Heres the question:
A plane leaves the airport in Galisto and flies $140$km at $68.0^∘$ east of north and then changes direction to fly $255$km at $48.0^∘$ south of east, after which it makes an immediate emergency landing in a pasture.
When the airport sends out a rescue crew, how far should this crew fly to go directly to this plane?
I tried using the pythagorean theorem, but he answer I got was incorrect. So I tried finding the components of the vectors and adding them, but that didn't work either.
Here is what I tried to do, but I have no idea if I was headed down the right track.
Components of $|\overrightarrow{A}|$ are:
$140sin(68^∘)=129.806$
$140cos(68^∘)=52.4449$
$|\overrightarrow{A}|=\sqrt{\left(52.4449)^2+(129.806)^2\right)} = 140$
Components of $|\overrightarrow{B}|$ are:
$255sin(48^∘)=189.502$
$255cos(48^∘)=170.628$
$|\overrightarrow{B}|=\sqrt{\left(189.502)^2+(170.628)^2\right)} = 255$
These weren't correct, so I greatly appreciate any guidance you can offer!
To the problem with vectors, we have to break the trip up into two parts, I'll call them $A$ (airport to change heading) and $B$ (change heading to emergency landing) like you did. From the picture we can see that the vector $A$ uses an angle of $68^{\circ}$ from the vertical. So $A = \langle 140\sin(68^{\circ}), 140\cos(68^{\circ})\rangle \approx \langle 129.806, 52.445\rangle$. You could also have done $A = \langle 140\cos(22^{\circ}), 140\sin(22^{\circ})\rangle$ if you wanted to work with the angle formed between $A$ and the horizontal. For vector $B$, we have the angle with the horizontal but it is below the horizontal. So the vertical component of $B$ is negative. $B = \langle 255\cos(48^{\circ}), -255\sin(48^{\circ})\rangle \approx \langle 170.628, -189.502\rangle$. The resultant is $R = A+B = \langle 300.434, -137.507\rangle$. $|R| = \sqrt{(300.434)^2+(-137.507)^2} \approx 330.220$ km. The angle they need to travel at is $\tan^{-1}\left(\frac{137.507}{300.434}\right) \approx 27.24^{\circ}$ below the horizontal or south of east.