1. **State the problem:** We start at point A (0,0). 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. **Convert bearings to standard angles:** Bearings are measured clockwise from North (0°). To use Cartesian coordinates, convert bearing $\theta$ to angle from positive x-axis (East) as $\alpha = 90^\circ - \theta$. - For 135°, $\alpha_B = 90^\circ - 135^\circ = -45^\circ$ (or 315°) - For 250°, $\alpha_C = 90^\circ - 250^\circ = -160^\circ$ (or 200°) 3. **Calculate coordinates of B:** Using $x = r \cos \alpha$, $y = r \sin \alpha$ with $r=7$ km, $$x_B = 7 \cos(-45^\circ) = 7 \times \frac{\sqrt{2}}{2} = 4.95$$ $$y_B = 7 \sin(-45^\circ) = 7 \times -\frac{\sqrt{2}}{2} = -4.95$$ So, $B = (4.95, -4.95)$ km. 4. **Calculate coordinates of C:** From B, travel 12 km at $\alpha_C = -160^\circ$: $$x_C = x_B + 12 \cos(-160^\circ) = 4.95 + 12 \times \cos(200^\circ)$$ $$y_C = y_B + 12 \sin(-160^\circ) = -4.95 + 12 \times \sin(200^\circ)$$ Calculate: $$\cos(200^\circ) = \cos(180^\circ + 20^\circ) = -\cos 20^\circ = -0.94$$ $$\sin(200^\circ) = \sin(180^\circ + 20^\circ) = -\sin 20^\circ = -0.34$$ So, $$x_C = 4.95 + 12 \times (-0.94) = 4.95 - 11.28 = -6.33$$ $$y_C = -4.95 + 12 \times (-0.34) = -4.95 - 4.08 = -9.03$$ Thus, $C = (-6.33, -9.03)$ km. 5. **Calculate distance from C to A:** Distance $d = \sqrt{(x_A - x_C)^2 + (y_A - y_C)^2} = \sqrt{(-6.33)^2 + (-9.03)^2}$ $$d = \sqrt{40.07 + 81.54} = \sqrt{121.61} = 11.03 \text{ km}$$ 6. **Calculate bearing from C to A:** Vector from C to A is $(0 - (-6.33), 0 - (-9.03)) = (6.33, 9.03)$. Angle from East axis: $$\theta = \arctan\left(\frac{9.03}{6.33}\right) = 55.3^\circ$$ Convert to bearing from North: $$\text{bearing} = 90^\circ - 55.3^\circ = 34.7^\circ$$ Since both $x$ and $y$ are positive, bearing is $34.7^\circ$. **Final answer:** Distance from C to A is approximately 11.03 km. Bearing from C to A is approximately 35° (rounded).