Hello, Stealth!

Sorry, your answers are off . . .

A Boeing 747 Jumbo Jet maintains an airspeed of 550 mph in a SW direction.

The velocity of the jet stream is a constant 80 mph from the direction of 260°.

Find the actual speed and direction of the plane. Code:

N
:
:
W - - - - - - - - oA
C o * :
*45°:
* :
* :
* :
* :
* :
* :
o :
B S

Draw segments $\displaystyle AC$ and $\displaystyle BC$.

. . (Hope you can follow my labeling.)

The plane is flying from $\displaystyle A$ to $\displaystyle B$: .$\displaystyle AB = 550,\;\angle BAS = 45^o$

The wind blows from $\displaystyle C$ to $\displaystyle A$: .$\displaystyle CA = 80$

Major angle $\displaystyle NAC = 260^o\quad\Rightarrow\quad \angle WAC = 10^o \quad\Rightarrow\quad \angle CAB = 35^o$

In $\displaystyle \Delta CAB\!:\;CB^2 \:=\:CA^2 + AB^2 - 2(CA)(AB)\cos(\angle CAB) $

. . . . . . . .$\displaystyle CB^2\:=\: 80^2 + 550^2 - 2(80)(55)\cos35^o \:=\:236814.6201$

Therefore: .$\displaystyle \boxed{CB \;\approx\;486.6\text{ mph}}$

We have: .$\displaystyle \cos(\angle ACB) \:=\:\frac{CA^2 + CB^2 - AB^2}{2(CA)(CB)} \:=\:\frac{80^2+486.6^2 - 550^2}{2(80)(486.6)} \:=\:-0.761925092$

Hence: .$\displaystyle \angle ACB \;\approx\;139.6^o$

Therefore, the bearing of $\displaystyle CB$ is about: .$\displaystyle \boxed{219.6^o}$