1. **Problem statement:** Find the bearing of the journey from point C to point A in the triangle formed by points A, B, and C. 2. **Given data:** - Bearing from A to B is 075°. - Distance AB = 12 km. - Bearing from B to C is 165°. - Distance CA = 20 km. 3. **Understanding bearings:** Bearings are measured clockwise from the north direction. 4. **Step 1: Find coordinates of B relative to A.** Using bearing 075° and distance 12 km: $$x_B = 12 \times \sin(75^\circ)$$ $$y_B = 12 \times \cos(75^\circ)$$ 5. **Step 2: Find coordinates of C relative to B.** Using bearing 165° and unknown distance BC = $d$ km: $$x_C = x_B + d \times \sin(165^\circ)$$ $$y_C = y_B + d \times \cos(165^\circ)$$ 6. **Step 3: Use distance CA = 20 km to find $d$.** Coordinates of A are (0,0), so: $$ (x_C - 0)^2 + (y_C - 0)^2 = 20^2 $$ Substitute $x_C$ and $y_C$: $$ (12 \sin 75^\circ + d \sin 165^\circ)^2 + (12 \cos 75^\circ + d \cos 165^\circ)^2 = 400 $$ 7. **Step 4: Calculate numerical values:** $$\sin 75^\circ \approx 0.9659, \cos 75^\circ \approx 0.2588$$ $$\sin 165^\circ \approx 0.2588, \cos 165^\circ \approx -0.9659$$ 8. **Step 5: Substitute and expand:** $$ (12 \times 0.9659 + d \times 0.2588)^2 + (12 \times 0.2588 + d \times (-0.9659))^2 = 400 $$ $$ (11.5908 + 0.2588 d)^2 + (3.1056 - 0.9659 d)^2 = 400 $$ 9. **Step 6: Expand squares:** $$ (11.5908)^2 + 2 \times 11.5908 \times 0.2588 d + (0.2588 d)^2 + (3.1056)^2 - 2 \times 3.1056 \times 0.9659 d + (0.9659 d)^2 = 400 $$ 10. **Step 7: Calculate constants:** $$ 134.34 + 6.002 d + 0.067 d^2 + 9.646 - 6.000 d + 0.933 d^2 = 400 $$ 11. **Step 8: Combine like terms:** $$ (0.067 + 0.933) d^2 + (6.002 - 6.000) d + (134.34 + 9.646) = 400 $$ $$ 1.0 d^2 + 0.002 d + 143.986 = 400 $$ 12. **Step 9: Simplify:** $$ d^2 + 0.002 d + 143.986 - 400 = 0 $$ $$ d^2 + 0.002 d - 256.014 = 0 $$ 13. **Step 10: Solve quadratic equation:** $$ d = \frac{-0.002 \pm \sqrt{(0.002)^2 - 4 \times 1 \times (-256.014)}}{2} $$ $$ d = \frac{-0.002 \pm \sqrt{0.000004 + 1024.056}}{2} $$ $$ d = \frac{-0.002 \pm 32.0009}{2} $$ 14. **Step 11: Choose positive root:** $$ d = \frac{-0.002 + 32.0009}{2} = 15.9995 \approx 16 \text{ km} $$ 15. **Step 12: Find coordinates of C:** $$ x_C = 11.5908 + 0.2588 \times 16 = 11.5908 + 4.1408 = 15.7316 $$ $$ y_C = 3.1056 - 0.9659 \times 16 = 3.1056 - 15.4544 = -12.3488 $$ 16. **Step 13: Find bearing from C to A:** Vector from C to A is: $$ \Delta x = 0 - 15.7316 = -15.7316 $$ $$ \Delta y = 0 - (-12.3488) = 12.3488 $$ 17. **Step 14: Calculate angle from north:** $$ \theta = \arctan \left( \frac{|\Delta x|}{|\Delta y|} \right) = \arctan \left( \frac{15.7316}{12.3488} \right) = 51.3^\circ $$ 18. **Step 15: Determine quadrant:** Since $\Delta x$ is negative and $\Delta y$ is positive, vector points northwest. Bearing is measured clockwise from north, so: $$ \text{Bearing} = 360^\circ - 51.3^\circ = 308.7^\circ $$ **Final answer:** The bearing of the journey from C to A is approximately **309°**.