Introductory Trigonometry

[nicholas]

An airplance is traveling on a 500km trip. After 150km the pilot notes that the plance is 20km off course. How much should the heading be corrected to then fly straight to the destination? Assume there is no wind during the plane's flight.
A
.............. *....................................
.........* |20km *....................
B *-------------|-----------------------------* C
-----150---- -------------350------------
---------------500km-----------------------

Is the graph I drew above is right and the angle I need to find is angle A ???

 

[tkhunny]

Your 350 km is no good. The 150 km was off course. Had it been on course, then there would be 350 km left. There is now more than that.

You travelled, in relation to the correct course, only \(\displaystyle \sqrt{150^{2} - 20^{2}}\) km. Do you see why?

 

[nicholas]

I tried many ways but couln't get it because the answer from my textbook is 28degree.Please help.Thank you.

 

[soroban]

Hello, nicholas!

Your diagram is wrong . . . and I don't agree with their answer.

An airplance is traveling on a 500 km trip.
After 150 km the pilot notes that the plance is 20 km off course.
How much should the heading be corrected to then fly straight to the destination?

                  P     500-x
                  o - - - - - - - * E
               *  :   *   @       :
        150 *     :20     *       :20
         *        :           *   :
    A o - - - - - + - - - - - - - o B
      :     x     Q     500-x     :
      : - - - - -  500  - - - - - :

\(\displaystyle \text{The plane intends to fly from }A\text{ to }B\!:\;AB = 500\)

\(\displaystyle \text{Instead, it flies 150 km to }P\text{, where }PQ = 20\)

\(\displaystyle \text{Let }x \,=\,AQ.\)

\(\displaystyle \text{We want angle }\theta \,=\,\angle EPB.\)


\(\displaystyle \text{In right triangle }PQA\!:\;x^2 + 20^2 \:=\:150^2 \quad\Rightarrow\quad x \:=\:10\sqrt{221}\)

\(\displaystyle \text{Then: }\:QB \:=\:500-x \:=\E\)


\(\displaystyle \text{In right triangle }PEB\!:\;\tan\theta \:=\:\frac{EB}{PE} \:=\:\frac{20}{500-10\sqrt{221}} \:=\:0.056925027\)

. . \(\displaystyle \text{Therefore: }\:\theta \;\approx\;3.258^o\)

 

[Denis]

                  P     
                  o 
               *  :   *           
        150 *     :20     *       
         *        :           *   
    A o - - - - - + - - - - - - - o B
            x     Q     500-x

Keeping it simple:

QB = 500 - sqrt(150^2 - 20^2) = ~351.34

PB = sqrt(QB^2 + 20^2) = ~351.91

anglePBQ = asin(20 / PB) = ~3.258