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anonymous

  • one year ago

Is anyone good with Geometric Proofs? I need help! Ill attach the paper

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @Hero

  3. e.mccormick
    • one year ago
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    First off, you know the principals relating to angles of lines that intersect?

  4. anonymous
    • one year ago
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    ok

  5. e.mccormick
    • one year ago
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    Not sure if you know them or not. I mean this stuff: |dw:1440021663729:dw|

  6. anonymous
    • one year ago
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    I know a little, but looking at the 2nd problem why does it say PC = QB There is no P or Q

  7. e.mccormick
    • one year ago
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    Probably a typo. I beat they meant \(AD\cong BE\).

  8. e.mccormick
    • one year ago
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    bet...

  9. e.mccormick
    • one year ago
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    On the first one, you can use the fact that the intersections of lines must add up to 180 degrees to get a generic measure for the angles at points B abd C. Then use the givens that the inner angles of B and C are the same to show that those outers are using 180-(the same number) to show thart they are the same. I just do not remember the names of the theroems off hand to show that, but you should have a list somewhere in their book or on ylur class web site of theorems you are using.

  10. e.mccormick
    • one year ago
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    Something to do with a transversal making supplementary angles...

  11. e.mccormick
    • one year ago
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    But yah, on the second one, I would just ask the teacher if it is supposed to be what it is, since those other points do not exist.

  12. anonymous
    • one year ago
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    thank you

  13. e.mccormick
    • one year ago
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    Ah, here, found this: Linear Pair Postulate If two angles form a linear pair, then the measures of the angles add up to 180°. That looks like the one you need to use in the first one. And lets see... Triangle Exterior Angle Postulate could also do it, which says that "The measure of an exterior angle of a triangle is equal to the sum of the measures of the remote interior angles." Either of those can get you in the correct direction. I found a PDF with the different postulates and theorems. http://web.cerritos.edu/dford/SitePages/Math_70_F13/PostulatesandTheorems.pdf

  14. anonymous
    • one year ago
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    oh great thank you so much!!

  15. e.mccormick
    • one year ago
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    This PDF might also help... but the visual queues did not show up =( http://www.riverview.wednet.edu/parade/teachers/mcclintict/Geometry-1/Geo-PostulatesTheorems-List-2011.pdf

  16. anonymous
    • one year ago
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    You are so much help

  17. e.mccormick
    • one year ago
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    With proofs, it cames down to finding a set of rules, one after the other, that will let you say "This works!" or "This never works!" in the proper order and with the proper name. So I can tell you that when one segment touches a line, the angles on each side of where it touces add up to 180 degrees... but it does not mean much. On the other hand, if I say Linear Pair Postulate, it means exactly that but is using words that a math teacher likes. Hehe.

  18. anonymous
    • one year ago
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    I'll have to remember that

  19. anonymous
    • one year ago
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    what did I do wrong? :'( @e.mccormick

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  20. e.mccormick
    • one year ago
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    They want you to prove they are conguent segments.

  21. anonymous
    • one year ago
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    ive tried fixing it and I can quite figure it out where I went wrong ? was it because I put dc in cd and ec instead ce

  22. e.mccormick
    • one year ago
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    The DC = EC in step B is wrong. You want to prove that int he end. You could say AC = BC in B.

  23. anonymous
    • one year ago
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    so b) would be ac = bc , cd = ce?

  24. e.mccormick
    • one year ago
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    No, you are given AC cong BC, and some mistake... I am betting that becaus it says you need to prove CD cong CE that the mistake is AD cong BE. If the given is: AC cong BC, AD cong BE then: AC = BC, AD = BE by Def. of Cong. Segments

  25. anonymous
    • one year ago
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    so then c is ac + ad = bc + be for addition property

  26. anonymous
    • one year ago
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    i thought i knew what i was doing then i messed up and know im 100% lost again

  27. e.mccormick
    • one year ago
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    OK, well, the addition property is... hmmm... not sure if you need that at all or just the segment addition postulate. See, AD + DC = AC, which is adding up the segments to make the line. You can do the same with the other side, so BE + EC = BC. Then you can say by substitutain that: AD + EC = BC For that substitution to be valid, EC = DC. Does that make some sense?

  28. e.mccormick
    • one year ago
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    Oh, and on the AD + EC = BC you can make it clearer by saying: AD + EC = BC Subsitutuion of AD for BE. Basically you are saying, "If I can subsitute the upper line segment, then I must also be able to substitute the lower line segment, therefore the lower ones are equal and equal means congruent."

  29. anonymous
    • one year ago
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    okay let me see if i got this right a) Ac = bc, ad=be b)ac=bc, ad = be c)ac +ad = bc +be d) ad + dc =ac be + ec = bc e)ec =dc f) dc = ce

  30. anonymous
    • one year ago
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    @e.mccormick

  31. e.mccormick
    • one year ago
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    c, as you have it, really does not need to exists. c) ad + cd = ac, be + ec = bc, segment addition postulate d) ad + ce = bc, subsitution of ad for be e) ad + ce = ac, subsitution of ac for bc Hmmm..... at that point, I know it is saying CD = CE beacuse they are in the same place int he same equation.. so probably by substitution too....

  32. e.mccormick
    • one year ago
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    Or perhaos the converse of subsitution.

  33. e.mccormick
    • one year ago
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    ARGH! I know what subsitution I meant! d) ad + ce = bc, subsitution of ad for be e) ad + ce = ac, subsitution of ac for bc f) ad + ce = ad + cd, subsitution of ad + cd for ac g) ce = cd... this is one of the basic addition ones. Do you remember the name of it?

  34. e.mccormick
    • one year ago
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    Additive equality.

  35. anonymous
    • one year ago
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    So then h) CD = ec

  36. e.mccormick
    • one year ago
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    Yah, then you do the cd cong ce because of definition of congruent.

  37. e.mccormick
    • one year ago
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    I hope that makes more sense now. Even I went a little sideways on that one for a moment. Well... there is usually more than one way to do it, but on this one a subsitution chain is clear. Being clear is the intent of proofs. Clear and detailed baby steps that lead to the answers is basically what a proof is.

  38. anonymous
    • one year ago
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    Thank hopefully it's all good

  39. e.mccormick
    • one year ago
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    And this is the full name of that additive thing... http://www.mathwords.com/a/additive_property_of_equality.htm

  40. anonymous
    • one year ago
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    Okay sweet

  41. anonymous
    • one year ago
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    @e.mccormick My teacher said it was Ac=bc , Dc= ec not ad=be im still confused

  42. anonymous
    • one year ago
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    @e.mccormick

  43. anonymous
    • one year ago
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    i got it!!!

  44. e.mccormick
    • one year ago
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    Well, that would not make much differrence, and it seems you were able to apply what was done yesterday to fix it.

  45. anonymous
    • one year ago
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    A) AC≅BC, DC≅EC B) AC = CB, DC=CE C) AC + CB = DC = CE D) DC + AC = DC CB + CE =CE E) DC = CE F) DC ≅ CE

  46. anonymous
    • one year ago
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    if that isnt right then i give up on this assignment

  47. e.mccormick
    • one year ago
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    Wait.... the teacher said it is DC≅EC at the top in the givens?

  48. anonymous
    • one year ago
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    yes with ac = bc

  49. e.mccormick
    • one year ago
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    Then it is stupidly simple. A) DC≅EC B) DC = CD, EC = CE: defenition of a line segment C) CD≅CE: substitution of CD for DC and CE for EC. I think your teacher made a mistake, and what we did yesterday is probably right. I do not think they would ask something that simple. See, what I did here is the line segment from D to C is the same as from C to D based on the definition of a line segment.

  50. anonymous
    • one year ago
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    okay so other than that part its okay?

  51. e.mccormick
    • one year ago
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    No, I am saying that what your teacher gave you as a "correcton" means it can be done in those 3 steps. If I go from point D to C, it is the same as if I go from C to D. Your teacher is saying that the answer is a given... which makes no sense for a proof.

  52. anonymous
    • one year ago
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    im so confused

  53. anonymous
    • one year ago
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    i thought my teacher wanted me to do both not just the dc=ec

  54. e.mccormick
    • one year ago
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    Well, what we did yesterday works well... but your teacher is saying that the correction to the typo is for what you want to prove to be is as a given. That just does not make sense.

  55. anonymous
    • one year ago
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    i dont know ill just figure it out i guess ..

  56. e.mccormick
    • one year ago
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    What I am saying is that your teacher seems to have messed up in fixing it. It is like saying, "Prove the the road from New York to New Jersey is the same length as the road from New Jersey to New York." Your teacher said: Given AC≅BC, DC≅EC Prove CD≅CE Well duh! DC≅EC = CD≅CE If that is NOT what your teacher meant, then the thing I went over yesterday is the correct answer.

  57. anonymous
    • one year ago
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    Oh wow im sorry i just fully understood that ill have to ask her about that

  58. e.mccormick
    • one year ago
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    LOL. Yah. I think you got it yesterday and your teacher is still confused. Yesterday we basically did: a + b = c and x + y = z Given a = x and c = z, prove that b = y. That is the short form, and we did all that using the geometric notation for it.

  59. anonymous
    • one year ago
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    well ill talk to her and let know

  60. anonymous
    • one year ago
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    let you know *

  61. e.mccormick
    • one year ago
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    I am the dyslexic, poor typist here... so a missing word or two is no big deal. =)

  62. anonymous
    • one year ago
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    haha she said it sounded good i just need the reasons next to those 3

  63. anonymous
    • one year ago
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    which i got from you thank you um do you think you can help me with the other problem on that paper?

  64. e.mccormick
    • one year ago
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    The top one?

  65. anonymous
    • one year ago
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    yes

  66. e.mccormick
    • one year ago
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    The Linear Pair Postulate I pointed out yesterday, and some substitution because the inner angles are given as equal should do it.

  67. anonymous
    • one year ago
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    my teacher said the 2 angles are given to be equal, this means it is an isosceles triangle. This should make your proof easier

  68. e.mccormick
    • one year ago
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    Well, I do not know of any isosceles triangle specific rules that apply with how I was saying you could do it. I was just pointing out that the inner angles match and can be substituted for each other, which means you can make it so the outer ones must also subsitute... it is actually kind of similar to the second one when we did it the last way we did yesterday.

  69. anonymous
    • one year ago
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    so A) is m<xbc=m<xbc then for b you would xcb+ xbc=xcd+ xba?

  70. e.mccormick
    • one year ago
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    On A, notice the C and B swap positions. A) m∠XBC = m∠XCB, Given For B I would do: m∠XBA = 180 - m∠XBC, and m∠XCD = 180 - m∠XCB: Linear Pair Postulate

  71. anonymous
    • one year ago
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    what do we do for the next step im starting to get a little confused

  72. e.mccormick
    • one year ago
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    A) m∠XBC = m∠XCB, Given B) m∠XBA = 180 - m∠XBC, and m∠XCD = 180 - m∠XCB: Linear Pair Postulate Two substitutions and it is done. With what you know there, what can you subsitute in: m∠XBA = 180 - m∠XBC

  73. anonymous
    • one year ago
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    this problem has me completely lost

  74. e.mccormick
    • one year ago
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    Perhaps it is all the m∠XBA, m∠XBC etc stuff. Lets try something to make that simpler. I will use Bi for the XBC anf Bo for XBA, that is because they are point B's inner and outer angles. Similar things can be done for point C. |dw:1440119011210:dw| Given: Bi = Ci Prove: Bo = Co A) Bi = Ci: given B) Bo = 180 - Bi and Co = 180 - Ci : Linear Pair Postulate Do you see where this is going? Two substitutions and it is done.

  75. anonymous
    • one year ago
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    yes i do understand what you did

  76. anonymous
    • one year ago
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    @e.mccormick im just having a hard time figuring out what each step like i dont know whats next

  77. anonymous
    • one year ago
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    i cant wrap my head around it to know what to do next but i do understand whats going on

  78. e.mccormick
    • one year ago
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    Well, at this point it is algebra to make them look alike. You just need to do a substitution. See, the line A tells you what substitutions can be done in B. You were given something. You use a rule to show something. Then you use the given and a new rule to substitute for the next step.

  79. anonymous
    • one year ago
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    oh okay

  80. e.mccormick
    • one year ago
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    Once you do that sub, then you will have one last sunstitution. The result of the first substitution and the other half of B will make for a second subsitution! Or you could do it a little longer, just putting something on seperate lines: Given: Bi = Ci Prove: Bo = Co A) Bi = Ci: given B) Bo = 180 - Bi : Linear Pair Postulate C) Co = 180 - Ci : Linear Pair Postulate D) subsitution involving lines B) and A) E) subsitution involving lines C) and D) And it will be done! That is one operation per line, which is the normal in formal proofs.

  81. anonymous
    • one year ago
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    okay hold im trying to figure it out lol

  82. anonymous
    • one year ago
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    D) M<xbc-m<xcb=m<xba=180-m<xbc E) m<xba=180-m<xbc=m<xbc-m<xcb=m<xba=180-m<xbc

  83. anonymous
    • one year ago
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    i think i totally messed that up @e.mccormick

  84. e.mccormick
    • one year ago
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    The goal is to get m∠XCD = m∠XBA So that is going to be step E no matter what. The question is, how do you get there through two substitutions. A) m∠XBC = m∠XCB, Given B) m∠XBA = 180 - m∠XBC: Linear Pair Postulate C) m∠XCD = 180 - m∠XCB: Linear Pair Postulate OK, so your next steps were: D) m∠XBC-m∠XBC=m∠XBA=180-m∠XBC E) ∠XBA=180-∠XBC=∠XBC-∠XCB=∠XBA=180-∠XBC Hmm.... I am not sure how you went that way. Let me give a little more detail. When you do a substitution, it is based on what you know. I said it involved these two lines: A) m∠XBC = m∠XCB, Given B) m∠XBA = 180 - m∠XBC: Linear Pair Postulate Because of the rules of equality, only four possibilities can be true: 1) m∠XBC can be replaced by m∠XCB 2) m∠XCB can be replaced by m∠XBC 3) m∠XBA can be replaced by (180 - m∠XBC) 4) (180 - m∠XBC) can be replaced by m∠XBA Now, while I can only replace (180 - m∠XBC) with m∠XBA, the m∠XBC INSIDE (180 - m∠XBC) can be replaced by m∠XCB. Take a look at where that leads.

  85. e.mccormick
    • one year ago
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    And remember, the last step is: E) m∠XCD = m∠XBA by some sort of substitution. So D must set up for that. It involves C and D. Well, by C: C) m∠XCD = 180 - m∠XCB: Linear Pair Postulate 1) m∠XCD can be replaced by (180 - m∠XCB) 2) (180 - m∠XCB) can be replaced by m∠XCD So with just a little logic, you can see that D must use at least one of those so it can become the other.

  86. anonymous
    • one year ago
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    so D) is m∠XBC = m∠XCB, m∠XBA = 180 - m∠XBC

  87. anonymous
    • one year ago
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    @e.mccormick

  88. e.mccormick
    • one year ago
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    I was off... the joys of work. C) m∠XCD = 180 - m∠XCB: Linear Pair Postulate If you swap that m∠XCB for ... hmmm.... did I write those wrong one of these times? Argh.... Let me use the shorthand one. Given: Bi = Ci Prove: Bo = Co A) Bi = Ci: given B) Bo = 180 - Bi : Linear Pair Postulate C) Co = 180 - Ci : Linear Pair Postulate D) Bo = 180 - Ci : Substitute Ci for Bi. That works because in A you say Bi is Ci. E) Bo = Co : Substitute Co for 180 - Ci That works becaus in C you say those are the same. There... Now you just need to replace the shorthand with the m∠XCB stuff. I hope you can understand why it works.

  89. anonymous
    • one year ago
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    I got D) m<XBC =180-m<XCB ?

  90. anonymous
    • one year ago
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    @e.mccormick

  91. anonymous
    • one year ago
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    D is the only im having trouble with

  92. e.mccormick
    • one year ago
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    It would be one of the otuer ones. m<XBC is inner... D) Bo = 180 - Ci : Substitute Ci for Bi. Becomes: D) m<XBA = 180 - m<XCB : Substitute m<XCB for m<XBC.

  93. anonymous
    • one year ago
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    oh okay i think i get it now

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is replying to Can someone tell me what button the professor is hitting...

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