[THOUGHT] I need to compute the area of triangle ABC given AB = 5, AC = 6, and cos(∠BAC) = -3/5. The area can be found using the formula: (1/2) * AB * AC * sin(∠BAC). First, I'll calculate sin(∠BAC) using the identity sin²θ + cos²θ = 1. Since the cosine is negative, the angle is obtuse, but sine is always non-negative for triangle angles, so sin(∠BAC) = sqrt(1 - (-3/5)²) = sqrt(1 - 9/25) = sqrt(16/25) = 4/5. Then, area = (1/2) * 5 * 6 * (4/5) = 12. I'll write a Python script to verify this calculation. [/THOUGHT] [PYTHON] import math

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

Calculate sin(bac) as positive square root

sin_angle_bac = math.sqrt(1 - cos_angle_bac**2)

Calculate area

area = 0.5 * AB * AC * sin_angle_bac

Print as integer if it's whole number

if area.is_integer(): print(int(area)) else: print(area) [/PYTHON] 12

[THOUGHT] The Python code output confirmed the area is 12, matching the calculated result from trigonometric identities. The negative cosine indicates an obtuse angle, but the positive sine value is correctly used in the area formula. The computation is verified as integer 12 with no decimal or fractional parts. [/THOUGHT] \boxed{12}