An aeroplane flew from city G to city H on a bearing of 24degree. The distance G and H is 250km. It then flew a distance of 180km to city J on a bearing of 055degree. calculate (a) the distance from G to J (b) how far north of H is J

Answer 1

No, the plane flew on a heading.
The bearing of H from G is 24°, so that's why they set the plane's heading to 24°

But enough of that. Draw a diagram. ∡GHJ = 149°
(a) using the law of cosines, we get the distance d is
d^2 = 250^2 + 180^2 - 2*250*180cos149°
d = 414.78km
(b) Now we need to extract the x- and y-displacements from the polar forms.
If we set G at (0,0) then J is at (249.1,331.6)

Answer 2

All angles are measured CW from +y-axis.
a. d = GH + HJ = 250km[24o] + 180km[55o].
X = 250*sin24 + 180*sin55 = 249.1 km.
Y = 250*Cos24 + 180*Cos55 = 331.6 km.
d = 249.1 + 331.6i = 414.7km[36.9o].

b. Y1 - Y2 = 250*Cos24 - 180*Cos55 = 228.4 - 103.2 = 125.2 km.
My calculation shows H 125km higher than J.