FLIGHT Two planes take off from the same airstrip. The first plane flies west for 150 miles and then flies \( 30^{\circ} \) south of west for 220 miles. The second plane flies east for 220 miles and then flies \( x^{\circ} \) south of east for 150 miles. If \( x<30 \), which plane is farther from the airstrip after the second leg? Justify your answer.
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To compare the distances of the two planes from the airstrip, we can use some trigonometry to determine their final positions. First plane: - It flies 150 miles west: Position = (-150, 0). - Next, it flies 220 miles at \( 30^{\circ} \) south of west. We can break this down: - The westward component: \( 220 \cdot \cos(30^{\circ}) = 220 \cdot \frac{\sqrt{3}}{2} \approx 190.52 \) miles. - The southward component: \( 220 \cdot \sin(30^{\circ}) = 220 \cdot \frac{1}{2} = 110 \) miles. Combining these, the new position of the first plane is: - \( x \) position: -150 - 190.52 = -340.52 - \( y \) position: -110 So, the coordinates are roughly (-340.52, -110). Second plane: - It flies 220 miles east: Position = (220, 0). - Next, it flies 150 miles at \( x^{\circ} \) south of east. Breaking this down: - The eastward component: \( 150 \cdot \cos(x^{\circ}) \) - The southward component: \( 150 \cdot \sin(x^{\circ}) \) Thus, the new position of the second plane is: - \( x \) position: \( 220 + 150 \cdot \cos(x^{\circ}) \) - \( y \) position: \( -150 \cdot \sin(x^{\circ}) \) To find the distances from the origin to each plane's position, we use the distance formula \( \sqrt{x^2 + y^2} \). Distance of the first plane: \[ D_1 = \sqrt{(-340.52)^2 + (-110)^2} \approx \sqrt{116,955.79 + 12,100} \approx \sqrt{129,055.79} \approx 359.64 \text{ miles} \] Distance of the second plane: \[ D_2 = \sqrt{(220 + 150 \cdot \cos(x))^2 + (150 \cdot \sin(x))^2} \] Since \( x < 30 \), \( \cos(x) \) will remain positive and \( 150 \cdot \cos(x) \) will be a relatively larger adjustment on the eastward position compared to its negative southward displacement. By checking the maximum distance when \( x = 0 \) and the angle increases (to 30) yields, - At \( x = 0 \), the second plane's distance is 370 miles or more when x increases from 0 to slightly lower than 30, - The distance moves towards less than second plane’s maximum adjustment south, and thus remains within those bounds. Therefore, the first plane leads on initial leg distance, but with the second leg closer on adjustments, the second's overall distance begins further beyond the limits. Thus, the second plane is farther from the airstrip after the second leg.