Question

two airplanes leave an airport at the same time. after one hour, airplane a is 315 kilometers away from the airport, airplane b is 405 kilometers away from the airport, and the airplanes are 285 kilometers apart. what is the approximate angle between the flight paths of the airplanes?

a. 30.1°

b. 84.8°

c. 50.8°

d. 44.5°

Explanation:

Step1: Identify triangle sides

Let airport be $O$, $OA=315$ km, $OB=405$ km, $AB=265$ km. Use Law of Cosines: $AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cdot \cos\theta$

Step2: Rearrange for $\cos\theta$

$\cos\theta = \frac{OA^2 + OB^2 - AB^2}{2 \cdot OA \cdot OB}$

Step3: Calculate numerator

$315^2 + 405^2 - 265^2 = 99225 + 164025 - 70225 = 193025$

Step4: Calculate denominator

$2 \cdot 315 \cdot 405 = 258300$

Step5: Compute $\cos\theta$ and $\theta$

$\cos\theta = \frac{193025}{258300} \approx 0.747$, so $\theta \approx \arccos(0.747) \approx 41.6^\circ$? Wait, recalculate: $315^2=99225$, $405^2=164025$, sum=263250; $265^2=70225$; 263250-70225=193025. Denominator=2315405=258300. 193025/258300≈0.747. $\arccos(0.747)≈41.6^\circ$? Wait, check options: closest is D (44.5°)? Wait, maybe miscalculation: $265^2=70225$, $315^2=99225$, $405^2=164025$. $99225+164025=263250$. 263250-70225=193025. 2315405=258300. 193025/258300≈0.747. $\arccos(0.747)≈41.6$, but option D is 44.5. Wait, maybe I mixed up sides? No, Law of Cosines is correct for angle at O. Alternatively, maybe approximate: 0.747 is close to cos41°≈0.7547, cos44°≈0.7193. 0.747 is between cos41 and cos44, so ~42°, but closest option is D (44.5°). Maybe rounding differences: let's compute more accurately: 193025/258300=0.74728997. $\arccos(0.74729)≈41.6°$, but since options don't have that, maybe typo? Wait no, wait the problem says "approximate", so D is the closest.

Answer:

D. 44.5°