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Math Help - Find x !

  1. #1
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    Find x !

    Hii !

    Can You Help me For This Exercice :



    Question is : Find x !
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  2. #2
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    Quote Originally Posted by Perelman View Post
    Hii !

    Can You Help me For This Exercice :



    Question is : Find x !
    Note that you have two triangles and the angles in each much add up to 180

    darn, never mind >.<
    Last edited by mr fantastic; February 5th 2010 at 02:26 AM. Reason: m --> r
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  3. #3
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    x = 22,5 ?
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  4. #4
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    Quote Originally Posted by Perelman View Post
    x = 22,5 ?
    It's not a right triangle :~:
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  5. #5
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    first we have B = 180 - A - C = 180 - 3x - x
    then we also find that B = B1+B2 (2 small triangle from left to right)
    B1 = 180 - (A + D1) = 180 - (45 + 3x) = 135 - 3x
    B2 = 180 - D2 - C = 180 - (180 - 45) - x = 45 - x
    since B = B1 + B2 = 135 -3x + 45 - x = 180 - 3x - x
    look up on the first line ( is this look familiar?)
    result B = B1 + B2 = A + C

    Now we got
    B = A + C
    and B = 180 - A - C
    so A + C = 180 -A - C
    => A + C + A + C = 180
    => 2A + 2C = 180
    plug in the real value A = 3x and C = x
    => 2(3x) + 2x = 180
    => 6x + 2x = 180
    => 8x = 180
    => x = 22.5
    The End!
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  6. #6
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    Quote Originally Posted by Perelman View Post
    Hii !

    Can You Help me For This Exercice :



    Question is : Find x !
    From the attached sketch

    3x+b=180^o-45^o=135^o=3(45^o)

    x+a=180^o-135^o=45^o

    Therefore 3x+b = 3(x+a) = 3x+3a

    Hence b = 3a.

    When the third apex is on the circle, x=a and 3x=b=3a, so 2x=45 degrees, so x=22.5 degrees.

    When it isn't...

    If...

    (3x+z)=3(x+e)

    then, as

    (x+e)+c=45^o

    (3x+z)+y=135^o=3(45^o)

    this means

    y=3c

    The third apex of triangles ABD and ABC may be anywhere on the red line,
    above the horizontal.
    Therefore x ranges from 0 to 45 degrees.
    However, the interior triangles may no longer touch at B.

    To find out if x can be any other angle than 22.5^o...

    we can use the Law of Sines.

    \frac{Sin3x}{|BD|}=\frac{Sinb}{k},\ \frac{Sinx}{|BD|}=\frac{Sina}{k}

    |BD|=\frac{kSin3x}{Sinb}=\frac{kSinx}{Sina}

    b=3a

    \frac{Sin3x}{Sin3a}=\frac{Sinx}{Sina}

    Sin3x=3Sinx-4Sin^3x

    \frac{3Sinx-4Sin^3x}{3Sina-4Sin^3a}=\frac{Sinx}{Sina}

    \frac{Sinx\left(3-4Sin^2x\right)}{Sina\left(3-4Sin^2a\right)}=\frac{Sinx}{Sina}

    3-4Sin^2x=3-4Sin^2a

    Sin^2x=Sin^2a

    Sinx={\pm}Sina

    As x and "a" are acute, Sinx=Sina, x=a.

    Thus x=22.5 degrees.
    Attached Thumbnails Attached Thumbnails Find x !-x.jpg  
    Last edited by Archie Meade; February 5th 2010 at 03:26 PM.
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  7. #7
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    Quote Originally Posted by Archie Meade View Post
    From the attached sketch

    3x+b=180^o-45^o=135^o=3(45^o)

    x+a=180^o-135^o=45^o

    Therefore 3x+b = 3(x+a) = 3x+3a

    Hence b = 3a.

    When the third apex is on the circle, x=a and 3x=b=3a, so 2x=45 degrees, so x=22.5 degrees.

    When it isn't...

    If...

    (3x+z)=3(x+e)

    then, as

    (x+e)+c=45^o

    (3x+z)+y=135^o=3(45^o)

    this means

    y=3c

    The third apex of the triangle may be anywhere on the red line,
    above the horizontal.
    Therefore x ranges from 0 to 45 degrees.
    That argument assumes that the two green lines intersect at a point on the red line. But that is not the case. If z = 3e ≠ 0 then the two green lines will not intersect on the red line. Alternatively, if the two green lines do intersect on the red line then it is impossible to have z = 3e (or y = 3c) unless z = e = 0.

    The only solution to the problem is x = 22.5.
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  8. #8
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    Yes,
    i forgot to maintain the base dimensions equal, sorry!

    Corrections are applied.
    Last edited by Archie Meade; February 5th 2010 at 03:27 PM.
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