. By the Law of Cosines at angle $C$ , $\cos C=\frac{AC^{2}+BC^{2}-AB^{2}}{2\,AC\cdot BC}=\frac{4AB^{2}}{2\,AC\cdot BC}.$

Step 3. Using the Law of Sines $AB=2R\sin C$ , $AC=2R\sin B$ , $BC=2R\sin A$ , we get $\cos C=\frac{2\sin^{2}C}{\sin A\sin B}\Rightarrow \sin A\sin B\cos C=2\sin^{2}C.$

Step 4. Therefore $\frac{\tan C}{\tan A}+\frac{\tan C}{\tan B}=\frac{1}{2}$ .

Final answer

$\frac{1}{2}$

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 28: compute