Ship Bearing 5Cd02C

1. **Problem Statement:** A ship travels $80m$ east from the port, then turns northeast (which is $45^\circ$ from east) and travels another $60m$. The ship is now $139m$ from the port. We need to find: a) The bearing from the port to the ship. b) The bearing from the ship back to the port. 2. **Step 1: Define the coordinate system and vectors** - Let the port be at the origin $(0,0)$. - The ship first travels $80m$ east, so its position after first leg is $(80,0)$. - Then it travels $60m$ northeast, which means $45^\circ$ from east. 3. **Step 2: Calculate the displacement vector of the second leg** The components of the second leg are: $$x_2 = 60 \cos 45^\circ = 60 \times \frac{\sqrt{2}}{2} = 30\sqrt{2}$$ $$y_2 = 60 \sin 45^\circ = 60 \times \frac{\sqrt{2}}{2} = 30\sqrt{2}$$ 4. **Step 3: Calculate the final position of the ship** $$x_{final} = 80 + 30\sqrt{2}$$ $$y_{final} = 0 + 30\sqrt{2}$$ 5. **Step 4: Verify the distance from port to ship** Distance $d$ is given by: $$d = \sqrt{x_{final}^2 + y_{final}^2}$$ Substitute values: $$d = \sqrt{(80 + 30\sqrt{2})^2 + (30\sqrt{2})^2}$$ Calculate: $$(80 + 30\sqrt{2})^2 = 80^2 + 2 \times 80 \times 30\sqrt{2} + (30\sqrt{2})^2 = 6400 + 4800\sqrt{2} + 1800$$ $$(30\sqrt{2})^2 = 30^2 \times 2 = 900 \times 2 = 1800$$ So, $$d^2 = 6400 + 4800\sqrt{2} + 1800 + 1800 = 10000 + 4800\sqrt{2}$$ Numerically, $$4800\sqrt{2} \approx 4800 \times 1.414 = 6787.2$$ So, $$d^2 \approx 10000 + 6787.2 = 16787.2$$ $$d \approx \sqrt{16787.2} \approx 129.56m$$ This is less than $139m$ given, so the problem's given distance $139m$ suggests the ship's position is slightly different or the direction is approximate. We will proceed to find the bearing using the given final distance $139m$. 6. **Step 5: Calculate the bearing from port to ship** Bearing is the angle clockwise from north to the line connecting port to ship. First, find the angle $\theta$ from east axis to the ship's position vector: $$\theta = \tan^{-1} \left( \frac{y_{final}}{x_{final}} \right) = \tan^{-1} \left( \frac{30\sqrt{2}}{80 + 30\sqrt{2}} \right)$$ Calculate denominator: $$80 + 30\sqrt{2} \approx 80 + 42.43 = 122.43$$ Calculate numerator: $$30\sqrt{2} \approx 42.43$$ So, $$\theta = \tan^{-1} \left( \frac{42.43}{122.43} \right) = \tan^{-1} (0.3465) \approx 19.1^\circ$$ This angle is from east towards north. Bearing from north clockwise is: $$\text{Bearing} = 90^\circ - \theta = 90^\circ - 19.1^\circ = 70.9^\circ$$ So the bearing from port to ship is approximately $70.9^\circ$. 7. **Step 6: Calculate the bearing from ship back to port** The bearing back is the opposite direction, which is $180^\circ$ opposite the vector from port to ship. So, $$\text{Bearing back} = (70.9^\circ + 180^\circ) \mod 360^\circ = 250.9^\circ$$ **Final answers:** - Bearing from port to ship: approximately $70.9^\circ$ - Bearing from ship back to port: approximately $250.9^\circ$