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Math Help - Triangle Trouble

  1. #1
    Newbie
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    Exclamation Triangle Trouble

    In a triangle ABC, X is a point on AC such that AX/XC=3 and the angle AXB = 70 degrees. Find the length of AC if the angle of ABC = 110 degrees and BC = 6 centimetres.

    thanks so much
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  2. #2
    Junior Member
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    Mesnil, Mauritius
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    By Similar Triangles,

    Angle ABC = Angle BXC
    Angle ACB = Angle XCB
    Angle BAC = Angle XBC

    XC=1/4*AC

    Solve for AC in AC/6=6/XC = \sqrt{144}

    Hope that helps.
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  3. #3
    MHF Contributor
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    Quote Originally Posted by googa View Post
    In a triangle ABC, X is a point on AC such that AX/XC=3 and the angle AXB = 70 degrees. Find the length of AC if the angle of ABC = 110 degrees and BC = 6 centimetres.

    thanks so much
    In straight angle AXC,
    angle BXC = 180 -70 = 110 degrees

    In triangle BXC,
    Let angle XBC = theta
    (theta +angle C) = 180 -110 = 70 deg

    In big triangle ABC
    (angle A +angle C) = 180 -110 = 70 deg also

    So,
    theta +angle C = angle A +angle C
    Hence,
    angle A = theta

    ---------------
    If AX /XC = 3,
    then side AC is divided into 4 equal parts.
    Let w = length of AC,
    So, XC = w/4

    In triangle BXC, by Law of Sines,
    BC /sin(angle BXC) = XC /sin(angle XBC)
    6 /sin(110deg) = (w/4) /sin(theta)
    Cross multiply,
    6sin(theta) = (w/4)sin(110deg)
    sin(theta) = w*sin(110deg) /24 ------**

    In triangle ABC, by Law of Sines,
    AC /sin(angle ABC) = BC /sin(angle A)
    w /sin(110deg) = 6 /sin(theta)
    Cross multiply,
    w*sin(theta) = 6sin(110deg)
    sin(theta) = 6sin(110deg) /w -----**

    sin(theta) = sin(theta)
    w*sin(110deg) /24 = 6sin(110deg) /w
    w/24 = 6/w
    w^2 = 6*24
    w = sqrt(144) = 12

    Therefore, AC = 12 cm ----------answer.
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  4. #4
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    Lexington, MA (USA)
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    Hello, googa!

    A slight variation of ticbol's solution . . .


    In \Delta ABC,\:X is a point on AC such that \frac{AX}{XC}=3

    and \angle AXB = 70^o,\;\;\angle ABC = 110^o,\;\;BC = 6 cm.

    Find the length of AC.
    Code:
                                  B
                                  o
                               *   **
                            *       * *
                         *           * β*  6
                      *               *   *
                   *                   *    * 
                *                       *     *
             * β                     70 *    α *
          o   *   *   *   *   *   *   *   o   *   o
          A               3a              X   a   C

    We have: . \angle ABC = 110^o,\;\;\angle AXB = 70^o,\;\;BC = 6
    Let: a = XC,\;AX = 3a


    Consider \Delta ABC and \Delta BXC

    . . They share \angle \alpha = \angle BCX

    Since \angle AXB = 70^o, then \angle BXC = 110^o

    . . They have equal angles: . \angle ABC \:=\:\angle BXC \:=\:110^o

    Hence, their third angles are equal: . \angle BAC \:=\:\angle CBX \:=\:\beta


    In \Delta ABC\!:\;\;\frac{\sin\beta}{6} \:=\:\frac{\sin110^o}{4a} \quad\Rightarrow\quad \sin\beta \:=\:\frac{3\sin110^o}{2a} .[1]

    In \Delta BXC\!:\;\;\frac{\sin\beta}{a} \:=\:\frac{\sin110^o}{6} \quad\Rightarrow\quad \sin\beta \:=\:\frac{a\sin110^o}{6} .[2]


    Equate [1] and [2]: . \frac{3\sin110^o}{2a} \:=\:\frac{a\sin110^o}{6} \quad\Rightarrow\quad 18 \:=\:2a^2 \quad\Rightarrow\quad a \:=\:3


    Therefore: . AC \:=\:4a \:=\:\boxed{12\text{ cm}}

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