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Thread: Help With Geometry

  1. #1
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    Help With Geometry

    The following link will take you to a file that has a screenshot of my problem. I'm not allowed to use trig.
    https://drive.google.com/file/d/0B9q...=sharing(Rofl)
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  2. #2
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    Re: Help With Geometry

    First label the vertex at the 'x' $F$.
    $\left\{ \begin{array}{l}m(\angle AEB=40^o)\\m(\angle ADB=50^o)\\m(\angle DFE=70^o)\end{array} \right.$

    Now leave it to you the explain those values and finish the question.
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  3. #3
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    Re: Help With Geometry

    Thank you so much for your help, but I don't know where to go from there.
    Last edited by Ineedhelpbruh; May 16th 2017 at 10:02 AM.
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  4. #4
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    Re: Help With Geometry

    Did your teacher teach nothing?
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  5. #5
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    Re: Help With Geometry

    I'm not sure why you would "label the vertex at the 'x' F" when it is already labeled "E". Oh, I see. DenisB is not referring to the "x degrees" that you are to determine, he is referring to the point where the two line segments, AE and BD, intersect.

    I would use the fact that the measures of the angles in a triangle sum to 180 degrees repeatedly. You are given that angle BAE has measure 60 degrees and angle ABD has measure 50, angle AFB has measure 180- 60- 50= 70 degrees. By the 'vertical angle theorem', angle DFE is also 70 degrees. We have that angle ABE has measure 50+ 30= 80 degrees so angle 180- 60- 80= 40 degrees. There are many more triangles- keep doing that.
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  6. #6
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    Re: Help With Geometry

    I did that, and this is where I got. https://docs.google.com/a/tsc.k12.in...it?usp=sharing
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  7. #7
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    Re: Help With Geometry

    Lol I'm kinda bad at geometry cause I started a month and a half ago but heres my steps
    1. Find Triangle CEA, we know that two of the angles are 20 each so that leaves us with the last angle being 140 we'll call this angle <E+X

    2. Find Triangle CDB, we have two angles which are 20 and 30, 180-50 is 130 (the angle that is 130 is <CDE and <CDB)

    3. Let's call angle EDB, "Y"

    4. When we compare Triangle CEA and Triangle CDB, we want to focus on the third sides, so 130 and 140
    CEA is comprised of X + __ = 140
    CDB is comprised of Y + __ = 130

    5. Then I realized that you can do substitution, because the difference between 140 and 130 is ten, Y= X-10

    6. Let us focus on the center triangle (the one with X in it) it's Y + X + 70 = 180
    Substitute Y for X-10,
    X-10+X+70= 180 -> 2x-10 = 110 -> 2x=120 -> X=60

    7. I know I'm late and you probably already discussed this question in your class already :/
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  8. #8
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    Re: Help With Geometry

    Quote Originally Posted by Zammyz View Post
    4. When we compare Triangle CEA and Triangle CDB, we want to focus on the third sides, so 130 and 140
    CEA is comprised of X + __ = 140
    CDB is comprised of Y + __ = 130

    5. Then I realized that you can do substitution, because the difference between 140 and 130 is ten, Y= X-10
    You can't do that Zammy: it assumes x = y !

    Try again. Hint: x = 30. See if you can get that...

    Also, capital letters are used to represents points,
    NOT values; hence my use of x and y, NOT X and Y.
    Last edited by DenisB; Jul 9th 2017 at 08:21 PM.
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    Re: Help With Geometry

    Can you write down the steps that you did cause I initially went on the forums to look at other people's solutions and work to get better
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  10. #10
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    Re: Help With Geometry

    Quote Originally Posted by Zammyz View Post
    Can you write down the steps that you did cause I initially went on the forums to look at other people's solutions and work to get better
    First label $\overline{DB}\cap\overline{AE}=\{F\}$.
    You tell us why each of the following is true.
    $\left\{ \begin{array}{l}m(\angle AEB=40^o)\\m(\angle ADB=50^o)\\m(\angle DFE=70^o)\end{array} \right.$

    If you cannot the explain why you cannot.
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  11. #11
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    Re: Help With Geometry

    AEB = 40 because 60+30++50= 140, 180-140 = 40
    ADB = 50 because 60+20+50= 130, 180-130 = 50
    DFE = 70 because 60+50 = 110, <F is 70 and by vertical angles it also = 70

    What other reasoning's did you use?

    The only other way I've seen to get 30 is get triangle BDC, assume <C is 20, 180-50 = 130 and subtract 50 to get 80, now angle D in triangle DEF is 80+70+X = 180
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  12. #12
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    Re: Help With Geometry

    Consider the bisector AH of angle EAB in triangle EAB, H on EB

    Show AH is parallel to DE using CD/DA = CE/EH

    To show CD/DA = CE/EH, consider point G on AB extended where BG = BC

    Triangles GCA and EAB are similar
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