1. **Stating the problem:** Given a triangle HAL with points H (harbour), A (buoy), and L (lighthouse), bearings and distances are provided; we need to find bearings, distances, area, shortest distance from boat sailing on HAB to L, and the boat's departure time. 2. **Part (a): Bearings** (i) Bearing of H from A: - Given bearing of A from H is 042°. - Bearings are reciprocal: to find bearing from A to H, add 180° to 042° if less than 180°, otherwise subtract 180°. - Since 042° + 180° = 222°, bearing of H from A is $222^\circ$. (ii) Bearing of L from A: - Given angle $\angle HÂL = 115^\circ$ between HA and AL. - Bearing of A from H is 042°, so line HA is at 042° from North. - Since $\angle HÂL = 115^\circ$ at A, the line AL makes $115^\circ$ to HA clockwise. - Bearing from A to L = 042° (bearing of HA) + 115° = 157°. 3. **Part (b): Distances and area** (i) Calculate HL using cosine rule: $$HL^2 = HA^2 + AL^2 - 2\times HA \times AL \times \cos(\angle HAL)$$ - Angle $\angle HAL = 180^\circ - 115^\circ = 65^\circ$ (since angles at A and between H, A, L) - Substitute: $$HL^2 = 4.5^2 + 2.8^2 - 2 \times 4.5 \times 2.8 \times \cos65^\circ$$ - Calculate: $4.5^2=20.25$, $2.8^2=7.84$, $\cos65^\circ \approx 0.4226$ - So: $$HL^2 = 20.25 + 7.84 - 2 \times 4.5 \times 2.8 \times 0.4226 = 28.09 - 10.68 = 17.41$$ - Therefore, $$HL = \sqrt{17.41} \approx 4.17 \text{ km}$$. (ii) Area of triangle HAL using sine rule: - $$\text{Area} = \frac{1}{2} \times HA \times AL \times \sin(\angle HAL)$$ - $\sin65^\circ \approx 0.9063$ - Substitute: $$\text{Area} = 0.5 \times 4.5 \times 2.8 \times 0.9063 = 5.72\text{ km}^2$$ approximately. 4. **Part (c): Boat sailing on HAB** (i) Shortest distance from line HAB to L (perpendicular distance from L to line HAB): - Since HAB is a straight line and A and B are on it, use vector cross product area formula: - Area of triangle HAL = 0.5 * base * height where base is HL and height is distance from L to HAB - Rearranged: $$\text{height} = \frac{2 \times \text{Area}}{HA + AB}\text{ but AB unknown}\n- Alternatively, use $$\text{distance} = AL \times \sin(\angle HAL) = 2.8 \times \sin65^\circ = 2.8 \times 0.9063 = 2.54 \text{ km}$$ (ii) Time boat left harbour: - Speed = 3 m/s = 0.003 km/s - Distance HA = 4.5 km - Time taken to sail HA: $$\frac{4.5}{0.003} = 1500 s = \frac{1500}{60} = 25 \text{ minutes}$$ - Given boat reached A at 07:15, it left harbour at: - 07:15 minus 25 minutes = 06:50 Final answers: (a)(i): $222^\circ$, (a)(ii): $157^\circ$ (b)(i): $4.17$ km, (b)(ii): $5.72$ km$^2$ (c)(i): $2.54$ km, (c)(ii): 06:50