1. **Problem statement:** Two planes are at the same altitude. One is 100 km away at direction N60°E, the other 160 km away at S50°E. We need to find the distance between the two planes. 2. **Understanding directions:** - N60°E means 60° east of north. - S50°E means 50° east of south. 3. **Convert directions to standard angles:** - North is 0°, east is 90°, so N60°E corresponds to 60° from north towards east, i.e., 60°. - South is 180°, so S50°E is 180° - 50° = 130° from north clockwise. 4. **Represent positions as vectors:** Let the observer be at origin O. - Plane A vector: $\vec{A} = 100(\cos 60^\circ, \sin 60^\circ)$ - Plane B vector: $\vec{B} = 160(\cos 130^\circ, \sin 130^\circ)$ 5. **Calculate components:** - $\cos 60^\circ = 0.5$, $\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$ - $\cos 130^\circ = \cos(180^\circ - 50^\circ) = -\cos 50^\circ \approx -0.6428$ - $\sin 130^\circ = \sin(180^\circ - 50^\circ) = \sin 50^\circ \approx 0.7660$ 6. **Calculate vector coordinates:** - $\vec{A} = (100 \times 0.5, 100 \times 0.866) = (50, 86.6)$ - $\vec{B} = (160 \times -0.6428, 160 \times 0.7660) = (-102.85, 122.56)$ 7. **Distance between planes:** $$ \text{Distance} = |\vec{A} - \vec{B}| = \sqrt{(50 - (-102.85))^2 + (86.6 - 122.56)^2} $$ 8. **Simplify inside the square root:** $$ = \sqrt{(50 + 102.85)^2 + (-35.96)^2} = \sqrt{152.85^2 + 35.96^2} $$ 9. **Calculate squares:** $$ 152.85^2 = 23363.12, \quad 35.96^2 = 1293.12 $$ 10. **Sum and square root:** $$ \sqrt{23363.12 + 1293.12} = \sqrt{24656.24} \approx 157.0 $$ **Final answer:** The planes are approximately 157 km apart.