1. **State the problem:** We are given a triangle with points A, B, and C. We know |AB| = 8 km, |BC| = 13 km, and the bearing of B from C is 230°. We need to find: (a) the distance AC (b) the bearing of C from A (c) how far east of B, C is 2. **Understand bearings and angles:** Bearing is measured clockwise from North. A bearing of 230° means the line from C to B points 230° clockwise from North. 3. **Find angle at C:** Since the bearing of B from C is 230°, the angle between North and line CB is 230°. The angle between East and North is 90°, so the angle between East and CB is $230^\circ - 180^\circ = 50^\circ$ (given in the problem). This confirms the angle at C is 50°. 4. **Find angle at A:** Given angle at A is 40° between AB and East. 5. **Calculate angle at B:** The sum of angles in triangle ABC is 180°, so $$\angle B = 180^\circ - 40^\circ - 50^\circ = 90^\circ.$$ 6. **Use the Law of Cosines to find AC:** $$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle B)$$ Substitute values: $$AC^2 = 8^2 + 13^2 - 2 \times 8 \times 13 \times \cos(90^\circ)$$ Since $\cos(90^\circ) = 0$: $$AC^2 = 64 + 169 - 0 = 233$$ So, $$AC = \sqrt{233} \approx 15.264$$ km 7. **Find bearing of C from A:** - The bearing of B from A is $40^\circ$ (angle between AB and East). - Since angle at A is 40°, and AB is along bearing 40°, AC makes an angle of $40^\circ + 90^\circ = 130^\circ$ from East (because angle at B is 90°, so AC is perpendicular to AB). - Bearing is measured from North clockwise, so bearing of C from A is: $$90^\circ - 130^\circ = -40^\circ$$ Since negative, add 360°: $$320^\circ$$ 8. **Calculate how far east of B, C is:** - From point B, line BC is 13 km at bearing 230°. - East component of BC is: $$13 \times \sin(230^\circ) = 13 \times (-0.7660) = -9.958$$ km - Negative means west of B, so C is approximately 9.96 km west of B. **Final answers:** (a) Distance AC $\approx 15.3$ km (b) Bearing of C from A $= 320^\circ$ (c) C is approximately 9.96 km west of B (or 9.96 km east of B is negative)