Question

Two airplanes leave an airport at the same time. The velocity of the first plane is 720 m/h at a heading 48.5°. The velocity of the second is 570 m/h at a heading of 148°. How far apart are they after 1.9 h ?
Answer in units of m.

Answer 1

Distance of the first plane $720\cdot 1.9=1368(m)$

Distance of the second plane $570\cdot 1.9=1083(m)$

The angle between them $148°-48.5°=99.5°$

Using the law of cosine $d^2=1368^2+1083^2-2\cdot1368\cdot1083\cdot \cos99.5° \to$

$d\approx1880(m)$