1. **State the problem:** We have points O, A, and B with distances OA = 20 km and OB unknown, but AB = 50 km. Bearings from O are 334° to A and 54° to B. We want the bearing of A from B. 2. **Convert bearings to angles relative to the east-west axis:** Bearings are measured clockwise from North. To work with vectors, convert bearings to standard angles measured counterclockwise from the positive x-axis (East). - Bearing of A from O is 334°, so angle from East is $$\theta_A = 90^\circ - 334^\circ = -244^\circ = 116^\circ$$ (adding 360° to get positive angle). - Bearing of B from O is 54°, so angle from East is $$\theta_B = 90^\circ - 54^\circ = 36^\circ$$. 3. **Find coordinates of A and B relative to O:** - Coordinates of A: $$A_x = 20 \cos 116^\circ, \quad A_y = 20 \sin 116^\circ$$ - Coordinates of B: Let $$OB = r$$ (unknown). Coordinates of B: $$B_x = r \cos 36^\circ, \quad B_y = r \sin 36^\circ$$ 4. **Use distance AB = 50 km to find r:** Distance formula: $$AB = \sqrt{(B_x - A_x)^2 + (B_y - A_y)^2} = 50$$ Substitute: $$\sqrt{(r \cos 36^\circ - 20 \cos 116^\circ)^2 + (r \sin 36^\circ - 20 \sin 116^\circ)^2} = 50$$ Square both sides: $$ (r \cos 36^\circ - 20 \cos 116^\circ)^2 + (r \sin 36^\circ - 20 \sin 116^\circ)^2 = 2500 $$ Expand and simplify: $$ r^2 (\cos^2 36^\circ + \sin^2 36^\circ) - 2r \times 20 (\cos 36^\circ \cos 116^\circ + \sin 36^\circ \sin 116^\circ) + 20^2 (\cos^2 116^\circ + \sin^2 116^\circ) = 2500 $$ Since $$\cos^2 \alpha + \sin^2 \alpha = 1$$, this reduces to: $$ r^2 - 40r \cos(116^\circ - 36^\circ) + 400 = 2500 $$ Calculate $$\cos(80^\circ) \approx 0.1736$$: $$ r^2 - 40r (0.1736) + 400 = 2500 $$ Simplify: $$ r^2 - 6.944r + 400 = 2500 $$ $$ r^2 - 6.944r - 2100 = 0 $$ 5. **Solve quadratic for r:** $$ r = \frac{6.944 \pm \sqrt{6.944^2 + 4 \times 2100}}{2} $$ Calculate discriminant: $$ 6.944^2 + 8400 = 48.22 + 8400 = 8448.22 $$ Square root: $$ \sqrt{8448.22} \approx 91.91 $$ Solutions: $$ r = \frac{6.944 \pm 91.91}{2} $$ Positive root: $$ r = \frac{6.944 + 91.91}{2} = \frac{98.854}{2} = 49.43 $$ Negative root is discarded as distance cannot be negative. 6. **Find coordinates of B:** $$ B_x = 49.43 \cos 36^\circ = 49.43 \times 0.8090 = 39.99 $$ $$ B_y = 49.43 \sin 36^\circ = 49.43 \times 0.5878 = 29.05 $$ 7. **Find vector from B to A:** $$ \overrightarrow{BA} = (A_x - B_x, A_y - B_y) $$ Calculate coordinates of A: $$ A_x = 20 \cos 116^\circ = 20 \times (-0.4384) = -8.77 $$ $$ A_y = 20 \sin 116^\circ = 20 \times 0.8988 = 17.98 $$ So: $$ \overrightarrow{BA} = (-8.77 - 39.99, 17.98 - 29.05) = (-48.76, -11.07) $$ 8. **Calculate bearing of A from B:** Find angle $$\phi$$ of vector $$\overrightarrow{BA}$$ relative to North (y-axis), clockwise. First, find angle $$\alpha$$ relative to East (x-axis): $$ \alpha = \arctan \left( \frac{y}{x} \right) = \arctan \left( \frac{-11.07}{-48.76} \right) = \arctan(0.227) = 12.9^\circ $$ Since both x and y are negative, vector is in the third quadrant, so add 180°: $$ \alpha = 12.9^\circ + 180^\circ = 192.9^\circ $$ Bearing is measured clockwise from North, so: $$ \text{bearing} = 90^\circ - \alpha = 90^\circ - 192.9^\circ = -102.9^\circ $$ Add 360° to get positive bearing: $$ 257.1^\circ $$ 9. **Final answer:** The bearing of A from B is approximately $$\boxed{257.1^\circ}$$ to one decimal place.