Question

An aircraft flies round a triangular course. The fist leg is 200km on a bearing of 115degrees .the second leg is 150km on a bearing of 230degrees. How long is the third leg of the course and what bearing must The aircraft fly

Answer 1

make your sketch
On mine, using the cosine law, I get
x^2 = 200^2 + 150^2 - 2(200)(15)cos65°
..x = 192.72.. km

then by the sine law,
sinø/150 = sin65/192.72..
ø = 44.86..°
I will let you translate that angle into your bearing notation

other way.... using vectors and using standard trig notation
bearing of 115degrees ---> 25°
bearing of 230degrees ---> 220°
vector r = (200cos-25, 200sin-25) + (150cos220, 150sin220)
= (66.3548..., -180.9417...)
magnitude = √(66.3548...)^2 + (-180.9417...)^2 ) = 192.72 , just as before
angle...
tanx = -180.9417.../66.3548...
x = -68.86° translate that into your bearing notation

Answer 2

All angles are measured CW from +y-axis.
D = 200[115o]+150[230o]
D = (200*sin115+150*sin230)+(200*cos115+150*cos230)i
D = 66.4-181i = 193km[20o].

Bearing(direction) = 180+20 = 200o.

Answer 3

Thanks I really appreciate