Vectors help!

[ndpndntfilm]

Argh! Don't you hate it when your teacher gives you all this information but none of it is useful on the homework?


A plane flies 600 km on a course of 300o. It then flies south for a while and finally flies on a course of 040o to return to its starting point. Find the total distance traveled.

Okay so, I drew a picture and I got a little help from my teacher.

But after that, I', not sure what to do. Help?

 

[tkhunny]

1) Mark a vector from the Origin with heading 40º.
2) Mark a vector from "600 km" with heading 180º
3) How far south should it go? Extend that 40º vector in the opposite direction through the origin into Quadrant III. When it intersects your Southern flight, you are done.

 

[soroban]

Hello, ndpndntfilm!

Did you draw a complete sketch?

\(\displaystyle \text{A plane flies 600 km on a course of }300^o.\)
\(\displaystyle \text{It then flies south for a while, and finally flies on a course of }040^o\text{ to return to its starting point.}\)
\(\displaystyle \text{Find the total distance traveled.}\)

\(\displaystyle \text{The plane flies from }A\text{ to }B\text{ on a course of }300^o\text{ for 600 km.}\)
. . \(\displaystyle AB = 600,\;\;\;\angle NAB \,=\, 60^o \quad\Rightarrow\quad \angle ABC \,=\,60^o\)

\(\displaystyle \text{Then it flies directly south for }x\text{ km until }\angle BCA = 40^o \quad\Rightarrow\quad \angle CAS = 40^o.\)
. . \(\displaystyle \text{Hence: }\:\angle BAC = 80^o.\)

\(\displaystyle \text{Let: }y = CA.\)

    B @           N
      |  * 600    :
      |60o  *  60o:
      |        *  :
      |      80o  @ A
    x |         * :
      |       *   :
      |40o  *  40o:
      |   * y     :
      | *         S
    C @

\(\displaystyle \text{Use the Law of Sines to determine }x\text{ and }y.\)

. . \(\displaystyle \frac{x}{\sin80^o} \:=\:\frac{600}{\sin40^o}\quad\text{ and }\quad \frac{y}{\sin60^o} \:=\:\frac{600}{\sin40^o}\)