posted by bubba .
two airplanes leave an airport at the same time. the velocity of the first airplane is 700 m/h at a heading of 52.2degrees . the velocity of the second is 600 m/h at a heading of 97 degrees . how far apart are they after 3.1 h? answer in units of m.
The angle between them is the difference of the two bearings, each has a leg of velocity*time
Using the law of cosines...
c^2=leg1^2 + leg2^2 -2*leg1*lleg2*cosC
where angle C is the difference of headings.
not sure if i calculated right but is 1562 m the right answer : /
what were your legs,and the angle?
I will check those.