# Math Help - [SOLVED] Problem having to do with bearing and planes

1. ## [SOLVED] Problem having to do with bearing and planes

Hi Everybody!

I have this math problem and I solved it but I'm not sure if I solved it correctly as finding the angle of the bearing always messes me up!:

A Boeing 747 Jumbo Jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant 80 mph from the direction of 260 degrees. Find the actual speed and direction of the plane.

Airplane Speed: 617.24 mph and the angle at which it is flying is 49.345 degrees.

Any help is appreciated,
Stealth

2. Hello, Stealth!

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 $AC$ and $BC$.
. . (Hope you can follow my labeling.)

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

The wind blows from $C$ to $A$: . $CA = 80$
Major angle $NAC = 260^o\quad\Rightarrow\quad \angle WAC = 10^o \quad\Rightarrow\quad \angle CAB = 35^o$

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

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

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

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

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

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

3. Oh ok. I see what you did. But than what did I do wrong?:

I tried to imagine a picture as you did, but than I solved it like this:

V1 = 550(cos(225), cos(225))
V2 = 80(cos(190), cos(190))

V1 + V2 = (-467.6933499, -402.8005839)

squareroot(V1squared + V2squared) = 617.24 mph

and than for the angle I did it like this:

Tanx = -402.800539/-467.6933499 = arctan(1.16) = 49.345 degrees.

Can you explain what I may have done wrong because I'm fairly sure my set up is correct.

Thanks,
Stealth