Hello, jordangiscool!
A plane flies 810 miles from A to B with a bearing of N 75° E.
Then it flies 648 miles from B to C with a bearing of N 32° E.
Find the straight-line distance and bearing from C to A Code:
Q o C
: *
:32°*
: *
: *
P 105° |*
: o B
: * :
: * 75° :
: 75° * :
: * R
A o
The plane flies from $\displaystyle A$ to $\displaystyle B\!:\;\;AB \,=\,810$
. . $\displaystyle \angle PAB \,=\, 75^o \,=\,\angle ABR \quad\Rightarrow\quad \angle ABQ \,=\,105^o$
Then it flies from $\displaystyle B$ to $\displaystyle C\!:\;\;BC \,=\,648$
. . $\displaystyle \angle QBC \,=\,32^o \quad\Rightarrow\quad \angle ABC \,=\,137^o$
Draw line segment $\displaystyle AC.$
We have .$\displaystyle \Delta ABC\!:\;\;AB = 810,\;BC = 648,\;\angle B = 137^o$
Law of Cosines: .$\displaystyle AC^2 \;=\;AB^2 + BC^2 - 2(AB)(BC)\cos B$
Got it?