[THOUGHT] I need to find the area of triangle ABC given AB = 5, AC = 6, and cos(∠BAC) = -3/5.

The area of a triangle can be calculated using the formula: Area = (1/2) × AB × AC × sin(∠BAC)

First, I need to find sin(∠BAC). Since cos(∠BAC) = -3/5, I can use the identity: sin²(θ) + cos²(θ) = 1

So sin²(∠BAC) = 1 - cos²(∠BAC) = 1 - 9/25 = 16/25 Therefore sin(∠BAC) = 4/5 (positive since angle is between 0 and π in a triangle)

Let me calculate this with Python. [/THOUGHT] [PYTHON] import math

Given values

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

Calculate sin(BAC) using sin^2 + cos^2 = 1

sin_BAC_squared = 1 - cos_BAC**2 sin_BAC = math.sqrt(sin_BAC_squared) # positive since angle is in (0, pi)

print(f"sin(BAC) = {sin_BAC}")

Calculate area using formula: Area = (1/2) * AB * AC * sin(BAC)

area = 0.5 * AB * AC * sin_BAC

print(f"Area = {area}")

Let's also verify with fractions for exact answer

from fractions import Fraction

cos_BAC_frac = Fraction(-3, 5) sin_BAC_squared_frac = 1 - cos_BAC_frac**2 print(f"sin^2(BAC) = {sin_BAC_squared_frac}")

sin(BAC) = 4/5

sin_BAC_frac = Fraction(4, 5)

area_frac = Fraction(1, 2) * AB * AC * sin_BAC_frac print(f"Area (exact) = {area_frac}") [/PYTHON]

[THOUGHT] The calculation confirms that:

• sin(∠BAC) = 4/5 (since sin²(∠BAC) = 16/25)
• Area = (1/2) × 5 × 6 × (4/5) = (1/2) × 24 = 12

The area of triangle ABC is 12. [/THOUGHT]

\boxed{12}