Hello, sam314159265!

An airplane is flying at 550 km/hr on a bearing of 80°.

The wind is blowing at 60 km/hr from a bearing of 120°.

1) Draw a vector diagram

2) Find the ground velocity of the airplane Code:

C Q
P o *
* * * 60 :
: * * :
: * *60°:
: * * :
: * 40° o B
: * * 80°:
:* * 550 :
A o *
R

The plane flies from $\displaystyle A$ to $\displaystyle B\!:\;AB \,=\, 550,\;\angle PAB \,=\, 80^o \,=\,\angle ABR$

The wind is blowing from the southeast: .$\displaystyle BC \,=\,60,\;\angle QBC \,=\,60^o$

Hence: .$\displaystyle \angle ABC \,=\,40^o$

Law of Cosines: .$\displaystyle AC^2 \;=\;550^2 + 60^2 - 2(550)(60)\cos40^o \;=\;255,\!541,0668$

. . Hence: .$\displaystyle AC \;=\;505.6510699 \;\approx\;505.5 $ km/hr.

Law of Cosines: .$\displaystyle \cos(\angle CAB) \;=\;\frac{550^2 + 505.5^2 - 60^2}{2(550)(505.5)} \;=\;0.997087043$

. . Hence: .$\displaystyle \angle CAB \;=\;4.374216288^o \;\approx\;4.4^o$

Then: .$\displaystyle \angle PAC \;=\;\angle PAB - \angle CAB \;=\;80^o - 4.4^o \;=\;75.6^o$

Therefore, the plane is flying at 505.5 km/hr at a bearing of 75.6°.