An ant starts at vertex $A$ in equilateral triangle $\triangle ABC$ and walks around the perimeter of the triangle from $A$ to $B$ to $C$ and back to $A$. When the ant is $42$ percent of its way around the triangle, it stops for a rest. Find the percent of the way from $B$ to $C$ the ant is at that point
1. First, we need to understand the total distance the ant travels around the perimeter of the equilateral triangle $\triangle ABC$. Since the ant starts at vertex $A$ and walks around the perimeter back to $A$, the total distance is the perimeter of the triangle.

2. Let the side length of the equilateral triangle be $s$. The perimeter of the triangle is $3s$.

3. The ant travels $42\%$ of the total perimeter. Therefore, the distance traveled by the ant is:
   \[
   0.42 \times 3s = 1.26s
   \]

4. Each side of the triangle represents $\frac{1}{3}$ or approximately $33.33\%$ of the total perimeter. Therefore, the ant travels one full side from $A$ to $B$ and part of the next side from $B$ to $C$.

5. The distance from $A$ to $B$ is $s$, which is $33.33\%$ of the total perimeter. The remaining distance the ant travels is:
   \[
   1.26s - s = 0.26s
   \]

6. The remaining distance $0.26s$ is the distance the ant travels from $B$ towards $C$. To find the percentage of the way from $B$ to $C$, we calculate:
   \[
   \frac{0.26s}{s} \times 100\% = 26\%
   \]

Therefore, the ant is $26\%$ of the way from $B$ to $C$ when it stops for a rest.

The final answer is $\boxed{26\%}$