Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook