Two airplanes leave an airport at the same

Question

time. The velocity of the first airplane is
730 m/h at a heading of 44.6 degrees.

The velocity
of the second is 610 m/h at a heading of 100 degrees.

How far apart are they after 2.3 h?
Answer in units of m.

Answer 1

remember the law of cosines?
the angle between the two headings is 55.4°

So, figuring distance = speed * time, we have your distance apart (z) after 2.3 hours is

z^2 = (730*2.3)^2 + (610*2.3)^2 - 2(730*2.3)(610*2.3)cos55.4°
z = 1453 m

I suspect you meant mi (miles), not m (meters), since no plane could get off the ground at 600 meters/hr.

Of course, in that case, they are flying faster than any commercial airliners in use today!

Answer 2

Thank you, but it did actually say meters haha