[THOUGHT] I will calculate $\sin(A)$ from the given $\cos(A)$ and then use the formula Area $= \frac{1}{2} b c \sin(A)$ to find the area of the triangle.

Given: $AB = 5$ $AC = 6$ $\cos(A) = -3/5$

Steps:

1. Calculate $\sin(A) = \sqrt{1 - \cos^2(A)}$.
2. Calculate Area $= 0.5 \times AB \times AC \times \sin(A)$. [/THOUGHT] [PYTHON] import math

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

Calculate sin(A)

Since A is an angle of a triangle, sin(A) is positive

sin_A = math.sqrt(1 - cos_A**2)

Calculate Area

area = 0.5 * AB * AC * sin_A print(f"{area=}") [/PYTHON]

[THOUGHT] The area of a triangle is given by the formula $\text{Area} = \frac{1}{2} bc \sin A$. Given $AB = c = 5$, $AC = b = 6$, and $\cos A = -\frac{3}{5}$. Since $A$ is an angle in a triangle, $\sin A > 0$. Using the identity $\sin^2 A + \cos^2 A = 1$: $\sin^2 A + \left(-\frac{3}{5}\right)^2 = 1$ $\sin^2 A + \frac{9}{25} = 1$ $\sin^2 A = \frac{16}{25}$ $\sin A = \frac{4}{5}$.

Now, calculate the area: $\text{Area} = \frac{1}{2} (5)(6) \left(\frac{4}{5}\right) = \frac{1}{2} (30) \left(\frac{4}{5}\right) = 15 \left(\frac{4}{5}\right) = 12$. The Python output confirms the area is 12.0. [/THOUGHT]

\boxed{12}