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Thread: 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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    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.
    $\displaystyle A = \frac{1}{2}acsin(B)$
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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:
    $\displaystyle a^2=b^2+c^2-2bc cosA $

    transpose it since we want to find angles:

    $\displaystyle a^2 + 2bcCosA=b^2+c^2 $

    $\displaystyle =2bcCosA=b^2+C^2-a^2 $

    $\displaystyle =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.

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

    $\displaystyle CosA= \frac {206}{270} $

    $\displaystyle CosA = 0.7629 $

    $\displaystyle A= Cos^-1 (0.7629) $

    $\displaystyle 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 $\displaystyle \mathrm{e}^{\mathrm{i}\pi}$, 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 $\displaystyle \alpha$.
    Now the law of cosine says that
    $\displaystyle a^2=b^2+c^2-2\cdot b\cdot c\cdot\cos(\alpha)$
    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 $\displaystyle \alpha$, are given, you can simply solve for $\displaystyle \alpha$, like this:
    $\displaystyle \alpha = \cos^{-1}\frac{b^2+c^2-a^2}{2\cdot b\cdot c}=\cos^{-1}\frac{10^2+9^2-15^2}{2\cdot 10\cdot 9}=\cos^{-1}\left(-\tfrac{11}{45}\right)\approx 104.15^\circ$
    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.

    $\displaystyle A = \frac{1}{2}acsin(B)
    $

    transpose it to make SinB by itself:

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

    ok all done.

    $\displaystyle \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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