[THOUGHT] To find the area of triangle ABC with sides AB = 5, AC = 6, and cos(angle BAC) = -3/5, we use the formula:
Area = (1/2) * AB * AC * sin(angle BAC).
First, compute sin(angle BAC) using the Pythagorean identity:
sin²θ + cos²θ = 1.
Given cosθ = -3/5, sinθ = √(1 - (9/25)) = √(16/25) = 4/5 (since angle BAC is obtuse but sine is positive in [0, π]).
Substitute into the area formula:
Area = (1/2) * 5 * 6 * (4/5) = 12.
Verify using the Law of Cosines to find BC and Heron's formula for cross-checking.
[/THOUGHT]
[PYTHON] import math

Given values

AB = 5 AC = 6 cos_theta = -3/5

Calculate sin(theta)

sin_theta = math.sqrt(1 - cos_theta**2)

Calculate area using (1/2)absin(theta)

area = 0.5 * AB * AC * sin_theta print(int(area) if area.is_integer() else area) [/PYTHON]

[THOUGHT] The Python code output confirms the area is 12. This matches the derived solution using the sine of the angle and the formula for the area of a triangle. The result is consistent and correct. [/THOUGHT] \boxed{12}