1. **Problem:** An airplane travels 2.5 hours at 300 mph, then turns 20° left and flies 1 hour at the same speed. Find the distance from the departure point. 2. **Formula:** Use vector components and the Law of Cosines. 3. **Step 1:** Calculate distances traveled: - First leg: $d_1 = 2.5 \times 300 = 750$ miles - Second leg: $d_2 = 1 \times 300 = 300$ miles 4. **Step 2:** The angle between the two legs is 20°. 5. **Step 3:** Use Law of Cosines to find the resultant distance $D$: $$D = \sqrt{d_1^2 + d_2^2 - 2 d_1 d_2 \cos(20^\circ)}$$ 6. **Step 4:** Calculate: $$D = \sqrt{750^2 + 300^2 - 2 \times 750 \times 300 \times \cos(20^\circ)}$$ 7. **Step 5:** Evaluate: - $750^2 = 562500$ - $300^2 = 90000$ - $2 \times 750 \times 300 = 450000$ - $\cos(20^\circ) \approx 0.9397$ 8. **Step 6:** Substitute: $$D = \sqrt{562500 + 90000 - 450000 \times 0.9397} = \sqrt{652500 - 422865} = \sqrt{229635}$$ 9. **Step 7:** Final distance: $$D \approx 479.2 \text{ miles}$$ **Answer:** The airplane is approximately 479.2 miles from its departure point.