1. **State the problem:** Anthony is in a hot-air balloon and wants to find its height. He measures the angle of depression to a landmark east of the balloon twice: first at 54°, then after moving 100 feet east, at 61°. 2. **Set up the scenario:** Let the height of the balloon be $h$ feet. Let the horizontal distance from the balloon's first position to the landmark be $x$ feet. 3. **Use the tangent of the angle of depression:** The angle of depression equals the angle between the horizontal line from the balloon and the line of sight to the landmark. Since the landmark is on the ground, we use right triangle trigonometry: $$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{d}$$ where $d$ is the horizontal distance from the balloon to the landmark. 4. **Write equations for each position:** - At the first position: $$\tan(54^\circ) = \frac{h}{x} \implies h = x \tan(54^\circ)$$ - After moving 100 feet east, the horizontal distance to the landmark is $x - 100$: $$\tan(61^\circ) = \frac{h}{x - 100} \implies h = (x - 100) \tan(61^\circ)$$ 5. **Set the two expressions for $h$ equal:** $$x \tan(54^\circ) = (x - 100) \tan(61^\circ)$$ 6. **Solve for $x$:** $$x \tan(54^\circ) = x \tan(61^\circ) - 100 \tan(61^\circ)$$ $$x \tan(54^\circ) - x \tan(61^\circ) = -100 \tan(61^\circ)$$ $$x (\tan(54^\circ) - \tan(61^\circ)) = -100 \tan(61^\circ)$$ $$x = \frac{-100 \tan(61^\circ)}{\tan(54^\circ) - \tan(61^\circ)}$$ 7. **Simplify by canceling negative signs:** $$x = \frac{100 \tan(61^\circ)}{\tan(61^\circ) - \tan(54^\circ)}$$ 8. **Calculate numerical values:** $$\tan(54^\circ) \approx 1.37638$$ $$\tan(61^\circ) \approx 1.80405$$ $$x = \frac{100 \times 1.80405}{1.80405 - 1.37638} = \frac{180.405}{0.42767} \approx 421.7$$ 9. **Find the height $h$:** $$h = x \tan(54^\circ) = 421.7 \times 1.37638 \approx 580.3$$ **Final answer:** Anthony is approximately 580 feet high.