1. **State the problem:** An airplane flies from airport A to B on a bearing of 035° for 1.5 hours at 600 km/h, then from B to C on a bearing of 130° for 1.5 hours at 400 km/h. We need to find the bearing from C back to A. 2. **Calculate distances:** Distance AB = speed × time = $600 \times 1.5 = 900$ km Distance BC = $400 \times 1.5 = 600$ km 3. **Convert bearings to standard angles:** Bearing is measured clockwise from north. To use trigonometry, convert bearing to angle from the positive x-axis (east) counterclockwise: - Bearing 035° means angle from north clockwise 35°, so angle from east is $90° - 35° = 55°$ - Bearing 130° means angle from north clockwise 130°, so angle from east is $90° - 130° = -40°$ or equivalently $320°$ 4. **Find coordinates of B relative to A:** Using east as x-axis and north as y-axis: $$x_B = 900 \times \cos 55°$$ $$y_B = 900 \times \sin 55°$$ Calculate: $$x_B = 900 \times 0.5736 = 516.24$$ $$y_B = 900 \times 0.8192 = 737.28$$ 5. **Find coordinates of C relative to B:** $$x_C = 600 \times \cos (-40°) = 600 \times 0.7660 = 459.6$$ $$y_C = 600 \times \sin (-40°) = 600 \times (-0.6428) = -385.68$$ 6. **Find coordinates of C relative to A:** $$x_{C/A} = x_B + x_C = 516.24 + 459.6 = 975.84$$ $$y_{C/A} = y_B + y_C = 737.28 - 385.68 = 351.6$$ 7. **Find vector from C to A:** Vector $\overrightarrow{CA} = (x_A - x_C, y_A - y_C) = (0 - 975.84, 0 - 351.6) = (-975.84, -351.6)$ 8. **Calculate bearing from C to A:** Calculate angle $\theta$ from east axis: $$\theta = \arctan \left( \frac{y}{x} \right) = \arctan \left( \frac{-351.6}{-975.84} \right) = \arctan(0.36) = 19.8°$$ Since both x and y are negative, vector is in the third quadrant, so add 180°: $$\theta = 19.8° + 180° = 199.8°$$ 9. **Convert angle to bearing:** Bearing is measured clockwise from north: $$\text{bearing} = 90° - \theta = 90° - 199.8° = -109.8°$$ Add 360° to get positive bearing: $$-109.8° + 360° = 250.2°$$ 10. **Final answer:** Bearing from C to A is approximately $\boxed{250°}$ to the nearest degree.