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Side lengths of inscribed triangle

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  1. Apr 8, 2015 #1
    1. The problem statement, all variables and given/known data
    The corners A, B and C of a triangle lies on a circle with radius 3. We say the triangle is inscribed in the circle. ∠A is 40° and ∠B is 80°.

    Find the length of the sides AB, BC and AC.

    2. Relevant equations


    3. The attempt at a solution
    I found out the arc AB is 2π, arc BC is 4π/3 and arc AC is 8π/3. Don't know if it's relevant, but I did it nonetheless. The book says nothing about finding the arc or finding the lengths, the question came out of nowhere. It's a shitty book to say the least. Don' really know where to go from here. Is there a constant ratio between length of sides of a central angle and something else?
     
  2. jcsd
  3. Apr 8, 2015 #2
    Hint:
    1-What's the sum of angle in a triangle ?
    2-Does this sound familiar BC/angle(A) = AC/angle(B) = AB/angle(C) = ?
    Edit: [= 2R], R is the radius ;)
     
  4. Apr 8, 2015 #3
    Oh my god, of course, thank you so much!
     
  5. Apr 8, 2015 #4
    Welcome ;)
     
  6. Apr 9, 2015 #5

    LCKurtz

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    Not to me. It just sounds wrong.
     
  7. Apr 9, 2015 #6
    I was thinking (s)he meant a/sin A = b / sin B = c/sin C, but I didn't have a single side length so it wasn't much help either way.
     
  8. Apr 9, 2015 #7

    LCKurtz

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    I supposed that too. So, have you solved the problem?
     
  9. Apr 9, 2015 #8
    No :-( Any ideas?
     
  10. Apr 9, 2015 #9

    LCKurtz

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    Hint: Draw the radii to the three vertices. There is a relation between the central angles and the angles of the triangle.
     
  11. Apr 9, 2015 #10
    This is how far I've gotten, but I haven't the slightest clue what to do next. Unless I can somehow find the one other angle in one of the triangles.

    Screenshot_2015-04-09-20-55-16.png
     
  12. Apr 9, 2015 #11

    LCKurtz

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    That's good so far. So for each interior triangle, you have two equal sides and need the third side. The law of Sines wasn't helping you. But there is another law...
     
  13. Apr 9, 2015 #12
    A/sin(a) = B/sin(b) = C/sin(c) = 2R
    Solve for lengh, you hava the radius and the angles,
     
  14. Apr 9, 2015 #13
    Cosines :-D Jesus, never would have thought of that, we haven't had any trig in this class. Out of sight, out of mind. Thank you! Is it correct?
     

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  15. Apr 9, 2015 #14
    I'm sorry, but I don't know what you mean.
     
  16. Apr 9, 2015 #15
    That can only be true if you know two length and one angle, in your case no lenght, i'm gonna break the law and tell you the solution,
     

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  17. Apr 9, 2015 #16
    I solved it above using the law of cosines, but where did the 2r come from? How do you know the ratio is 2×radius of circumscribed circle? :-)
     
  18. Apr 9, 2015 #17
    You just messed it up and wrongly used the cosine law (al-kashi), just if you read my first post, the sine law is a/sin(A) = b/sin(B) = c/sin(C) = 2R where r is the radius of the circle you are talking about, take it easy .
     
  19. Apr 9, 2015 #18
    Take it easy? Don't know what you're talking about, but okay. What was wrong with it? What does al-kashi mean? Haha. What I'm wondering is how you know the ratio A/sin A = 2r? :-)
     
  20. Apr 9, 2015 #19
    What i mean is that you over complexed the problem and its wrong , from where did the 3 came of, wrong use of cos law and check any site or your textbook to check the esuation i gave you,
     
  21. Apr 9, 2015 #20

    SammyS

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    As Maddie asked: Where did you get the relation, A/sin(A) = 2R ?

    It seems LCKurtz didn't consider it to be something commonly known, and I agree with him.
     
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