An aircraft covers a distance of 400km; 60 east of south and then flew 500km, 45 east of north. Calculate the approximate resultant displacement

Answer 1

60 east of south ---> S60°E ----> 330° in standard math notation
45 east of north ---> N45°E ----> 45° in standard math notation

using vectors
R = 400(cos330°, sin330°) + 500(cos45°,sin45°)
= (346.41.., -200) + (353.5534,353.5534)
= (699.9636, 153.5534)
|R| = appr 716.6 km
angle of R
tanØ = 153.5534/699.9636
= .2193...
Ø = appr 12.37°
or N 77.6° E

or by cosine law, after making a sketch and calculating the angles,
R^2 = 400^2 + 500^2 - 2(400)(500)cos105°
= 513,527.618
R = √513,527.618 = 716.6 km , just like above

Answer 2

All angles are measured CW from +y-axis.
Disp. = 400km[120o] + 500km[45o].
X = 400*sin120 + 500*sin45 = 700 km.
Y = 400*Cos120 + 500*Cos45 = 154 km.

Disp. = 700 + 154i = 717km[78o] CW.