1. **Problem 1: Angle of Depression from Hot Air Balloon** A hot air balloon is 40 ft above the ground and 70 ft away horizontally from a farm. We need to find the angle of depression from the balloon to the farm. 2. **Formula and Explanation:** The angle of depression is the angle between the horizontal line from the observer (balloon) and the line of sight to the object (farm). We can use the tangent function in a right triangle: $$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$ where: - opposite side = height difference = 40 ft - adjacent side = horizontal distance = 70 ft 3. **Calculation:** $$\theta = \tan^{-1}\left(\frac{40}{70}\right) = \tan^{-1}(0.5714)$$ Using a calculator: $$\theta \approx 29.74^\circ$$ 4. **Interpretation:** The angle of depression from the balloon to the farm is approximately $29.74^\circ$. 5. **Problem 2: Distance Between Airports A and B** An airplane travels 50 miles North 35° West (N35°W) from airport A, then 70 miles South 20° West (S20°W) to reach airport B. We need to find the straight-line distance between airports A and B. 6. **Approach:** We will use vector components to find the resultant displacement. - For N35°W: - North component: $50 \times \cos(35^\circ)$ - West component: $50 \times \sin(35^\circ)$ - For S20°W: - South component: $70 \times \cos(20^\circ)$ - West component: $70 \times \sin(20^\circ)$ 7. **Calculate components:** - First leg: - North: $50 \times 0.8192 = 40.96$ miles - West: $50 \times 0.574 = 28.7$ miles - Second leg: - South: $70 \times 0.9397 = 65.78$ miles - West: $70 \times 0.3420 = 23.94$ miles 8. **Net displacement components:** - North-South: $40.96 - 65.78 = -24.82$ miles (southward) - West: $28.7 + 23.94 = 52.64$ miles 9. **Distance between airports:** $$d = \sqrt{(-24.82)^2 + (52.64)^2} = \sqrt{616.5 + 2771.5} = \sqrt{3388} \approx 58.22 \text{ miles}$$ 10. **Final answer:** The distance between airport A and airport B is approximately $58.22$ miles.