[THOUGHT] We have triangle ABC with sides AB = 5, AC = 6, and cos(∠BAC) = -3/5. The angle at A is ∠BAC, so side BC is opposite angle A. We can use the law of cosines to find BC, then use Heron's formula or another method to find the area.

Alternatively, area = (1/2)ABACsin(∠BAC). We know cos(∠BAC) = -3/5, so sin(∠BAC) = sqrt(1 - cos^2) = sqrt(1 - 9/25) = sqrt(16/25) = 4/5 (positive because angle between 0 and π, and sin is positive for angles between 0 and π). So area = (1/2)56(4/5) = (1/2)56*4/5 = (1/2)64 = 12. Wait, check: (1/2)56 = 15, times (4/5) = 12. So area = 12.

But let's verify using Python to be sure.

We'll compute sin(angle) = sqrt(1 - cos^2) = sqrt(1 - 9/25) = sqrt(16/25) = 4/5. Then area = 0.5 * AB * AC * sin(angle). We'll do that in Python.

Also, we can compute BC using law of cosines: BC^2 = AB^2 + AC^2 - 2ABACcos(angle) = 25 + 36 - 256(-3/5) = 61 - 60*(-3/5) = 61 + 36 = 97. So BC = sqrt(97). Then using Heron's formula: s = (5+6+sqrt(97))/2. Compute area = sqrt(s(s-a)(s-b)(s-c)). That should give the same area.

Let's verify both methods in Python. [/THOUGHT] [PYTHON] import math

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

Method 1: area = 0.5 * AB * AC * sin(A)

sin_A = math.sqrt(1 - cos_A**2) # positive root since angle in (0, pi) area1 = 0.5 * AB * AC * sin_A print("Method 1 area:", area1)

Method 2: law of cosines for BC, then Heron's formula

BC = math.sqrt(AB2 + AC2 - 2ABAC*cos_A) s = (AB + AC + BC) / 2 area2 = math.sqrt(s * (s - AB) * (s - AC) * (s - BC)) print("Method 2 area:", area2)

Check sin_A value

print("sin(A) =", sin_A) print("cos(A) =", cos_A) print("cos^2 + sin^2 =", cos_A2 + sin_A2) [/PYTHON]

[THOUGHT] The Python output shows the area is 12.0 (or 12.000000000000002 due to floating-point precision). The problem asks for the area, which should be an integer. Since the exact area is 12, we output 12. [/THOUGHT] \boxed{12}