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Math Help - Cosine Rule

  1. #1
    Newbie
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    Abingdon, Oxon.
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    Cosine Rule

    I need to calculate the permeter of a triangle ABC
    Side AB 9cm, Side BC 15cm and Angle B 110

    Using the Cosine Rule I Know
    b = a+c-2accos B
    b = 15+9-2x15x9xcos110
    It is claimed that
    b=398.345
    b=20cm approx, a scaled drawing proves this to be the case.

    My problem is, I want to make it

    b = 15+9-2x15x9xcos110
    b = 225+81-2x15x9x.3420
    b = 225+81-92.34
    b = 306-92.34
    b = 213.66
    b = 14.62 approx which is wrong!!

    If I add 92.34 to 306 I get the correct answer but I do not know where I have gone wrong. Can anybody help me?

    berniecole
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  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
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    Hello, berniecole!

    It looks like you aren't using your calculator well . . .


    Calculate the perimeter of a triangle ABC:
    c= 9,\;a = 15,\;\angle B = 110^o

    Using the Cosine Rule:

    . b^2\:=\:a^2+c^2-2ac\cos B \;=\;15^2+9^2-2(15)(9)\cos110^o \;=\; 398.3454387

    . . b\:=\:19.9585931 \:\approx\:20



    My problem is, I want to make it

    . . b^2 \;=\;15^2+9^2-2(15)(9)\cos110^o .
    . . . I don't see any difference

    . . . . =\;225+81-(2)(15)(9)\underbrace{({\color{red}0.3420})}_{\tex  t{No!}}

    \cos110^o \;=\;{\color{red}-}0.342020143

    110^o is in Quadrant 2, where cosine is negative.

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