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Homework Help: Trig Problem

  1. Dec 28, 2004 #1
    I just read about the cosine law and the sine law.

    I have a practise problem and know to use the cosine law but what ever answer I get gives me a math error in my calculator.

    the sides are a=4.3 b=5.2 c=7.5 I need to solve the triangle so find the 3 angles within.

    [tex] a^2=b^2+c^2-2bc cos(A)\longrightarrow

    18.49=27.04+56.25-78 cos(A) [/tex]

    I tried bringing everything to the left side with the exception of cos(A) and then doing [tex] cos^-1 [/tex] but it just wont work in my calculator I keep getting a math error. :cry:
     
    Last edited: Dec 28, 2004
  2. jcsd
  3. Dec 28, 2004 #2
    Are you sure you're on degree mode? Trig is usually only used with radians, but this one should involve degrees.

    Isolate the cosA

    -64.8 = -78(cosA)
    Divide by -78.

    cos^-1(.83) should yield the answer, but if you're on radians, it probably won't work.

    I got 33.82 as an answer.
     
    Last edited: Dec 28, 2004
  4. Dec 28, 2004 #3
    Must just be an arithmetic error. I don't think it is a radian/degree issue.
     
  5. Dec 28, 2004 #4

    VietDao29

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    Homework Helper

    Hi,
    Dextercioby, haven't you realised that your post is completely wrong?
    "Has it ever occured to u that one side (viz."c") is exactly the sum of the other two...???I guess not,else u have realized that your triangle is not a regular one.It's a degenerate triangle.It has one angle of 180° and the other of 0°.The three summits are on the same line."
    With a=4.3 b=5.2 c=7.5
    Are you trying to say that c = a + b? And therefore 7.5 = 4.3 + 5.2 ?!
    So... A = 33.82 is the answer.
    Bye bye,
    Viet Dao,
     
  6. Dec 28, 2004 #5
    Try rearranging the cosine equation:

    [tex]a^2=b^2+c^2-2bccosA[/tex]

    => [tex]a^2+2bccosA=b^2+c^2[/tex]

    => [tex]2bccosA=b^2+c^2-a^2[/tex]

    => [tex]cosA=\frac{b^2+c^2-a^2}{2bc}[/tex]

    Now try adding the numbers to this:

    [tex]cosA = \frac{(5.2^2)+(7.5^2)-(4.3^2)}{(2 \times 5.2 \times 7.5)}[/tex]

    [tex]cosA = \frac{27.04+56.25-18.49}{78}[/tex]

    [tex]cosA = \frac{64.8}{78}[/tex]

    [tex]cosA = 0.8307[/tex]

    [tex]A = cos^-^10.8307[/tex]

    [tex]A = 33.82[/tex]

    Just do that for the rest (but with a = 5.2, b = 7.5 and c = 4.3 etc...) and you will have three angles for the triangle.

    Hope that helps. :smile:

    The Bob (2004 ©)
     
  7. Dec 28, 2004 #6

    BobG

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    Science Advisor
    Homework Helper

    Just out of curiosity, for [tex]cos^{-1}[/tex], are you hitting:

    2ND key, COS key

    or are you hitting:

    COS key, 2ND key, [tex]x^{-1}[/tex]

    [tex]cos^{-1}[/tex] is an abbreviated term for ARCCOS. Above your COS key, you should either have ARCCOS or [tex]cos^{-1}[/tex].

    The only other possible problem is if you entered the equation into your calculator wrong. You can't take the arcosine of a number larger than 1 and that will also give you an error.
     
    Last edited: Dec 28, 2004
  8. Dec 28, 2004 #7
    I got it I was subtracting [tex] 78 [/tex] instead of dividing both sides and isolating [tex] cosA [/tex]

    My answers for the angles are <A=34 <B=42 and <C=104
     
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