calculation of largest angles

Discussion in 'Math' started by Biggsy100, Jul 23, 2014.

  1. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    88
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    Having three struts measuring 5.6m, 3.7m, 2.9m to form a triangular joist section, use appropriate formulae to calculate largest angle.

    So by the numbers given I know this as an hypotonese triangle.

    If I delegate each length a letter A B C I can get each angle by simply square rooting to letters? So for instance, a= square root of b*2 + C*2?
     
  2. amilton542

    Active Member

    Nov 13, 2010
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    64


    2.9^2 + 3.7^2 not equal to 5.6^2 m^2


    Use sine/cosine rules.
     
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  3. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    88
    1
    can you please clarify what neq means please?
     
  4. amilton542

    Active Member

    Nov 13, 2010
    494
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    Sorry, I tried to use Latex editor but it failed. It was code, but I changed it.
     
  5. amilton542

    Active Member

    Nov 13, 2010
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    Do you know sine/cosine rules?
     
  6. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    88
    1
    So it seems a little long winded but I have designated side a to 2.9 metres: hence

    2.9^2=3.7^2 +5.6^" - 2.3.7.56 Cos( α)

    8.41 = 13.69 +31.36 - 41.44. Cos( α)

    41.44. cos(α ) = 13.69 +31.36 -8.41

    41.44. cos( α) = 36.64

    cos (α) =0.88416988417

    (a) = 27.85 degrees?

    *** This is the part you tell me it can be done in under 5 seconds :D
     
  7. amilton542

    Active Member

    Nov 13, 2010
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    Yep, you got it. Five seconds is a bit of an over statement, but it speeds up with practice.

    Maths is fun!
     
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  8. amilton542

    Active Member

    Nov 13, 2010
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    @Biggsy100

    I knew that angle seemed tiny, I wasn't paying attention. The largest angle is opposite the longest side.

    You need to recalculate!
     
  9. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,802
    832

    I am lost at the end of this equation. 5.6 is raised to what? And what does 2.3.7.56 represent? It's not a number…
     
  10. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    88
    1
    I will have to work on this and give some more thought as I am not sure how to get the angle with just the strut measuremenst
     
  11. Blackbull

    Well-Known Member

    Jul 26, 2008
    70
    6
    The Cosine Rule: In any triangle the square of one side equals the sum of the squares of the other 2 sides minus twice the product of the same two sides and the cosine of their included angle:
    a^2 = b^2 + c^2 - 2bc Cos A
    or Cos A = b^2 + c^2 - a^2 / 2bc
    and Cos B = a^2 + c^2 - b^2 / 2ac
     
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  12. Chalma

    Member

    May 19, 2013
    54
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    When you need help always remember the old Indian tribe, SOHCAHTOA. Have fun.
     
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  13. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    88
    1
    Is this better?

    a^2 = b^2 + c^2 - 2.b.c cos(a)

    Substituting a = 5.6, b = 3.7,c=2.9

    5.6^2= 3.7^2 +2.9^2-2.37.2.9 cos(a)

    31.36 = 13.69 +8.41 -21.46. cos (a)

    21,46= cos(a) =13.69+8.41-31.36

    21.46.cos (a) = -9.26

    cos(a) = - 0.431

    a = arcoss (0.431)

    (a) = 115.56
     
  14. Blackbull

    Well-Known Member

    Jul 26, 2008
    70
    6
    That's what I get Biggsy - SOHCAHTOA, I was told it was a mountain in Japan with sides of ratio 3,4,5. It would have been easier if our maths teacher, at school, had told us this, I didn't come across it till later. I can remember the Cosine Rule in words rather than the equation.
     
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  15. amilton542

    Active Member

    Nov 13, 2010
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    SOH-CAH-TOA does not apply to an obtuse triangle. It holds for right-angled triangles.
     
  16. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    88
    1
    Have I got it correct this time?
     
  17. amilton542

    Active Member

    Nov 13, 2010
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    Yep, largest angle is opposite the longest side.
     
  18. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    88
    1
    b^2= a^2 + C^2 - 2.a.c Cos(B)

    After substitution b = 3.7, a =5.6, c= 2.9

    b^2=5.6^2 + 2.9^2 - 2. 5.6 .2.9 Cos B

    13.69 = 31.36 +8.41 - 32.48. Cos B

    32.48. Cos(B) = 31.36 +8.41 - 13.69

    32.48 Cos(B) = 26.08

    Cos (B) = 0.802

    (B) = arcos (0.802)

    (B) = 36.58
     
  19. amilton542

    Active Member

    Nov 13, 2010
    494
    64
    Are you trying to solve for all angles?

    Yep it sure is, but try substituting your numbers at the very end. It's less error prone and you can see what you're looking at before you make the substitution.

    b^2 = a^2 + c^2 - 2ac cos(B)

    2ac cos(B) = a^2 + c^2 - b^2

    Angle B = Arc cos [(a^2 + c^2 - b^2)\(2ac)]

    Angle B = Arc cos [(5.6^2 + 2.9^2 - 3.7^2)\(2)(5.6)(2.9)]

    Angle B = 36° 35' 12.19"

    Optional method of expressing the angle using the base-60 Sexagesimal system, the (° ' ") on your calculator. It's an exact plane angle and you can retrieve it in pure decimal format if you wish.

    Angle B = 36 + (35\60) + (12.19\60^2) = 36.58671944° ≈ 36.59°
     
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  20. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    88
    1
    Thank you for that! Should I apply the same methods to a base plate calculation? I know I need to split the calculation or visualise the base plate split into two?

    The problem I have is:

    A design brief indicates three holes A,B,C to be drilled in a plate, with a=73mm, b=117mm and a =37 degrees

    I need to find the possible value for the dimension of C? Should I simply split into 2 triangles and measure accordingly?
     
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