1. **Problem Statement:** Find the value of $x$ in the first right triangle where the hypotenuse is $x$, one side is $7\sqrt{6}$, and the angles are $45^\circ$ and $60^\circ$. 2. **Recall the properties of a right triangle:** The sum of angles in a triangle is $180^\circ$. Given one right angle ($90^\circ$), the other two angles are $45^\circ$ and $60^\circ$ which is inconsistent because $45^\circ + 60^\circ = 105^\circ$ and $105^\circ + 90^\circ = 195^\circ$ which is impossible. However, assuming the triangle is right angled with angles $45^\circ$ and $45^\circ$ or $30^\circ$ and $60^\circ$, we proceed with the given data. 3. **Using the Law of Sines or special right triangle ratios:** Since the triangle has angles $45^\circ$ and $60^\circ$, the third angle is $75^\circ$ (not right angle). But the problem states a right triangle, so we consider the right angle is $90^\circ$ and the other angles are $45^\circ$ and $45^\circ$ or $30^\circ$ and $60^\circ$. Given the side $7\sqrt{6}$ opposite $60^\circ$, we use the sine ratio: $$\sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{7\sqrt{6}}{x}$$ 4. **Calculate $x$:** $$x = \frac{7\sqrt{6}}{\sin(60^\circ)}$$ Recall $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, so: $$x = \frac{7\sqrt{6}}{\frac{\sqrt{3}}{2}} = 7\sqrt{6} \times \frac{2}{\sqrt{3}} = 14 \times \frac{\sqrt{6}}{\sqrt{3}}$$ Simplify the radical: $$\frac{\sqrt{6}}{\sqrt{3}} = \sqrt{\frac{6}{3}} = \sqrt{2}$$ So: $$x = 14 \sqrt{2}$$ 5. **Final answer:** $$\boxed{14 \sqrt{2}}$$ --- **Summary:** Given the side opposite $60^\circ$ is $7\sqrt{6}$, the hypotenuse $x$ is $14\sqrt{2}$.