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Sine rule

  1. Nov 20, 2008 #1
    I have read the sine rule:
    It states-->
    sin A/a=sin B/b=sin C/c = 1/2R

    where R is circumradius.
    a=2Rsin A
    b=2Rsin B
    c=2Rsin C

    For a triangle R is fixed.
    In an obtuse angled triangle, the side opposite largest angle is the longest(geomtrically)

    But the sine of an obtuse angle can be less than that of an acute angle. How is this possible? Is it possible that the side opposite obtuse angle isn't the largest? Or is the sin rul giving a wrong stand in this case??
  2. jcsd
  3. Nov 20, 2008 #2


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    Hi ritwik06! :smile:

    Yes, I see what you mean …

    if A is obtuse, then as A increases, sinA decreases, but a (the side a) actually increases though it should decrease!

    hmm :rolleyes:

    ah, but that's assuming that b and c are constant, in which case R is not constant, so everything sorts itself out.

    But if you keep R constant, so that, say, the "C" end of side a slides round the circle, then as A increases, the "C" end will get closer, and therefore a will actually decrease, (and so will b) …

    there isn't actually a problem. :smile:
  4. Nov 20, 2008 #3
    Suppose the angles of a triangle are 30,45, 105!
    what will be the ratio of their sides?

    as I get by sin rule. Isnt it?

    does that mean that the side opposite 30 is greater than the one opposite 45???
  5. Nov 20, 2008 #4


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    No, the ratio of the sides is 1/2 : 1/√2 : (√3 + 1)/2√2,

    or √2 : 2: √3 + 1 :smile:
  6. Nov 20, 2008 #5
    I am sorry! Thanks for pointing out my misconception. Thanks a lot!
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