1. **State the problem:** A plane is flying east at 150 mi/h, and there is a crosswind blowing north at 30 mi/h. We need to find the plane's actual speed and direction. 2. **Formula used:** The actual velocity is the vector sum of the plane's velocity and the wind velocity. The magnitude (speed) is given by the Pythagorean theorem: $$v = \sqrt{v_x^2 + v_y^2}$$ where $v_x = 150$ mi/h (east) and $v_y = 30$ mi/h (north). The direction angle $\theta$ north of east is given by: $$\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)$$ 3. **Calculate the magnitude:** $$v = \sqrt{150^2 + 30^2} = \sqrt{22500 + 900} = \sqrt{23400}$$ 4. **Simplify the magnitude:** $$v = \sqrt{23400} \approx 153.0$$ 5. **Calculate the direction angle:** $$\theta = \tan^{-1}\left(\frac{30}{150}\right) = \tan^{-1}(0.2)$$ 6. **Evaluate the angle:** $$\theta \approx 11^\circ$$ 7. **Interpretation:** The plane's actual speed is approximately 153 mi/h, and its direction is $11^\circ$ north of east. **Final answer:** 153 mi/h; N 11° E