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Math Help - Little problem to keep sharp

  1. #1
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    Little problem to keep sharp

    Came across this problem today, and thought it was pretty nice. Most of the kids in my class solved it in about 1-2 minutes, how quickly can you solve it?

    Little problem to keep sharp-math-problem1.png

    I drew it in paint, so pardon sloppiness (I don't want to mess with latex today).

    Given:
    m||n;
    All lines are coplanar;

    Find x

    Spoiler:
    x=110
    Last edited by fireballs619; October 21st 2010 at 06:02 PM.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    I think that something is missing... are the two points make a perpendicular to m (n) ? is it matter? (asking myself)
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  3. #3
    MHF Contributor
    skeeter's Avatar
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    nothing missing ... sketch an auxiliary line parallel to m and n thru the vertex of angle x
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  4. #4
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    Hello, fireballs619!

    Given: PQ \parallel RS,\;\angle PAB = 115^o,\;\angle RCB = 135^o

    Find: x \:=\:\angle ABC


    Code:
                A
        P * - - * - - - - - - - - * Q
              115 *
                    *
                      *
                        * 
                          *
                          x * B
                           *
                          *
                         *
                        *
                   135 *
        R * - - - - - * - - - - * S
                      C

    Extend CB to intersect PQ at \,D.


    Code:
                A                 D
        P * - - * - - - - - - - - * - Q
              115 * 65        45 /
                    *           /
                      *        /
                        *  70 / 
                          *  /
                          x * B
                           *
                          *
                         *
                        *
                   135 * 45
        R * - - - - - * - - - - * S
                      C

    \angle PAB = 115^o \quad\Rightarrow\quad \angle QAB = 65^o

    \angle RCB = 135^o \quad\Rightarrow\quad \angle BCS = 45^o = \angle ADB

    \text{Hence: }\:\angle ABD \:=\:180^o - 65^o - 45^o \;=\;70^o

    \text{Therefore: }\:x \;=\;\angle ABC \;=\;180^o - 70^o \;=\;110^o

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  5. #5
    MHF Contributor Unknown008's Avatar
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    I just make use of the parallel lines and the time to grab my calculator, it takes less than 20 seconds.



    So, x = \alpha + \beta = (180 - 115) + (180 - 135) = 110^o
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