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Math Help - Calculating an angle of a none right angle triangle

  1. #1
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    Calculating an angle of a none right angle triangle

    I am very new to maths so I would appreciate if posters are patient with me as this is a learning curve

    I have a triangle with no equal sides and I am looking to find angle B. The length of the sides I will make up as 12cm, 22cm and 28cm.

    I know all three sides and none of the angles and I am thinking that I should be using the cosine rule, therefore I have completed the following;

    C2 = a2 + b2 2ab cos

    122 = 222 + 282 2 22 28 cos
    144 = 484 + 784 1232 cos
    144 = 1268 1232 cos
    1232 cos = 1124
    Cos = 1124
    1232
    Cos = 0.91
    Cos -1 = 24.2 (1dp)

    Have I used the right rule, if not please advise why, and if I have would you please advise why I can't use the Sine Rule and what the difference is?

    Thanks

    David
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    Re: Calculating an angle of a none right angle triangle

    Quote Originally Posted by David Green View Post
    to find angle B. The length of the sides I will make up as 12cm, 22cm and 28cm.
    First you must know which is which.
    Say a=12,~b=22,~\&~c=28. If that is so then
    m(\angle B) = \arccos \left( {\frac{{a^2  + c^2  - b^2 }}{{2ac}}} \right)
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    Re: Calculating an angle of a none right angle triangle

    hi David Green

    you chose the right rule.the Sines rule wouldn't have worked because for that you need at least one angle.

    i am not sure about the last line.it should be:
    =cos-1 (0.91)
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    Re: Calculating an angle of a none right angle triangle

    This is confusing me now, if I use the first example I get the result 2.67, and if I use cos then I get 99.

    If I use theta cos -1 (0.91) I get 24.5 degrees

    Help
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    Re: Calculating an angle of a none right angle triangle

    Quote Originally Posted by David Green View Post
    This is confusing me now, if I use the first example I get the result 2.67, and if I use cos then I get 99.
    If I use theta cos -1 (0.91) I get 24.5 degrees
    Did you read reply #2?
    The angle B is opposite the side with length b.
    So it all depends on how you name the triangle.
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  6. #6
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    Re: Calculating an angle of a none right angle triangle

    hi plato

    i think that's what confused the OP.you named it differently than him and me.
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    Re: Calculating an angle of a none right angle triangle

    Quote Originally Posted by Plato View Post
    Did you read reply #2?
    The angle B is opposite the side with length b.
    So it all depends on how you name the triangle.
    Reply 2 is correct, A = 12, B = 22 and C = 28

    So are we saying that the angle is 2.67 degrees?

    2.67 = theta 99

    0.91 cos -1 = 99

    Are they not both the same then?
    Last edited by David Green; August 18th 2011 at 04:25 PM. Reason: comparing results
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    Re: Calculating an angle of a none right angle triangle

    Quote Originally Posted by anonimnystefy View Post
    i think that's what confused the OP.you named it differently than him and me.
    No that is the whole point.
    He did not name them at all.
    He talked about finding B and the uses
    c^2=a^2+b^2-2ab\cos(\theta)
    He is clearly confused by the notation.
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    Re: Calculating an angle of a none right angle triangle

    Quote Originally Posted by David Green View Post
    Reply 2 is correct, a = 12, b = 22 and c = 28
    If you say that a = 12,~ b = 22 ~\&~ c = 28
    then m(\angle B) = \arccos \left( {\frac{{12^2  + 28^2  - 22^2 }}{{2(12)(28)}}} \right)
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    Re: Calculating an angle of a none right angle triangle

    Looking deeper into this today it seems that I am using the wrong cosine rule?

    The book in the summary section only shows; c^2 = a^2 + b^2 - 2ab cos C

    I initially thought the book print had make an error where I thought that cos C should have read cos theta, which is what I then wrote on this forum.

    Using;

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

    angle B =

    28^2 = 12^2 + 22^2 - 2 x 22 x 12 cos B

    784 = 144 + 484 - 328 cos B

    - 528 cos B = 784 - 628

    = 156 / 528

    cos B = 0.29

    theta = cos -1 (0.29) = 73 degrees.

    This is what I think is right, any views welcome

    David
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    Re: Calculating an angle of a none right angle triangle

    You say:
    Quote Originally Posted by David Green View Post
    Reply 2 is correct, A = 12, B = 22 and C = 28
    But then you use
    Quote Originally Posted by David Green View Post
    b^2 = c^2 + a^2 - 2ac cos B

    28^2 = 12^2 + 22^2 - 2 x 22 x 12 cos B
    Why change?

    Do the calculations is reply #9 please.
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    Re: Calculating an angle of a none right angle triangle

    Quote Originally Posted by Plato View Post
    You say:


    But then you use

    Why change?

    Do the calculations is reply #9 please.
    The calculations in reply 9 would = 0.66 = 49 degrees

    The problem there was that I used the wrong cosine rule.

    if AB = 12 and BC = 22 and AC = 28, then I am asked to find the angle for B, then this is why I changed the cosine rule.

    David
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    Re: Calculating an angle of a none right angle triangle

    Quote Originally Posted by David Green View Post
    The calculations in reply 9 would = 0.66 = 49 degrees

    The problem there was that I used the wrong cosine rule.

    if AB = 12 and BC = 22 and AC = 28, then I am asked to find the angle for B, then this is why I changed the cosine rule.
    I think that I now understand you confusion.
    \angle B is opposite side \overline{AC} so b=28.
    \angle A is opposite side \overline{BC} so a=22.
    \angle C is opposite side \overline{AB} so c=12.

    Now use m\left( {\angle B} \right) = \arccos \left( {\frac{{a^2  + c^2  - b^2 }}{{2ac}}} \right)
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  14. #14
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    Re: Calculating an angle of a none right angle triangle

    Quote Originally Posted by Plato View Post
    I think that I now understand you confusion.
    \angle B is opposite side \overline{AC} so b=28.
    \angle A is opposite side \overline{BC} so a=22.
    \angle C is opposite side \overline{AB} so c=12.

    Now use m\left( {\angle B} \right) = \arccos \left( {\frac{{a^2 + c^2 - b^2 }}{{2ac}}} \right)
    Your right, the conclusion I get using the formula you wrote is 107 degrees?

    Now I am even more confused

    David

    Edited section below;

    I have had another look at the formula I used here;

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

    I can see in my transposition that I changed the negative sign to a positive when changing the integers from side to side because I thought ending up with a negative - 0.29 was incorrect, however after working out the angle I do indeed get 107 degrees, which is what I also get when using the formula as you wrote it, which I might add is a tidy method and proves that a less chance of making the mistake I made possible.

    I think we can agree now that the solution is 107 degrees and I do like the way you do the formula, which is better than in my maths book.

    Thanks

    David
    Last edited by David Green; August 19th 2011 at 02:42 PM. Reason: Re-worked through the formula
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    Re: Calculating an angle of a none right angle triangle

    Quote Originally Posted by David Green View Post
    Your right, the conclusion I get using the formula you wrote is 107 degrees? Now I am even more confused
    The c107 is correct. What are you confused about?
    If you solve b^2=a^2+c^2-2ac\cos(B) for \cos(B) we get cos(B)={\frac{{a^2  + c^2  - b^2 }}{{2ac}}}.
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