1. **Problem statement:** We have a right triangle with a hypotenuse $e$, one leg of length $3\sqrt{3}$, and an angle of $60^\circ$ adjacent to that leg. We want to find the length of $e$. 2. **Relevant formula:** In a right triangle, the leg adjacent to an angle $\theta$ is related to the hypotenuse by the cosine function: $$\cos(\theta) = \frac{\text{adjacent leg}}{\text{hypotenuse}}$$ 3. **Apply the formula:** Here, $\theta = 60^\circ$, the adjacent leg is $3\sqrt{3}$, and the hypotenuse is $e$. So, $$\cos(60^\circ) = \frac{3\sqrt{3}}{e}$$ 4. **Evaluate $\cos(60^\circ)$:** $$\cos(60^\circ) = \frac{1}{2}$$ 5. **Set up the equation:** $$\frac{1}{2} = \frac{3\sqrt{3}}{e}$$ 6. **Solve for $e$:** Multiply both sides by $e$ and then by 2: $$e \times \frac{1}{2} = 3\sqrt{3} \implies \cancel{\frac{1}{2}} e \times \cancel{2} = 3\sqrt{3} \times 2$$ $$e = 6\sqrt{3}$$ **Final answer:** $$e = 6\sqrt{3}$$