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Geometry: Finding The Representation For An Angle

  1. May 26, 2012 #1
    [itex]\stackrel{\rightarrow}{PQ}[/itex] bisects [itex]\angle{ABC}[/itex].
    If [itex]m\angle{ABD}[/itex] can be represented by [itex]3a + 10[/itex] and [itex]m\angle{DBC}[/itex] can be represented by [itex]5a - 6[/itex], what is [itex]m\angle{ABC}[/itex]?

    What I did was add the two representations together and multiplied by two, which would give me [itex]16a + 8[/itex]. Does that sound correct?
     
  2. jcsd
  3. May 26, 2012 #2
    Where is the point D? Also, is it supposed to read ##\vec{BD}## instead of ##\vec{PQ}##?

    Also, why would you multiply by 2?
     
  4. May 26, 2012 #3

    SammyS

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    What does [itex]\stackrel{\rightarrow}{PQ}[/itex] have to do with anything?

    Why multiply by 2 ?
     
  5. May 28, 2012 #4
    Sorry to both of you. It is not suppose to be PQ, but BD; I was looking at the wrong problem. Well, I was thinking, since it was a bisection of an angle, it would be the whole angle cut in half. But, know that I really consider it, that would be equivalent to [itex](1/2 + 1/2)2[/itex], which not represent the whole angle. Now, since I know that those two expressions represent the same angle measurement, could I set them equal to each other, solve for a, and finally multiply that value by two get get the un-bisected angle?
     
  6. May 28, 2012 #5
    Okay, that makes more sense! You almost have the right idea. You are absolutely correct in your method for solving for a. However, a by itself does not represent either angle, so multiply a by 2 would serve no purpose. However, plugging in a for either of the two equations (##3a+10## or ##5a-6##) will give you the equation for one half of the angle. Then you are correct, you would multiply THAT angle by two to get the entire angle.

    Alternatively, you can add ##3a + 10## and ##5a-6## together and plug in a to get the entire angle. Both methods work!
     
  7. May 28, 2012 #6
    Thank you for helping me.
     
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