1. **State the problem:** We have a right triangle with hypotenuse $13$, one leg (opposite the $60^\circ$ angle) labeled $a$, the other leg (adjacent to the $60^\circ$ angle) labeled $b$, and the vertical leg given as $4\sqrt{3}$. We need to find the lengths of $a$ and $b$. 2. **Identify the sides:** The side opposite the $60^\circ$ angle is $a$, the side adjacent to the $60^\circ$ angle is $b$, and the hypotenuse is $13$. The vertical side labeled $4\sqrt{3}$ corresponds to the side opposite or adjacent depending on orientation, but since $a$ is opposite $60^\circ$, and the vertical side is $4\sqrt{3}$, we conclude $a = 4\sqrt{3}$. 3. **Use the Pythagorean theorem:** For a right triangle, $$a^2 + b^2 = c^2$$ where $c$ is the hypotenuse. 4. **Plug in known values:** $$ (4\sqrt{3})^2 + b^2 = 13^2 $$ 5. **Simplify:** $$ 16 \times 3 + b^2 = 169 $$ $$ 48 + b^2 = 169 $$ 6. **Isolate $b^2$:** $$ b^2 = 169 - 48 $$ $$ b^2 = 121 $$ 7. **Take the square root:** $$ b = \sqrt{121} = 11 $$ 8. **Check with trigonometry:** Using the angle $60^\circ$, $$ \sin 60^\circ = \frac{a}{c} = \frac{4\sqrt{3}}{13} $$ which matches the known value $\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$, and $\frac{4\sqrt{3}}{13} \approx 0.866$, confirming correctness. **Final answers:** - $a = 4\sqrt{3}$ - $b = 11$