1. **Problem statement:** Find the bearing from camp site C to camp site A given the triangle with sides AB = 15 km, BC = 8 km, and AC = 9.5 km, where B is due north of A. 2. **Recall the bearing from B to C:** From previous calculation, angle at B (\(\alpha\)) is approximately 80.6°. 3. **Find angle at C (\(\beta\)) using the Law of Cosines:** \[\cos(\beta) = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC} = \frac{15^2 + 8^2 - 9.5^2}{2 \times 15 \times 8} = \frac{225 + 64 - 90.25}{240} = \frac{198.75}{240} = 0.828125\] \[\beta = \cos^{-1}(0.828125) \approx 34.2^\circ\] 4. **Find angle at A (\(\gamma\)) using triangle angle sum:** \[\gamma = 180^\circ - \alpha - \beta = 180^\circ - 80.6^\circ - 34.2^\circ = 65.2^\circ\] 5. **Determine bearing from C to A:** Since B is due north of A, the bearing from C to A is calculated by adding 180° (to reverse direction) and subtracting angle \(\beta\) from 180°: \[\text{Bearing from C to A} = 180^\circ + (180^\circ - \beta) = 180^\circ + 145.8^\circ = 325.8^\circ\] 6. **Final answer:** The bearing from camp site C to camp site A is approximately **325.8°** (correct to one decimal place).