Problem Analysis #30

[THOUGHT] The area of triangle ABC is given by (1/2) * AB * AC * sin(angle BAC). Given cos(angle BAC) = -3/5, compute sin(angle BAC) = sqrt(1 - cos^2(angle BAC)) = sqrt(1 - (9/25)) = sqrt(16/25) = 4/5. Then area = (1/2)56*(4/5) = 12. Verify exactly with sympy. [/THOUGHT] [PYTHON] import sympy as sp

cos_A = sp.Rational(-3, 5) sin_A = sp.sqrt(1 - cos_A**2) area = sp.Rational(1, 2) * 5 * 6 * sin_A print(area) [/PYTHON]

[THOUGHT] The Python output confirms the area is 12. [/THOUGHT] \boxed{12}

Explanation

The reasoning follows the standard formula for the area of a triangle with two sides and the included angle: area = (1/2) × AB × AC × sin(∠BAC). Given AB = 5, AC = 6, and cos(∠BAC) = -3/5, sin(∠BAC) is derived from the Pythagorean identity: sin(∠BAC) = √(1 - cos²(∠BAC)) = √(1 - (-3/5)²) = √(1 - 9/25) = √(16/25) = 4/5. Substituting these values yields: area = (1/2) × 5 × 6 × (4/5) = (1/2) × 30 × (4/5) = 15 × (4/5) = 12. The Python code using SymPy verified this exact computation, outputting 12, confirming the result as ground truth.