Problem Analysis #30

[THOUGHT] We are given triangle ABC with sides AB = 5, AC = 6, and cos(∠BAC) = -3/5. ∠BAC is the angle at vertex A between sides AB and AC. The area can be computed using the formula: Area = (1/2) * AB * AC * sin(∠BAC). We have cos(∠BAC) = -3/5. We can find sin(∠BAC) using sin²θ + cos²θ = 1. Since angle in a triangle is between 0 and π, sin is non-negative. So sin(∠BAC) = sqrt(1 - cos²(∠BAC)) = sqrt(1 - 9/25) = sqrt(16/25) = 4/5. Then area = (1/2) * 5 * 6 * (4/5) = (1/2) * 5 * 6 * 4/5 = (1/2) * 6 * 4 = 12. Let's verify with Python. [/THOUGHT] [PYTHON] import math cos_val = -3/5 sin_val = math.sqrt(1 - cos_val**2) print("sin(∠BAC) =", sin_val) area = 0.5 * 5 * 6 * sin_val print("Area =", area) [/PYTHON]

[THOUGHT] The Python output confirms sin(∠BAC) = 0.8 and area = 12.0. The area is an integer 12. [/THOUGHT] \boxed{12}