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Math Help - problem!!!!

  1. #1
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    problem!!!!

    Let ABC be an isosceles triangle with AB = AC. Suppose that the angle bisector
    of angle B meets AC at D and that BC = BD + AD. Determine angle A.
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  2. #2
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    Re: problem!!!!

    Hello, geniusgarvil!

    This reminds me of a classic problem.
    Is there a typo?


    ABC is an isosceles triangle with AB = AC.
    The bisector of angle B meets AC at D
    and BC \,=\, BD\;{\color{red}=}\;AD.
    Determine angle A.

    Code:
                  A
                  *
                 / \
                / θ \
               /     \
              /       * D
             /     * 2θ\
            / θ *       \
           / * θ      2θ \
        B * - - - - - - - * C
    Let \theta = \angle ABD = \angle DBC

    Since \Delta ADB is isosceles, \angle A \,=\,\theta

    \angle BDC is an exterior angle to \Delta ADB.
    . . Hence: \angle BDC \,=\,2\theta

    Since \Delta BDC is isosceles, \angle C \,=\,2\theta

    In \Delta BDC\!:\;\theta + 2\theta + 2\theta \:=\:180^o \quad\Rightarrow\quad 5\theta \:=\:180^o

    Therefore: . \angle A \:=\:\theta \:=\:36^o
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  3. #3
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    Re: problem!!!!

    Let the base angles be 2\theta, (later, angles in degrees), in which case the angle BDA=3\theta.
    Using the sine rule in triangles CBD and ABD, we have, respectively,

    \frac{BD}{\sin(2\theta)}=\frac{CB}{\sin(3\theta)}, in which case BD=\frac{CB\sin(2\theta)}{\sin(3\theta)},

    and

    \frac{AD}{\sin\theta}=\frac{BD}{\sin(4\theta)}, so AD=\frac{BD\sin\theta}{\sin(4\theta)}=\frac{CB\sin  \theta\sin(2\theta)}{\sin(4\theta)\sin(3\theta)}


    Substitute into the given equation AD+BD=CB, and cancel CB throughout,

    \frac{\sin\theta\sin(2\theta)}{\sin(4\theta)\sin(3  \theta)}+\frac{\sin(2\theta)}{\sin(3\theta)}=1.

    Multiply throuout by \sin(4\theta)\sin(3\theta),

    \sin\theta\sin(2\theta)+\sin(2\theta)\sin(4\theta)-\sin(4\theta)\sin(3\theta)=0.

    Now use the trig identity

     \cos A -\cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)

    \{\cos(3\theta)-\cos\theta\}+\{\cos(6\theta)-\cos(2\theta)\}-\{cos(7\theta)-\cos\theta\}=0.

    Rearrange,

    \{\cos(\6\theta)+\cos(3\theta)\}-\{\cos(7\theta)+\cos(2\theta)\}=0,

    and now use the identity

     \cos A + \cos B = 2\cos \left( \frac{A+B}{2}\right)\cos \left( \frac{A-B}{2}\right)

    so

    2\cos(9\theta/2)\cos(3\theta/2)-2\cos(9\theta/2)\cos(5\theta/2)=0,

    \cos(9\theta/2)\{\cos(3\theta/2)-\cos(5\theta/2)\}=0,

    and therefore

    \cos(9\theta/2)=0, since \cos(3\theta/2)-\cos(5\theta/2)\neq 0,

    Therefore

    9\theta/2=90,\quad \theta = 20 in whch case angle A=100.
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  4. #4
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    Re: problem!!!!

    thanks it is really awesome
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