1. **State the problem:** We start at point A, travel 7 km on a bearing of 135° to point B, then from B travel 12 km on a bearing of 250° to point C. We need to find the distance and bearing of A from C. 2. **Set up the coordinate system:** Take north as the positive y-axis and east as the positive x-axis. 3. **Convert bearings and distances to coordinates:** - From A to B: distance $7$ km, bearing $135^\circ$. Bearing is measured clockwise from north, so the angle from the positive y-axis is $135^\circ$. Coordinates of B relative to A: $$x_B = 7 \times \sin(135^\circ) = 7 \times \frac{\sqrt{2}}{2} = 4.95$$ $$y_B = 7 \times \cos(135^\circ) = 7 \times (-\frac{\sqrt{2}}{2}) = -4.95$$ 4. **Coordinates of B:** $$B = (4.95, -4.95)$$ 5. **From B to C: distance 12 km, bearing 250°:** Angle from north is $250^\circ$. Calculate components: $$x_C = 12 \times \sin(250^\circ) = 12 \times (-0.9397) = -11.28$$ $$y_C = 12 \times \cos(250^\circ) = 12 \times (-0.3420) = -4.10$$ 6. **Coordinates of C relative to A:** $$C_x = x_B + x_C = 4.95 + (-11.28) = -6.33$$ $$C_y = y_B + y_C = -4.95 + (-4.10) = -9.05$$ 7. **Vector from C to A:** $$\vec{CA} = (0 - (-6.33), 0 - (-9.05)) = (6.33, 9.05)$$ 8. **Distance from C to A:** $$d = \sqrt{6.33^2 + 9.05^2} = \sqrt{40.07 + 81.90} = \sqrt{121.97} = 11.05 \text{ km}$$ 9. **Bearing of A from C:** Calculate angle $\theta$ from north (positive y-axis) clockwise: $$\theta = \arctan\left(\frac{6.33}{9.05}\right) = 34.5^\circ$$ Since $x$ and $y$ are positive, bearing is $34.5^\circ$. **Final answer:** Distance from C to A is approximately $11.05$ km. Bearing of A from C is approximately $35^\circ$.