1. **Problem statement:** An airplane leaves Manila and flies 300 km on a bearing N 55° W, then flies 200 km on a bearing S 25° W. We need to find the distance from Manila after the second leg and the bearing from Manila. 2. **Understanding bearings:** - Bearing N 55° W means starting north and turning 55° towards the west. - Bearing S 25° W means starting south and turning 25° towards the west. 3. **Convert bearings to standard angles:** - N 55° W corresponds to an angle of 360° - 55° = 305° from the east counterclockwise. - S 25° W corresponds to 180° + 25° = 205°. 4. **Calculate components of each leg:** For leg 1 (300 km, 305°): $$x_1 = 300 \cos 305^\circ = 300 \cos (360^\circ - 55^\circ) = 300 \cos 55^\circ \approx 300 \times 0.5736 = 172.08$$ $$y_1 = 300 \sin 305^\circ = 300 \sin (360^\circ - 55^\circ) = 300 (-\sin 55^\circ) \approx 300 \times (-0.8192) = -245.76$$ For leg 2 (200 km, 205°): $$x_2 = 200 \cos 205^\circ = 200 \cos (180^\circ + 25^\circ) = 200 (-\cos 25^\circ) \approx 200 \times (-0.9063) = -181.26$$ $$y_2 = 200 \sin 205^\circ = 200 \sin (180^\circ + 25^\circ) = 200 (-\sin 25^\circ) \approx 200 \times (-0.4226) = -84.52$$ 5. **Sum components to find resultant vector:** $$x = x_1 + x_2 = 172.08 - 181.26 = -9.18$$ $$y = y_1 + y_2 = -245.76 - 84.52 = -330.28$$ 6. **Calculate distance from Manila:** $$d = \sqrt{x^2 + y^2} = \sqrt{(-9.18)^2 + (-330.28)^2} = \sqrt{84.27 + 109084.58} = \sqrt{109168.85} \approx 330.5 \text{ km}$$ 7. **Calculate bearing from Manila:** - Find angle $\theta$ from east axis: $$\theta = \arctan \left( \frac{y}{x} \right) = \arctan \left( \frac{-330.28}{-9.18} \right) = \arctan(35.97) \approx 88.4^\circ$$ - Since $x<0$ and $y<0$, the vector is in the third quadrant. - Bearing is measured clockwise from north, so: $$\text{Bearing} = 180^\circ + (90^\circ - 88.4^\circ) = 180^\circ + 1.6^\circ = 181.6^\circ$$ - Convert to compass bearing: Since the vector points slightly west of south, bearing is S 1.6° W. **Final answers:** - Distance from Manila after second leg is approximately **330.5 km**. - Bearing from Manila is approximately **S 1.6° W**.