Question

1. two planes take off from the same airstrip. the first plane flies west for 150 miles and then flies 30° south of west for 220 miles. the second plane flies east for 220 miles and then flies x° south of east for 150 miles. if x < 30, which plane is farther from the airstrip after the second leg? justify your answer.

Explanation:

Step1: Define coordinate system

Set airstrip at $(0,0)$, east = +x, north = +y.

Step2: Calculate Plane 1 coordinates

First leg: west 150 miles: $(-150, 0)$
Second leg: $30^\circ$ south of west, 220 miles:
$\Delta x = -220\cos(30^\circ) = -220 \cdot \frac{\sqrt{3}}{2} = -110\sqrt{3}$
$\Delta y = -220\sin(30^\circ) = -220 \cdot \frac{1}{2} = -110$
Total coordinates: $(-150 - 110\sqrt{3}, -110)$

Step3: Plane 1 distance squared

Use distance formula $d^2 = x^2 + y^2$:
$d_1^2 = (-150 - 110\sqrt{3})^2 + (-110)^2$
Expand: $150^2 + 2 \cdot 150 \cdot 110\sqrt{3} + (110\sqrt{3})^2 + 110^2$
$= 22500 + 33000\sqrt{3} + 36300 + 12100$
$= 70900 + 33000\sqrt{3}$

Step4: Calculate Plane 2 coordinates

First leg: east 220 miles: $(220, 0)$
Second leg: $x^\circ$ south of east, 150 miles:
$\Delta x = 150\cos(x^\circ)$, $\Delta y = -150\sin(x^\circ)$
Total coordinates: $(220 + 150\cos(x^\circ), -150\sin(x^\circ))$

Step5: Plane 2 distance squared

$d_2^2 = (220 + 150\cos(x^\circ))^2 + (-150\sin(x^\circ))^2$
Expand: $220^2 + 2 \cdot 220 \cdot 150\cos(x^\circ) + (150\cos(x^\circ))^2 + (150\sin(x^\circ))^2$
Use $\cos^2\theta + \sin^2\theta = 1$:
$= 48400 + 66000\cos(x^\circ) + 150^2(\cos^2(x^\circ) + \sin^2(x^\circ))$
$= 48400 + 66000\cos(x^\circ) + 22500 = 70900 + 66000\cos(x^\circ)$

Step6: Compare $d_1^2$ and $d_2^2$

Since $x < 30^\circ$, $\cos(x^\circ) > \cos(30^\circ) = \frac{\sqrt{3}}{2}$
So $66000\cos(x^\circ) > 66000 \cdot \frac{\sqrt{3}}{2} = 33000\sqrt{3}$
Thus $d_2^2 = 70900 + 66000\cos(x^\circ) > 70900 + 33000\sqrt{3} = d_1^2$
Since distance is positive, $d_2 > d_1$

Answer:

The second plane is farther from the airstrip.