In $\triangle A B C, B C=A C-1$ and $A C=A B-1$. If $\cos (\angle B A C)=\frac{3}{5}$, determine the perimeter of $\triangle A B C$.
Let $A C=x$.

Thus $B C=A C-1=x-1$.

Since $A C=A B-1$, then $A B=A C+1=x+1$.

The perimeter of $\triangle A B C$ is $B C+A C+A B=(x-1)+x+(x+1)=3 x$.

By the cosine law in $\triangle A B C$,

$$
\begin{aligned}
B C^{2} & =A B^{2}+A C^{2}-2(A B)(A C) \cos (\angle B A C) \\
(x-1)^{2} & =(x+1)^{2}+x^{2}-2(x+1) x\left(\frac{3}{5}\right) \\
x^{2}-2 x+1 & =x^{2}+2 x+1+x^{2}-\frac{6}{5}\left(x^{2}+x\right) \\
0 & =x^{2}+4 x-\frac{6}{5}\left(x^{2}+x\right) \\
0 & =5 x^{2}+20 x-6\left(x^{2}+x\right) \\
x^{2} & =14 x
\end{aligned}
$$

Since $x>0$, then $x=14$.

Therefore, the perimeter of $\triangle A B C$, which equals $3 x$, is 42.

ANSWER: 42