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Math Help - I'm stuck with some easy trigonometry

  1. #1
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    I'm stuck with some easy trigonometry

    I cant figure out the answers to these questions:

    If this triangle has an area of 24cm^2, what is the value of angle D? http://i431.photobucket.com/albums/q...signmentq8.jpg

    ---

    Find the largest angle of this triangle:
    http://i431.photobucket.com/albums/q...ignmentq10.jpg

    ---

    Please show working in a simple way, I'm not too good at this and I really want to understand trigonometry.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by ibenot1337 View Post
    I cant figure out the answers to these questions:

    If this triangle has an area of 24cm^2, what is the value of angle D? http://i431.photobucket.com/albums/q...signmentq8.jpg

    ---

    Find the largest angle of this triangle:
    http://i431.photobucket.com/albums/q...ignmentq10.jpg

    ---

    Please show working in a simple way, I'm not too good at this and I really want to understand trigonometry.
    A = \frac{1}{2}acsin(B)
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  3. #3
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    Let's change this up... A is B, B is C and C is A
    Get the formula:
    a^2=b^2+c^2-2bc cosA

    transpose it since we want to find angles:

    a^2 + 2bcCosA=b^2+c^2

    =2bcCosA=b^2+C^2-a^2

    =CosA = \frac {b^2+c^2-a^2}{2bc}

    you with me?

    alright time to substitute. You should know that the side opposite the angle will be the "corresponding" letter.

     CosA = \frac {9^2 + 15^2 - 10^2}{2*9*15}

     CosA= \frac {206}{270}

     CosA = 0.7629

     A= Cos^-1 (0.7629)

     A= 40'16'

    Now just do that for B one. Add A and B then subtract from 180. You have all your angles.
    If anything is wrong here please correct me as I am in the stages of learning this too...
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by ibenot1337 View Post
    I cant figure out the answers to these questions:

    If this triangle has an area of 24cm^2, what is the value of angle D? http://i431.photobucket.com/albums/q...signmentq8.jpg
    You have already got a hint for this first problem from \mathrm{e}^{\mathrm{i}\pi}, so I won't elaborate on this one any further.

    As to your second problem:
    Find the largest angle of this triangle:
    http://i431.photobucket.com/albums/q...ignmentq10.jpg

    Please show working in a simple way, I'm not too good at this and I really want to understand trigonometry.
    Because the largest angle of a triangle is always opposite to the longest side, you are, in effect, asked to determine the angle at vertex A. Let me call this angle \alpha.
    Now the law of cosine says that
    a^2=b^2+c^2-2\cdot b\cdot c\cdot\cos(\alpha)
    where a, b, and c is the length of the side opposite to vertex A, B, and C, respectively.
    Since all values in this equation, with the sole exception of \alpha, are given, you can simply solve for \alpha, like this:
    \alpha = \cos^{-1}\frac{b^2+c^2-a^2}{2\cdot b\cdot c}=\cos^{-1}\frac{10^2+9^2-15^2}{2\cdot 10\cdot 9}=\cos^{-1}\left(-\tfrac{11}{45}\right)\approx 104.15^\circ
    and are done!
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  5. #5
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    For the first one use the formula above. I'll assume that B is D.

    A = \frac{1}{2}acsin(B)<br />

    transpose it to make SinB by itself:

    24=\frac {1}{2}acsin(B)
    \frac {48}{ac} = SinB

    ok all done.

     \frac {48}{56} = SinB

    You can do the rest..
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  6. #6
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    Cool Thanks

    Thank you for all your help.

    I now understand the rules I need to solve those sorts of problems.
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