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Math Help - Law of Cosines (finding side x in a triangle)

  1. #1
    Junior Member Melancholy's Avatar
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    Law of Cosines (finding side x in a triangle)

    Find x to the nearest tenth.


    Okay, so according to the Law of Cosines:
    a^2 = b^2 + c^2 - 2bc (cos A)
    b^2 = a^2 + c^2 - 2ac (cos B)
    c^2 = a^2 + b^2 - 2ab (cos C)

    This is what I have so far:
    a^2 = 32^2 + 23^2 - 2(32)(23) (cos A)
    a^2 = 1024 + 529 - 1472 (cos A)
    a^2 = 1553 - 1472 (cos A)
    a = [square root sign] 1553 - 1472 (cos A)

    First of all, is the math/formula I did/chose correct? Second, I'm having Algebra 1 issues: where the heck do I go from here? I tried putting this in my calculator: "[square root sign](cos)1553 - 1472" but it came out with the decimal 0.395517971. I have a TI-30XS calculator... and I really need help understand this... test tommorrow...
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Melancholy View Post
    Find x to the nearest tenth.


    Okay, so according to the Law of Cosines:
    a^2 = b^2 + c^2 - 2bc (cos A)
    b^2 = a^2 + c^2 - 2ac (cos B)
    c^2 = a^2 + b^2 - 2ab (cos C)

    This is what I have so far:
    a^2 = 32^2 + 23^2 - 2(32)(23) (cos A)
    a^2 = 1024 + 529 - 1472 (cos A)
    a^2 = 1553 - 1472 (cos A)
    a = [square root sign] 1553 - 1472 (cos A)

    First of all, is the math/formula I did/chose correct? Second, I'm having Algebra 1 issues: where the heck do I go from here? I tried putting this in my calculator: "[square root sign](cos)1553 - 1472" but it came out with the decimal 0.395517971. I have a TI-30XS calculator... and I really need help understand this... test tommorrow...
    first of all, type [tex]\sqrt{x}[/tex] to get \sqrt{x}

    second of all, you know the value for A (it is 82^o), so plug it in. you will then be able to solve for a
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