1. **State the problem:** We need to find the actual speed and direction of an airplane flying with a velocity of 580 mph headed S 55° W, while there is a wind blowing from the southeast at 80 mph. 2. **Set up the vectors:** - Airplane velocity relative to air: $580$ mph, direction S 55° W. - Wind velocity: $80$ mph from southeast, which means wind is blowing towards northwest (NW). 3. **Convert directions to vector components:** - For the airplane: S 55° W means 55° west of south. Using standard coordinate system (x east, y north): $$V_{plane} = 580(\sin 55^\circ, -\cos 55^\circ)$$ $$= 580(0.819, -0.574) = (474.02, -332.92)$$ - For the wind: from southeast means wind blows towards northwest (NW), which is 45° north of west. So wind vector points NW at 80 mph: $$V_{wind} = 80(-\cos 45^\circ, \sin 45^\circ) = 80(-0.707, 0.707) = (-56.57, 56.57)$$ 4. **Find the resultant velocity vector:** $$V_{result} = V_{plane} + V_{wind} = (474.02 - 56.57, -332.92 + 56.57) = (417.45, -276.35)$$ 5. **Calculate the magnitude (speed) of the resultant vector:** $$|V_{result}| = \sqrt{417.45^2 + (-276.35)^2} = \sqrt{174253 + 76300} = \sqrt{250553} = 500.55$$ mph (rounded to nearest thousandth) 6. **Calculate the direction:** Direction angle $\theta$ relative to east axis: $$\theta = \tan^{-1}\left(\frac{-276.35}{417.45}\right) = \tan^{-1}(-0.662) = -33.52^\circ$$ Since x-component is positive and y-component negative, vector is in the fourth quadrant, so direction is 33.52° south of east. 7. **Convert to compass bearing:** East 33.52° South is written as S 56.48° E (since 90° - 33.52° = 56.48°). **Final answer:** The actual speed of the plane is approximately **500.550 mph**. The direction is approximately **S 56.48° E**.