SOLUTION: A plane leaves the airport and travels S 46 degrees E at 440 mph. Another plane leaves the same airport half an hour later and travels N 28 degrees E at 475 mph. How far apart are
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Question 1105642: A plane leaves the airport and travels S 46 degrees E at 440 mph. Another plane leaves the same airport half an hour later and travels N 28 degrees E at 475 mph. How far apart are the planes 1.5 hours after the second plane leave the airport.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
plane leaves the airport and travels S 46 degrees E at 440 mph.
Another plane leaves the same airport half an hour later and travels N 28 degrees E at 475 mph.
How far apart are the planes 1.5 hours after the second plane leave the airport.
:
find the angle between the paths of the two planes.
Convert to true bearing, (ref to 0 degrees)
S 46E: 180 - 46 = 134 degree
N 28E: 28 degrees
134 - 28 = 106 degrees is angle between the paths of the two planes
:
1.5 + .5 = 2 hrs is the flying time of the 1st plane
then
2(440) = 880 mi flown by the 1st plane
and
1.5(475) = 712.5 mi flown by the 2nd plane
:
Use the law of cosines: a^2 = b^2 + c^2 - 2(b*c)*cos(A), where
A = 106 degrees
a = distance between the two planes
b = 880
c = 712.5
:
a^2 = 880^2 + 712^2 - 2(880*712.5)*cos(106)
I'll let you do the tedious math here, I got
a = 1275.8 mi apart
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