word problem

[swimmer3]

Two planes take off at the same time from an airport. The first plane is flying at 260 mph on a bearing of S45 degrees E. The second plane is flying in the direction S 45 degrees W at 280mph. If there are no wind currents blowing, how far apart are they after 2 hrs? What is the bearing of the second plane form the first after 2 hrs?

I know they are 764 miles apart after 2 hrs. How do I figure out the final bearing?

 

[soroban]

Hello, swimmer3!

Two planes take off at the same time from an airport.
The first plane is flying at 260 mph on a bearing of \(\displaystyle S\,45^oE\).
The second plane is flying at 280 mph on a bearing of \(\displaystyle S\, 45^oW\).
If there are no wind currents blowing, how far apart are they after 2 hrs?
What is the bearing of the second plane from the first after 2 hrs?

I know they are 764 miles apart after 2 hrs.
How do I figure out the final bearing?

After two hours, their positions look like this:

                        P
                        o           N
                      * : *  520    :
              560   *   :   *       :
                  * 45d : 45d * 45d :
                *       :       *   :
              *         :         * :
            *           S           o A
          *                 *       :
        *           *               :
      *     *        764            M
  B o

\(\displaystyle \text{The first plane flies from }P\text{ to }A.\)
. . \(\displaystyle \angle SPA \,=\,45^o \,=\,\angle PAN,\;\;PA \,=\,520\)

\(\displaystyle \text{The second plane flies from }P\text{ to }B.\)
. . \(\displaystyle \angle SPB \,=\,45^o,\;\;PB \,=\,560\)

\(\displaystyle \text{Since }\Delta APB\text{ is a right triangle: }\;AB \;=\;\sqrt{520^2 + 560^2} \;=\;764.198927 \;\approx\;764\text{ miles}\)


\(\displaystyle \text{We want }\angle MAB.\)

\(\displaystyle \text{In }\Delta PAB\text{, we have: }\:\cos A \;=\;\frac{764^2 + 520^2 - 560^2}{2(764)(520)} \;=\; 0.680245671\)

\(\displaystyle \text{Then: }\:A \;=\;47.13715641^o \;\approx\;47^o\)

. . \(\displaystyle \text{Hence: }\:\angle MAB \;=\;180^o - 45^o - 47^o \;=\;88^o\)


\(\displaystyle \text{The bearing of the second plane from the first is: }\:S\,88^o\,W\)