Two airplanes leave the airport at the same time. One plane is flying 700mph at a bearing of N46°W and the other plane is flying 895mph at a bearing of N43° E. After 3 hours, how far are the planes from one another?

9514 1404 393

  3380 miles

Step-by-step explanation:

The distance separating the planes can be found using the law of cosines. One plane has gone 3×700 = 2100 miles; the other has gone 3×895 = 2685 miles. The angle between their flight paths is 46°+43° = 89°. Then the separation (s) is ...

  s² = 2100² +2685² -2(2100)(2685)cos(89°) = 11,619,225 -11,277,000×0.01745

  s² ≈ 11,422,414.21

  s ≈ √11,422,414.21 ≈ 3379.7

The planes are separated by about 3380 miles after 3 hours.

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4

the arc length is found by finding the circumference of the full circle (2xpixradius) and then dividing by 4 to find just one quarter of the circle (90 degrees).