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Math Help - I need help on how to find variables for Isosceles Triangles

  1. #1
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    I need help on how to find variables for Isosceles Triangles

    Here is what I'm dealing with:



    I know y would be 49, but I don't know how to find the value of x. I'm a beginner at Geometry, so hopefully someone can explain this to me in a way that I understand.
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  2. #2
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    Quote Originally Posted by ladyk View Post
    Here is what I'm dealing with:



    I know y would be 49, but I don't know how to find the value of x. I'm a beginner at Geometry, so hopefully someone can explain this to me in a way that I understand.
    The angles opposite the congruent sides are congruent. (y is not 49)

    In classical (Euclidean) geometry, the sum of the measures of the angles in a triangle equals 180^{\circ}

    Use these two facts to solve for x and y.
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  3. #3
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    Ok, so I meant to put y would be 46, and therefore x would be 88. That's what I got
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  4. #4
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    Quote Originally Posted by ladyk View Post
    Ok, so I meant to put y would be 46, and therefore x would be 88. That's what I got
    x is not 88. You found the angle for the bigger triangle. What about the one whose measures are x, 90, 46?
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  5. #5
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    OK, so y is 46, and x is 90. Why would x be 90? I want to understand why that is.
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  6. #6
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    I labeled your triangle.

    Consider triangle BCD. I mentioned previously that the sum of the measures of the angles in a triangle equals 180. Therefore:
    x+46+90=180

    What you did was x+y+46=180, which is incorrect. The measure of angle B is not x. As you can see, there is an angle bisector that divides angle B into 2 congruent angles(<CBD and <DBA)
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  7. #7
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    Thank you so much Chop Suey. When it's explained simply, you realize how easy it actually is. Would x=44 though?
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  8. #8
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    Yes.
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  9. #9
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    Thank you!
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