1. **State the problem:** An aeroplane leaves airport A and flies on a bearing of 035° for 1.5 hours at 600 km/h to airport B. Then it flies on a bearing of 130° for 1.5 hours at 400 km/h to airport C. We need to calculate the distance from C to A. 2. **Calculate distances AB and BC:** - Distance AB = speed × time = $600 \times 1.5 = 900$ km - Distance BC = $400 \times 1.5 = 600$ km 3. **Find coordinates of B relative to A:** Using bearing 035°, convert to standard angle from east: $\theta = 90° - 35° = 55°$ (measured from east counterclockwise) Coordinates of B: $$x_B = 900 \cos 55°$$ $$y_B = 900 \sin 55°$$ Calculate: $$x_B = 900 \times 0.5736 = 516.24$$ $$y_B = 900 \times 0.8192 = 737.28$$ 4. **Find coordinates of C relative to B:** Bearing 130° means angle from north clockwise 130°, so from east: $$\theta = 90° - 130° = -40°$$ or equivalently 320° Coordinates of C relative to B: $$x_C = 600 \cos (-40°) = 600 \times 0.7660 = 459.6$$ $$y_C = 600 \sin (-40°) = 600 \times (-0.6428) = -385.68$$ 5. **Find coordinates of C relative to A:** $$x_C^{(A)} = x_B + x_C = 516.24 + 459.6 = 975.84$$ $$y_C^{(A)} = y_B + y_C = 737.28 - 385.68 = 351.6$$ 6. **Calculate distance from C to A:** $$d = \sqrt{(x_C^{(A)})^2 + (y_C^{(A)})^2} = \sqrt{975.84^2 + 351.6^2}$$ Calculate squares: $$975.84^2 = 952,252.5$$ $$351.6^2 = 123,623.0$$ Sum: $$952,252.5 + 123,623.0 = 1,075,875.5$$ Distance: $$d = \sqrt{1,075,875.5} \approx 1037.5$$ km 7. **Final answer:** The distance from C to A is approximately **1038 km** (to the nearest km).