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Thread: triangle altitude help

  1. #1
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    triangle altitude help

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  2. #2
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    Re: triangle altitude help

    Hey vegasgunner.

    Your best bet is to draw a diagram (like you did in your other question that you posted) as a starting point.
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  3. #3
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    Re: triangle altitude help

    Hello, vegasgunner!

    \text{Altitudes }AX\text{ and }BY\text{ of acute triangle }ABC\text{ intersect at }H.
    \text{If }\angle BAC = 43^o\text{ and }\angle ABC = 67^o,\text{ then what is }\angle HCA\,?

    Code:
                  C
                  o
                 ** *  X
              Y * *   o
               o  * .   *
              *   o       *
             *  . *H  .     *
            * .   *       .   *
           *43o   *           . *
        A o   *   o   *   *   *   o B
                  Z
    In a triangle, the three altitudes are concurrent.
    Hence, CHZ is an altitude.
    . . \angle CZA = 90^o

    In right triangle CZA,\;\angle ZAC = 43^o.

    Therefore: \angle HCA = \angle ZCA = 47^o.
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