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Question from trigonometry -- prove that the largest angle is greater than 120°

  1. Apr 17, 2016 #1
    • Thread moved from the technical forums, so no HH Template is shown.
    In a triangle whose sides are 3,4 and root 38 metres respectively, prove that the largest angle is greater than 120°
    • My answer: Here angle C > B.....(1) and C > A ......(2) adding (1) and (2) we get 2c > A+B, taking sine both side , sin2C = sin(A+B) = sin(C), therefore cosc > (1/2) therefore angle C> 60°
    • I am not getting C> 120°, what is the problem please help
     
  2. jcsd
  3. Apr 17, 2016 #2
    (180-60)=120 is also a solution, now you should think how to prove that particular solution
     
  4. Apr 17, 2016 #3
    Why not use the law of cosines? Find the angle between the short sides.

    ## 38 = 3^2 + 4^2 - 2(3)(4)cos(θ) ##

    Now solve for θ.
     
  5. Apr 18, 2016 #4

    SteamKing

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    There's a typo in the equation above. The 38 should also be squared.
     
  6. Apr 18, 2016 #5

    SammyS

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    No. It's ##\ \sqrt{38}\ ## which is being squared.
     
  7. Apr 18, 2016 #6

    SteamKing

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    Missed the root in the OP.
     
  8. Apr 18, 2016 #7

    haruspex

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    I guess you meant sin 2C > sin(A+B), but that does not follow. Once the angle exceeds 90 degrees the sine function is decreasing. Indeed, sin(2C) would be < sin(A+B) here.
    That's backwards. If sin(2C) were > sin(A+B), it would follow that C is less than 60.

    @mfig's method is fine, but you do not need to solve for theta. Just show that the longest side is too long for a 120 angle.
     
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