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.

To determine which plane is farther from the airstrip after the second leg of their flights, we will break down the movements of each plane into their respective components and calculate their final positions. ### Plane 1: 1. **First leg**: Flies west for 150 miles. - Position after first leg: \( (-150, 0) \) (in Cartesian coordinates, where west is negative x-direction and north is positive y-direction). 2. **Second leg**: Flies \( 30^{\circ} \) south of west for 220 miles. - The angle south of west means we need to find the x and y components: - \( x_2 = -220 \cos(30^{\circ}) \) - \( y_2 = -220 \sin(30^{\circ}) \) Using the values of cosine and sine: - \( \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \) - \( \sin(30^{\circ}) = \frac{1}{2} \) Therefore: - \( x_2 = -220 \cdot \frac{\sqrt{3}}{2} = -110\sqrt{3} \) - \( y_2 = -220 \cdot \frac{1}{2} = -110 \) Now, we can find the total position of Plane 1: - Total x-coordinate: \( -150 - 110\sqrt{3} \) - Total y-coordinate: \( -110 \) ### Plane 2: 1. **First leg**: Flies east for 220 miles. - Position after first leg: \( (220, 0) \). 2. **Second leg**: Flies \( x^{\circ} \) south of east for 150 miles. - The angle south of east means we need to find the x and y components: - \( x_2 = 150 \cos(x^{\circ}) \) - \( y_2 = -150 \sin(x^{\circ}) \) Therefore, the total position of Plane 2 is: - Total x-coordinate: \( 220 + 150 \cos(x^{\circ}) \) - Total y-coordinate: \( -150 \sin(x^{\circ}) \) ### Distance from the airstrip: Now we can calculate the distance from the airstrip for both planes using the distance formula \( d = \sqrt{x^2 + y^2} \). 1. **Distance for Plane 1**: \[ d_1 = \sqrt{(-150 - 110\sqrt{3})^2 + (-110)^2} \] 2. **Distance for Plane 2**: \[ d_2 = \sqrt{(220 + 150 \cos(x^{\circ}))^2 + (-150 \sin(x^{\circ}))^2} \] ### Comparison: To determine which plane is farther, we will compare \( d_1 \) and \( d_2 \). Let's calculate \( d_1 \) and \( d_2 \) using the expressions derived above. We will first calculate \( d_1 \) and then \( d_2 \) for \( x < 30 \). ### Calculating \( d_1 \): \[ d_1 = \sqrt{(-150 - 110\sqrt{3})^2 + (-110)^2} \] ### Calculating \( d_2 \): \[ d_2 = \sqrt{(220 + 150 \cos(x^{\circ}))^2 + (-150 \sin(x^{\circ}))^2} \] Now, let's compute these distances. Calculate the value by following steps: - step0: Calculate: \(\left(-150-110\sqrt{3}\right)^{2}+\left(-110\right)^{2}\) - step1: Simplify: \(\left(150+110\sqrt{3}\right)^{2}+110^{2}\) - step2: Evaluate the power: \(\left(150+110\sqrt{3}\right)^{2}+12100\) - step3: Evaluate the power: \(58800+33000\sqrt{3}+12100\) - step4: Add the numbers: \(70900+33000\sqrt{3}\) Calculate or simplify the expression \( (220 + 150*\cos(x))^2 + (-150*\sin(x))^2 \). Simplify the expression by following steps: - step0: Solution: \(\left(220+150\cos\left(x\right)\right)^{2}+\left(-150\sin\left(x\right)\right)^{2}\) - step1: Transform the expression: \(100\left(22+15\cos\left(x\right)\right)^{2}+22500\sin^{2}\left(x\right)\) - step2: Transform the expression: \(100\left(22+15\cos\left(x\right)\right)^{2}+22500-22500\cos^{2}\left(x\right)\) - step3: Transform the expression: \(70900+66000\cos\left(x\right)\) We have calculated the distances for both planes: 1. **Distance for Plane 1**: \[ d_1 = \sqrt{70900 + 33000\sqrt{3}} \] 2. **Distance for Plane 2**: \[ d_2 = \sqrt{70900 + 66000\cos(x)} \] ### Comparison of Distances: To determine which plane is farther from the airstrip, we need to compare \( d_1^2 \) and \( d_2^2 \) (since the square root function is increasing, comparing the squares will yield the same result). - \( d_1^2 = 70900 + 33000\sqrt{3} \) - \( d_2^2 = 70900 + 66000\cos(x) \) ### Analyzing the Comparison: We need to compare: \[ 70900 + 33000\sqrt{3} \quad \text{and} \quad 70900 + 66000\cos(x) \] This simplifies to comparing: \[ 33000\sqrt{3} \quad \text{and} \quad 66000\cos(x) \] ### Since \( x < 30^{\circ} \): - The maximum value of \( \cos(x) \) when \( x < 30^{\circ} \) is \( \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \). - Therefore, \( 66000\cos(x) \) can be at most \( 66000 \cdot \frac{\sqrt{3}}{2} = 33000\sqrt{3} \). ### Conclusion: Since \( 33000\sqrt{3} \) is equal to \( 66000\cos(x) \) when \( x = 30^{\circ} \) and \( \cos(x) \) decreases as \( x \) decreases from \( 30^{\circ} \), we can conclude that: - For \( x < 30^{\circ} \), \( 66000\cos(x) < 33000\sqrt{3} \). Thus, we find that: \[ d_1 > d_2 \] ### Final Answer: **Plane 1 is farther from the airstrip after the second leg.**