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Math Help - Finding the length

  1. #1
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    Finding the length

    Radar stations A and B are on an east-west line, 8.6 km apart. Station A detects a plane at C, on a bearing of 53 degree. Station B simultaneously detects the same plane, on a bearing of 323 degree. Find the distance from B to C.

    That is the question. And Why can I not use cosine to find the lenth?

    And I thank you in advance everyone who help me with this problem.
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  2. #2
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    Hello, Judi!

    What do you mean by "use cosine"?


    Radar stations A and B are on an east-west line, 8.6 km apart.
    Station A detects a plane at C, on a bearing of 53.
    Station B simultaneously detects the same plane, on a bearing of 323.
    Find the distance from B to C.

    Why can I not use cosine to find the length?
    Code:
                            C
                            *
          :              *   *    :
          :           *       *37:
          :        *           *  :
          : 53 *               * :
          :  * 37           53 *:
          * - - - - - - - - - - - *
          A          8.6          B

    If you mean "the Law of Cosines", we can't.
    . . The Law of Cosines requires two sides ... and we have only one.


    If you note that \angle C = 90^o, you can use cosine.

    . \cos B \:=\:\frac{BC}{AB} \quad\Rightarrow\quad BC \:=\:8.6\cos53^o \quad\hdots\:\text{etc.}

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  3. #3
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    Station bearing on plane posted by Judi

    the east and west station should be identifed by a letter and station. I assumed that A was east and Bwest.I do not get C being 90 deg.using the bearing at A 323 and the bearing at B 53 .the relative bearing at A is 63 deg north of west. This makes Abc an isosoles triangle.BC is obtained using the cosine function.


    bjh
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  4. #4
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    Quote Originally Posted by bjhopper View Post
    the east and west station should be identifed by a letter and station. I assumed that A was east and Bwest.I do not get C being 90 deg.using the bearing at A 323 and the bearing at B 53 .the relative bearing at A is 63 deg north of west. This makes Abc an isosoles triangle.BC is obtained using the cosine function.

    bjh
    If station A were east of station B then the bearing angles are out of phase by 180 degrees.


    Soroban's beautiful image depicts the facts.

    \angle CAB = 37^o
    \angle ABC = 53^o
    that leaves
    \angle BCA = 180^o - 37^o -53^o = 90^o

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  5. #5
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    Thank you for your drawing. Yes, I meant by the method 2 you showed. I know I can get the answer by using either cos or sin. But is it work every single time? I have encountered the problem where cos didn't work.

    please help,





    Quote Originally Posted by Soroban View Post
    Hello, Judi!

    What do you mean by "use cosine"?


    Code:
                            C
                            *
          :              *   *    :
          :           *       *37:
          :        *           *  :
          : 53 *               * :
          :  * 37           53 *:
          * - - - - - - - - - - - *
          A          8.6          B

    If you mean "the Law of Cosines", we can't.
    . . The Law of Cosines requires two sides ... and we have only one.


    If you note that \angle C = 90^o, you can use cosine.

    . \cos B \:=\:\frac{BC}{AB} \quad\Rightarrow\quad BC \:=\:8.6\cos53^o \quad\hdots\:\text{etc.}

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