Two small planes leave the Abbotsford airport at the same time. The first flies at 225km/h at a heading 320°, while the second flies at 190km/h at the heading of 70°. How far apart are they after 2 hours?

Answer 1

320 is 40 degrees west of north
north speed= +225 cos 40
east speed = -225 sin 40

70 is 70 degrees east of north
north speed= 190 cos 70
east speed = 190 sin 70

difference in north speed
=225cos40-190 sin70
=-6.18 km/h

difference in east speed
=-225sin40-190sin 70
=-323.17 km/h

magnitude of difference in speed
= sqrt (6.18^2+323^2)
= 323 km/h speed apart
* 2 hours
= 646 km

Answer 2

The angle between their flight paths is 110°
I sketched the triangle and used the cosine law.
x^2 = 450^2 + 380^2 - 2(4500)(380)cos110°
= 463870.889
x = √463870.889 = appr 681 km

or using vectors and the actual angles

let v = (450cos320, 450sin320) = (344.72,-289.25)
let u = (380cos70, 380sin70) = (129.97, 357.08)
let w be the vector joining the end of v to the start of w

then v + w - u = 0
w = u - v = (129.97, 357.08) - (344.72,-289.25)
= (-214.75, 646.34)

|w| = √(214.75^2 + 646.34^2) = appr 681 km , as before