1. **Problem statement:** We are given three points A, B, and C with C due west of B, the bearing of A from B is 241°, and AC = CB. We need to find the bearing of A from C. 2. **Understanding bearings:** Bearing is measured clockwise from north. A bearing of 241° from B to A means starting at north at B and rotating 241° clockwise to point towards A. 3. **Set up coordinate system:** Place B at the origin $(0,0)$. Since C is due west of B, let $C = (-d,0)$ for some positive $d$. 4. **Find coordinates of A relative to B:** Bearing from B to A is 241°, which is $241° - 180° = 61°$ west of south. Convert bearing to vector direction: $$x_A = r \sin(241°) = r \sin(241°)$$ $$y_A = r \cos(241°) = r \cos(241°)$$ Using approximate values: $$\sin(241°) = \sin(180° + 61°) = -\sin(61°) \approx -0.8746$$ $$\cos(241°) = \cos(180° + 61°) = -\cos(61°) \approx -0.4848$$ So, $$A = (r \times -0.8746, r \times -0.4848)$$ 5. **Use AC = CB:** Distance $CB = d$ (since $C=(-d,0)$ and $B=(0,0)$). Distance $AC = |A - C| = \sqrt{(r(-0.8746) + d)^2 + (r(-0.4848) - 0)^2} = d$ Square both sides: $$(r(-0.8746) + d)^2 + (r(-0.4848))^2 = d^2$$ Expand: $$r^2 (0.8746^2 + 0.4848^2) + 2 r d (-0.8746) + d^2 = d^2$$ Simplify: $$r^2 (0.7649 + 0.2350) - 1.7492 r d + d^2 = d^2$$ $$r^2 (1.0) - 1.7492 r d + d^2 = d^2$$ Subtract $d^2$ from both sides: $$r^2 - 1.7492 r d = 0$$ Factor: $$r (r - 1.7492 d) = 0$$ Since $r \neq 0$, we have: $$r = 1.7492 d$$ 6. **Find coordinates of A:** $$A = (1.7492 d \times -0.8746, 1.7492 d \times -0.4848) = (-1.529 d, -0.848 d)$$ 7. **Find bearing of A from C:** Vector from C to A is: $$\overrightarrow{CA} = A - C = (-1.529 d + d, -0.848 d - 0) = (-0.529 d, -0.848 d)$$ Calculate angle $\theta$ from north (positive y-axis) clockwise to vector $\overrightarrow{CA}$: First find angle $\alpha$ between vector and positive x-axis: $$\alpha = \arctan\left( \frac{y}{x} \right) = \arctan\left( \frac{-0.848 d}{-0.529 d} \right) = \arctan(1.603) \approx 58.0°$$ Since both $x$ and $y$ are negative, vector is in third quadrant, so angle from positive x-axis is: $$180° + 58° = 238°$$ Bearing is measured clockwise from north (positive y-axis). The angle between positive y-axis and vector is: $$\text{bearing} = 180° + 58° - 90° = 148°$$ Alternatively, bearing from north clockwise is: $$\theta = 360° - (238° - 90°) = 360° - 148° = 212°$$ Check carefully: Vector angle from positive x-axis is 238°, so from positive y-axis (north) clockwise is: $$\text{bearing} = (360° + 90° - 238°) \mod 360° = 212°$$ 8. **Final answer:** The bearing of A from C is **212°**.