Law of Sines

In Triangle ABC, a=15cm, c=9cm, and angle C= 35 degrees. Find b and angle b. (Recall sin theta= sin (180 degrees-theta). I'm not sure how to approach this.

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asked by Richard
  1. you have C and c and a, so you can find A using

    sinA/a = sinC/c

    Now, having A and C, you know that A+B+C=180, so you can easily find B.

    Then, use the law of sines again to get b:

    b/sinB = c/sinC = a/sinA

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    posted by Steve
  2. sin A/15 = sin 35/9 = .0637

    sin A = .956

    so A = 72.9 degrees or 180-72.9 = 107.1
    then A + B + C = 180
    72.9 + B + 35 = 180
    B = 72.1 degrees (One answer)
    sin B/b = .0637
    so b = 14.94
    or
    A = 107.1
    then B = 180 - 35 -107.1 = 37.9 degrees (alternate answer)

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    posted by Damon
  3. Thanks! I should have mentioned this in the question, but the answer sheet asks for two values of angle A rather than 1.

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    posted by Richard
  4. Oh nvm thanks Damon.

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    posted by Richard
  5. And steve sorry I didn't read the whole answer.

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    posted by Richard

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