1. **State the problem:** A light beam inside a fiber optic cable travels along the cable, bouncing off the walls at the same angle. The cable is 1,200 µm long. The beam strikes the wall after traveling 720 µm horizontally. We need to find the total distance $x + y$ the beam travels inside the cable. 2. **Use the Pythagorean theorem:** The beam forms a right triangle with vertical side 300 µm and horizontal side 720 µm. The hypotenuse $x$ is the distance the beam travels to the wall. $$x^2 = 300^2 + 720^2$$ 3. **Calculate $x$:** $$x^2 = 90000 + 518400 = 608400$$ $$x = \sqrt{608400} = 780 \text{ µm}$$ 4. **Find the remaining length of the cable:** The total cable length is 1,200 µm, so the remaining horizontal length after 720 µm is $$1200 - 720 = 480 \text{ µm}$$ 5. **Use similarity of triangles:** The beam forms two similar right triangles. Using the ratio of corresponding sides, $$\frac{y}{480} = \frac{420}{720}$$ 6. **Solve for $y$:** $$y = 480 \times \frac{420}{720}$$ Simplify the fraction: $$y = 480 \times \frac{7}{12}$$ $$y = 280 \text{ µm}$$ 7. **Calculate total distance traveled:** $$x + y = 780 + 280 = 1060 \text{ µm}$$ **Final answer:** The beam travels a total distance of **1060 µm** inside the fiber optic cable.