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Math Help - Triangle side length

  1. #1
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    Triangle side length

    Given \triangle ABC, where AD is perpendicular to BC, B=45 ^{\circ}, C=60 ^{\circ}, and AB=10, determine the length of AC.
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  2. #2
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    Hello, chengbin!

    Given \Delta ABC, where AD \perp BC,\;\;B=45^o,\;\;C=60^o,\;\;AB=10
    Determine the length of AC.
    Is there a typo?
    . . As written, altitude AD is irrelevant.
    Code:
                        A
                        *
                      *  *
                    *     *
               10 *        *
                *           *
              *              *
            *                 *
          * 45            60 *
      B * - - - - - - - - - - - * C

    Law of Sines: . \frac{AC}{\sin45^o} \:=\:\frac{10}{\sin60^o} \quad\Rightarrow\quad AC \:=\:\frac{10\sin45^o}{\sin60^o}


    Therefore: . AC \;=\;8.164965809

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  3. #3
    Senior Member
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    There is no typo.

    This was just the first page into the study of triangles.

    I didn't get to the law of sine yet. How do you do this without the law of sine?

    Can you also check this problem? I didn't get my solution book yet. This problem, and the problem in the first post, are all on the first page. So no advanced stuff yet.

    Express the following function in terms of A.

    \sin \frac{B+C}{2}

    These are my steps.

    = \sin (90-\frac {A}{2})

    = \cos \frac {A}{2}
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  4. #4
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    Hello again, chengbin!

    Okay, so there is no typo and we can't use the Law of Sines.
    . . Then that altitude is necessary.
    Code:
                        A
                        *
                      * |*
                    *   | *
               10 *     |  *
                *       |   *
              *         |    *
            *           |     *
          * 45         |  60 *
      B * - - - - - - - * - - - * C
                        D
    We have to work our way over to side AC.


    In right triangle ADB\!:\;\;\sin45^o \:=\:\frac{AD}{10} \quad\Rightarrow\quad AD \:=\:10\sin45^o \:\approx\:7.07

    In right triangle ADC\!:\;\;\sin60^o \:=\:\frac{AD}{AC} \quad\Rightarrow\quad AC \:=\:\frac{AD}{\sin60^o}

    . . Therefore: . AC \;=\;\frac{7.07}{\sin60^o} \;\approx\;8.16

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