Problem Analysis #30

[THOUGHT] We are given AB = 5, AC = 6, and cos(∠BAC) = -3/5. The area of the triangle is (1/2) * AB * AC * sin(∠BAC). Since sin²(∠BAC) = 1 - cos²(∠BAC) = 1 - (9/25) = 16/25, so sin(∠BAC) = 4/5 (taking the positive root because ∠BAC is an angle in a triangle, which is between 0° and 180°, and sine is positive in that range). Therefore, area = (1/2) * 5 * 6 * (4/5) = (1/2) * 24 = 12. [/THOUGHT] \boxed{12}

[THOUGHT] We are given AB = 5, AC = 6, and cos(∠BAC) = -3/5. To compute the area, we use the formula: area = (1/2) * AB * AC * sin(∠BAC). First, we find sin(∠BAC) using the identity sin²(θ) = 1 - cos²(θ). Then, we plug in the values. [/THOUGHT] [PYTHON] print(0.5 * 5 * 6 * (4/5)) [/PYTHON]

\boxed{12}