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Math Help - A hard question of finding angle

  1. #1
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    A hard question of finding angle

    could any smart person help solve this question?
    thank you so much.....

    Find x where AB=BC
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  2. #2
    Member pflo's Avatar
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    This is not a hard question, it is an impossible one. Not enough information here. Now I could be missing something, but I don't see how you can lock in an angle for x. If m<DAC = 0, then x=90. But if m<DAC = 30, then x = 60. And if m<DAC = 45, then x = 45. This is true regardless of what is going on with triangleABC. We need some other constraint on the problem.
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  3. #3
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    Quote Originally Posted by 3ASV444 View Post
    could any smart person help solve this question?
    thank you so much.....

    Find x where AB=BC
     \angle x = 90 - \angle EAF
     \angle x = \angle AEF
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  4. #4
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    It is a messy problem, but not impossible.
    To simplify notation, let m\left( {\angle DAC} \right) = y\;\& \,m\left( {\angle BAC} \right) = m\left( {\angle BCA} \right) = z
    The sum of measures of angles of a quadrilateral is [ 360^o.
    So x+y+2z+205=360. But x+y=90, so 2z+295=360.
    Now you can finish.
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  5. #5
    Member pflo's Avatar
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    But even if you can solve for z, you still have 2 variables and a single equation (or two equivelant equations anyway). Even if you know z, all you're left with is x+y=90.

    Isn't it? How would you finish from there?
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