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Given angle A and two sides, find Angle B.

  1. Jun 23, 2009 #1
    1. The problem statement, all variables and given/known data

    In triangle ABC, A=65 degrees, b=9, and a=10. Find B.


    2. Relevant equations



    3. The attempt at a solution

    I honestly don't know how to start this problem... I tried to use the law of cosines to find side C, and then again to find Angle B, but that answer was incorrect. I haven't been in Trig for two years, and can't think of how else to start it.

    Thanks for any help!
     
  2. jcsd
  3. Jun 23, 2009 #2
    Try the law of sines.
     
  4. Jun 23, 2009 #3
    6joz60.jpg

    First use the sine to find h, and then use again the sine to find B, using arcsin(B).

    [tex]\sin A = \frac{h}{b}\text{ and } \sin B = \frac{h}{a}.[/tex]

    Regards.
     
    Last edited: Jun 23, 2009
  5. Jun 23, 2009 #4

    HallsofIvy

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    so that h= b sin A= a sin B and, as a result, [itex]\frac{sin A}{a}= \frac{sin B}{b}[/itex]. That is the "sine law" Bohrok is talking about.
     
    Last edited: Jun 23, 2009
  6. Jun 23, 2009 #5
    Yes, you're right. I misspelled the words. I thought of "sine" and not "sine law".

    Anyway, the things are same, and he would come up with the same result.
     
  7. Jun 23, 2009 #6

    HallsofIvy

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    Yes, I wasn't criticizing, just pointing out to JacklinH that you were both giving the same advice!
     
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