Two airplanes leave an airport at the same time. the velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . the velocity of the second is 590 m/h at a heading of 102◦ . how far apart are they after 1.7 h? answer in units of m.

Plane will 741.6959 m apart after 1.7 hour                    

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is   =

Speed of first plane  = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, representing the distance between the planes, we see that:

r = 741.6959 m

i. like it i'm gonna edit it tomorrow

pretty sure it's false