Question

An aeroplane flew from city G to city H on a bearing of 145 degree. The distance between G and H is 280 km, it then flew a distance of 430 km to city J on a bearing of 60 degree. Calculate: (a) The distance of G to J (b) How far North of H is J (c) How far west of H is G
In triangle GHJ,
∠H = 95°, so
(a) the distance GJ = side h, so
h^2 = 280^2 + 430^2 - 2*280*430 cos95°

If G is at (0,0) then
H is at (160.6, -229.4)
J is at (533.0, -14.36)
now finish it off
Given: GH = 280km[145o], HJ = 430km[60o].

a. GJ = 260km[145o]+430km[60o]
GJ = (260*sin145+430*sin60)+(260*c0s145+430*cos60)i
GJ = 522+2i = 522[0o] CW from +y-axis.

b. 360-145 = 215o.

c. 270-145 = 125o.
a. Correction: GJ = 522km[90o].

b. 145-90 = 55o.

c.

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