1. **Problem Statement:** We need to find the final position of a fishing boat that traveled in two legs with given bearings and distances, then calculate: - The straight-line distance from the command post to the boat. - The bearing from the command post to the boat. - The bearing from the boat back to the command post. 2. **Given Data:** - First leg: 40 km on bearing N 35° E. - Second leg: 28 km on bearing E 50° S. 3. **Step 1: Convert bearings to standard angles and find components.** - Bearing N 35° E means 35° east of north. So angle from east axis (x-axis) is $90^\circ - 35^\circ = 55^\circ$. - First leg components: $$x_1 = 40 \times \cos 55^\circ$$ $$y_1 = 40 \times \sin 55^\circ$$ - Bearing E 50° S means 50° south of east. So angle from east axis is $-50^\circ$ (since south is negative y direction). - Second leg components: $$x_2 = 28 \times \cos (-50^\circ)$$ $$y_2 = 28 \times \sin (-50^\circ)$$ 4. **Step 2: Calculate numerical values of components.** - Using approximate values: $$\cos 55^\circ \approx 0.5736, \sin 55^\circ \approx 0.8192$$ $$\cos (-50^\circ) = \cos 50^\circ \approx 0.6428, \sin (-50^\circ) = -\sin 50^\circ \approx -0.7660$$ - First leg: $$x_1 = 40 \times 0.5736 = 22.944$$ $$y_1 = 40 \times 0.8192 = 32.768$$ - Second leg: $$x_2 = 28 \times 0.6428 = 18.0$$ $$y_2 = 28 \times (-0.7660) = -21.448$$ 5. **Step 3: Find total displacement components.** $$x = x_1 + x_2 = 22.944 + 18.0 = 40.944$$ $$y = y_1 + y_2 = 32.768 - 21.448 = 11.32$$ 6. **Step 4: Calculate straight-line distance from command post to boat.** Using Pythagoras theorem: $$d = \sqrt{x^2 + y^2} = \sqrt{40.944^2 + 11.32^2}$$ $$d = \sqrt{1676.9 + 128.1} = \sqrt{1805} \approx 42.5 \text{ km}$$ 7. **Step 5: Calculate bearing from command post to boat.** Bearing is angle from north clockwise to the line. - Calculate angle $\theta$ from east axis: $$\theta = \arctan \left( \frac{y}{x} \right) = \arctan \left( \frac{11.32}{40.944} \right) \approx 15.5^\circ$$ - Since $x > 0$ and $y > 0$, point is in first quadrant. - Bearing from north is: $$90^\circ - \theta = 90^\circ - 15.5^\circ = 74.5^\circ$$ - So bearing is N 74.5° E. 8. **Step 6: Calculate bearing from boat back to command post.** This is the reverse direction. - Reverse vector components: $$-x = -40.944, -y = -11.32$$ - Calculate angle from east axis: $$\phi = \arctan \left( \frac{-11.32}{-40.944} \right) = \arctan \left( 0.276 \right) = 15.5^\circ$$ - Since both components are negative, vector is in third quadrant. - Angle from east axis is $180^\circ + 15.5^\circ = 195.5^\circ$. - Bearing from north clockwise: $$195.5^\circ - 90^\circ = 105.5^\circ$$ - This corresponds to S 105.5° W, but bearings are usually given as S x° W where x is less than 90°. - Alternatively, from south axis: $$105.5^\circ - 90^\circ = 15.5^\circ$$ - So bearing is S 15.5° W. **Final answers:** - Total distance from command post to boat: **42.5 km** - Bearing from command post to boat: **N 74.5° E** - Bearing from boat back to command post: **S 15.5° W**