1. **Problem statement:** We have a triangle formed by points A, B, and L. The ship sails from A to B due North, so AB = 3 km vertically. The lighthouse L is observed from A at a bearing of N 42.7° E, meaning the angle between the north line and line AL is 42.7° to the east. The distance from B to L is 2.93 km. We need to find: (i) The angle $\angle ABL$ to one decimal place. (ii) The time taken to sail from A to B at 23 km/hr, rounded to the nearest minute. 2. **Understanding bearings and angles:** Bearing N 42.7° E means starting from north (line AB), rotate 42.7° towards east to get line AL. 3. **Set up the triangle:** - AB is vertical, length 3 km. - AL makes a 42.7° angle with AB. - BL = 2.93 km. 4. **Find coordinates for points:** Place A at origin (0,0). Since AB is due north, B is at (0,3). Point L is at distance AL from A, but AL is unknown. However, we can find AL using the Law of Cosines in triangle ABL. 5. **Use Law of Cosines to find AL:** In triangle ABL, sides AB = 3 km, BL = 2.93 km, and angle at A between AB and AL is 42.7°. Law of Cosines: $$ BL^2 = AB^2 + AL^2 - 2 \times AB \times AL \times \cos(42.7^\circ) $$ Rearranged to solve for AL: $$ AL^2 - 2 \times AB \times AL \times \cos(42.7^\circ) + AB^2 - BL^2 = 0 $$ Substitute values: $$ AL^2 - 2 \times 3 \times AL \times \cos(42.7^\circ) + 3^2 - 2.93^2 = 0 $$ Calculate constants: $$ 3^2 = 9 $$ $$ 2.93^2 = 8.5849 $$ $$ \cos(42.7^\circ) \approx 0.7357 $$ So: $$ AL^2 - 2 \times 3 \times AL \times 0.7357 + 9 - 8.5849 = 0 $$ $$ AL^2 - 4.4142 AL + 0.4151 = 0 $$ 6. **Solve quadratic equation for AL:** $$ AL = \frac{4.4142 \pm \sqrt{4.4142^2 - 4 \times 0.4151}}{2} $$ Calculate discriminant: $$ 4.4142^2 = 19.489 $$ $$ 4 \times 0.4151 = 1.6604 $$ $$ \sqrt{19.489 - 1.6604} = \sqrt{17.8286} = 4.222 $$ So: $$ AL = \frac{4.4142 \pm 4.222}{2} $$ Two solutions: $$ AL_1 = \frac{4.4142 + 4.222}{2} = \frac{8.6362}{2} = 4.3181 $$ $$ AL_2 = \frac{4.4142 - 4.222}{2} = \frac{0.1922}{2} = 0.0961 $$ Since AL must be longer than AB (3 km) to make sense with the given distances, choose: $$ AL = 4.3181 \text{ km} $$ 7. **Find coordinates of L:** From A (0,0), L is at distance 4.3181 km at 42.7° east of north. Coordinates: $$ x_L = AL \times \sin(42.7^\circ) = 4.3181 \times 0.6780 = 2.926 \text{ km} $$ $$ y_L = AL \times \cos(42.7^\circ) = 4.3181 \times 0.7357 = 3.177 \text{ km} $$ 8. **Find angle $\angle ABL$:** At point B (0,3), vectors: - BA = A - B = (0 - 0, 0 - 3) = (0, -3) - BL = L - B = (2.926 - 0, 3.177 - 3) = (2.926, 0.177) Use dot product formula: $$ \cos(\angle ABL) = \frac{BA \cdot BL}{|BA||BL|} $$ Calculate dot product: $$ BA \cdot BL = 0 \times 2.926 + (-3) \times 0.177 = -0.531 $$ Magnitudes: $$ |BA| = 3 $$ $$ |BL| = 2.93 \text{ (given)} $$ So: $$ \cos(\angle ABL) = \frac{-0.531}{3 \times 2.93} = \frac{-0.531}{8.79} = -0.0604 $$ Angle: $$ \angle ABL = \cos^{-1}(-0.0604) = 93.5^\circ $$ 9. **Time to sail from A to B:** Distance AB = 3 km Speed = 23 km/hr Time: $$ t = \frac{3}{23} = 0.1304 \text{ hours} $$ Convert to minutes: $$ 0.1304 \times 60 = 7.82 \text{ minutes} $$ Rounded to nearest minute: $$ 8 \text{ minutes} $$ **Final answers:** (i) $\angle ABL = 93.5^\circ$ (ii) Time = 8 minutes