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Math Help - Please help to find the angle

  1. #1
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    Please help to find the angle

    Hi
    Can you help me please with this question :
    ABC is a triangle where the point D is on Bc such that BAD = 60 , B=20 , AB = CD ...
    Can you tell me ACB = ?????
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  2. #2
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    Hello, OnlyMath!

    I did a lot of trig and came with a surprisingly simple answer.
    I suspect that there is an elegant geometric solution . . . somewhere.



    ABC is a triangle where the point D is on BC
    . . such that: . \angle BAD = 60^o,\;\;\angle B=20^o,\;\' AB = CD

    Find \angle C.
    Code:
          B
          o
           *  *
            * 20  *
             *        *   D
            a *       100 o
               *        *  80 *    a
                * 60  * x         *
                 *  *              @  *
                A o  *  *  *  *  *  *  *  o C

    We have: . \angle BAD = 60^o,\;\angle B = 20^o \quad\Rightarrow\quad \angle BDA = 100^o,\;\angle ADC = 80^o

    Let a = AB = CD,\;x = AD,\;\theta = \angle C \quad\Rightarrow\quad \angle DAC = 100-\theta


    \text{In }\Delta ABD\!:\;\dfrac{x}{\sin20^o} \:=\:\dfrac{a}{\sin100^o} \quad\Rightarrow\quad \dfrac{x}{a} \:=\:\dfrac{\sin20^o}{\sin100^o} .[1]

    \text{In }\Delta ADC\!:\;\dfrac{x}{\sin\theta} \:=\:\dfrac{a}{\sin(100^o-\theta)} \quad\Rightarrow\quad \dfrac{x}{a} \:=\:\dfrac{\sin\theta}{\sin(100^o-\theta)} .[2]


    Equate [1] and [2]:

    . . \dfrac{\sin20}{\sin100} \:=\:\dfrac{\sin\theta}{\sin(100-\theta)} \quad\Rightarrow\quad \sin20\sin(100-\theta) \;=\;\sin100\sin\theta<br />

    . . \sin20\bigg(\sin100\cos\theta - \cos100\sin\theta\bigg) \;=\;\sin100\sin\theta

    . . \sin20\sin100\cos\theta - \sin20\cos100\sin\theta \;=\;\sin100\sin\theta

    . . \sin100\sin\theta + \sin20\cos100\sin\theta \;=\;\sin20\sin100\cos\theta

    . . \bigg(\sin100 + \sin20\cos100\bigg)\sin\theta \;=\;\sin20\sin100\cos\theta

    . . . . . . . \dfrac{\sin\theta}{\cos\theta} \;=\;\dfrac{\sin20\sin100}{\sin100 + \sin20\cos100}


    Therefore: . \tan\theta \:=\:0.363970234 \quad\Rightarrow\quad \theta \:=\:20^o


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Fascinating!

    \Delta ABC is isosceles: . AB \,=\,AC\,=\,a

    Also, \Delta ADC is isosceles: . AC\,=\,CD\,=\,a

    Is there a more direct way to derive those facts?

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