1. **State the problem:** Two planes leave JFK airport at different times and directions. We want to find how far apart they are 40 minutes after the second plane leaves. 2. **Given data:** - Plane 1 speed: 570 mi/hr, traveling due west. - Plane 2 speed: 585 mi/hr, traveling 22° west of north. - Plane 2 leaves 20 minutes after Plane 1. - Find distance between planes 40 minutes after Plane 2 leaves. 3. **Calculate distances traveled by each plane:** - Time Plane 1 has traveled when Plane 2 has traveled 40 minutes = 40 + 20 = 60 minutes = 1 hour. - Distance Plane 1 traveled: $d_1 = 570 \times 1 = 570$ miles. - Distance Plane 2 traveled: $d_2 = 585 \times \frac{40}{60} = 585 \times \frac{2}{3} = 390$ miles. 4. **Set up the triangle:** - Plane 1 path is west (horizontal leg). - Plane 2 path is 22° west of north, so angle between Plane 1 path and Plane 2 path is 90° + 22° = 112°. 5. **Use Law of Cosines to find distance $D$ between planes:** $$ D^2 = d_1^2 + d_2^2 - 2 d_1 d_2 \cos(112^\circ) $$ 6. **Calculate:** $$ D^2 = 570^2 + 390^2 - 2 \times 570 \times 390 \times \cos(112^\circ) $$ 7. **Evaluate values:** - $570^2 = 324900$ - $390^2 = 152100$ - $\cos(112^\circ) \approx -0.3746$ 8. **Substitute:** $$ D^2 = 324900 + 152100 - 2 \times 570 \times 390 \times (-0.3746) $$ 9. **Calculate product:** $$ 2 \times 570 \times 390 = 444600 $$ 10. **Calculate last term:** $$ 444600 \times (-0.3746) = -166555.16 $$ 11. **Substitute:** $$ D^2 = 324900 + 152100 - (-166555.16) = 324900 + 152100 + 166555.16 = 643555.16 $$ 12. **Find $D$:** $$ D = \sqrt{643555.16} \approx 802.22 $$ 13. **Round to nearest ten miles:** $$ D \approx 800 \text{ miles} $$ **Final answer:** The planes are approximately 800 miles apart 40 minutes after the second plane leaves.