Problem 7 An airplane flies 220 miles with a bearing of \( 40^{\circ} \), and then flies 180 miles with a bearing of \( 170^{\circ} \). How far is the plane from its starting point, and at what bearing? Round answers to the nearest tenth.

To solve this problem, we first break down the movements of the airplane into their components using trigonometry. 1. **First Leg**: The airplane flies 220 miles at a bearing of \( 40^{\circ} \). - The angle with respect to the east (x-direction) is \( 40^{\circ} \). - The x-component: \( 220 \cos(40^{\circ}) \approx 220 \times 0.7660 \approx 168.5 \) miles. - The y-component: \( 220 \sin(40^{\circ}) \approx 220 \times 0.6428 \approx 141.4 \) miles. 2. **Second Leg**: The airplane then flies 180 miles at a bearing of \( 170^{\circ} \). - The angle with respect to the east is \( 180^{\circ} - 170^{\circ} = 10^{\circ} \). - The x-component: \( 180 \cos(10^{\circ}) \approx 180 \times 0.9848 \approx 177.3 \) miles. - The y-component: \( 180 \sin(10^{\circ}) \approx 180 \times 0.1736 \approx 31.2 \) miles. 3. **Total Components**: - Total x: \( 168.5 + (-177.3) \approx -8.8 \) miles (negative direction). - Total y: \( 141.4 + (-31.2) \approx 110.2 \) miles. 4. **Distance from Starting Point**: Using the Pythagorean theorem to find the overall distance: \[ d = \sqrt{(-8.8)^2 + (110.2)^2} \approx \sqrt{77.44 + 12144.04} \approx \sqrt{12221.48} \approx 110.6 \text{ miles}. \] 5. **Bearing from Starting Point**: The bearing is found using the arctangent: \[ \tan(\theta) = \frac{y}{-x} = \frac{110.2}{8.8} \Rightarrow \theta \approx \tan^{-1}(12.5) \approx 85.4^\circ. \] Since the x-component is negative and the y-component is positive, the angle from the north is: \[ 180^{\circ} - 85.4^{\circ} \approx 94.6^{\circ}, \text{ (or bear it in the second quadrant)} \] Thus, the airplane is approximately **110.6 miles** from the starting point at a bearing of approximately **94.6 degrees**.