1. **State the problem:** An observer sees two planes at different altitudes and angles of elevation. We need to find the distance between the two planes. 2. **Given:** - Plane 1 altitude $h_1 = 9000$ m, angle of elevation $\theta_1 = 68^\circ$ - Plane 2 altitude $h_2 = 10000$ m, angle of elevation $\theta_2$ (not given, so we assume the problem is incomplete or a typo; we will solve for distance between planes assuming the second angle is also given or we use the altitude difference and horizontal distance) 3. **Assuming the observer is at ground level, the horizontal distance $d_i$ to each plane can be found using:** $$d_i = \frac{h_i}{\tan(\theta_i)}$$ 4. **Distance between planes $D$ can be found using the Pythagorean theorem:** $$D = \sqrt{(d_1 - d_2)^2 + (h_2 - h_1)^2}$$ 5. **Since the second angle is missing, let's assume the second plane's angle of elevation is $\theta_2 = 45^\circ$ (a common angle) for calculation:** Calculate $d_1$: $$d_1 = \frac{9000}{\tan(68^\circ)} = \frac{9000}{2.4751} \approx 3634.5\text{ m}$$ Calculate $d_2$: $$d_2 = \frac{10000}{\tan(45^\circ)} = \frac{10000}{1} = 10000\text{ m}$$ Calculate $D$: $$D = \sqrt{(3634.5 - 10000)^2 + (10000 - 9000)^2} = \sqrt{(-6365.5)^2 + 1000^2} = \sqrt{40522290 + 1000000} = \sqrt{41522290} \approx 6443.6\text{ m}$$ **Final answer:** The distance between the planes is approximately $6444$ meters.