SOLUTION: Two planes leave an airport at the same time, one flying due west at 500 km/h and the other flying due southeast at 300 km/h. What is the distance between the planes two hours late

Question 970320: Two planes leave an airport at the same time, one flying due west at 500 km/h and the other flying due southeast at 300 km/h. What is the distance between the planes two hours later. What do I do if there is no angle because I have to use trig. Please help.
Found 2 solutions by Boreal, josgarithmetic:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
W+++Start
;;;;;;;;+
;;;;;;;;;;;;;+SE
The angle between SE and W is 135 degrees. SE is half way between E (90) and S (180)
Two hours later, the plane flying west has gone 1000 km; the one SE 600 km. If they were in opposite directions, the distance would be 1600 km; the same direction and it would be 400 km, so the answer is between these values and closer to 1600 km by the picture above.
The angle between them is 135 degrees
The side you want is c.
a=1000
b=600
the angle between is theta, and it is 135 degrees cos (135)=-(sqrt(2)/2))
C^2=a^2 +b^2-2ab cos theta LAW OF COSINES
c^2=1000000+360000+2(1000)(600)(.7071) ; .7071 is sqrt (2)/2. The last term is positive, because the cosine is negative, and there are two negatives.
c^2=1360000+848520=2208520.
c=1486.1 miles.
This makes intuitive sense.

Answer by josgarithmetic(39625) (Show Source): You can put this solution on YOUR website!
One plane goes due west; and the other plane goes southeast. The angle is well specified.

Put the vectors on a cartesian system with departure location at the origin. West is in the direction of the negative x-axis. Due southeast is in the middle between the positive x-axis and the negative y-axis.
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