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Math Help - equatios system

  1. #1
    Super Member dhiab's Avatar
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    equatios system

    Solve the equatios system:

    \left\{ \begin{array}{l}<br />
7\cos \alpha + 9\cos \beta = 10 \\ <br />
7\sin \alpha - 9\sin \beta = 0 \\ <br />
\end{array} \right.<br />
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dhiab View Post
    Solve the equatios system:

    \left\{ \begin{array}{l}<br />
7\cos \alpha + 9\cos \beta = 10 \\ <br />
7\sin \alpha - 9\sin \beta = 0 \\ <br />
\end{array} \right.<br />
    Note that you have

    \left\{ \begin{array}{l}<br />
7 \sin \alpha \cos \alpha + 9 \sin \alpha \cos \beta = 10 \sin \alpha \\ <br />
7\sin \alpha \cos \alpha - 9\sin \beta \cos \alpha = 0 \\ <br />
\end{array} \right.<br />

    (Do you see what I did?) Now subtract the second equation from the first. can you finish from there?
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  3. #3
    MHF Contributor red_dog's Avatar
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    Add two equations to the system:

    \left\{\begin{array}{l}7\cos\alpha+9\cos\beta=10\\  7\sin\alpha-9\sin\beta=0\\\sin^2\alpha+\cos^2\alpha=1\\\sin^2\  beta+\cos^2\beta=1\end{array}\right.

    \sin\beta=\frac{7\sin\alpha}{9}, \ \cos\beta=\frac{10-7\cos\alpha}{9}

    Replace them in the last equation.

    \frac{49\sin^2\alpha}{81}+\frac{100-140\cos\alpha+49\cos^2\alpha}{81}=1\Rightarrow \cos\alpha=\frac{17}{35}\Rightarrow\cos\beta=\frac  {77}{105}

    Now find \sin\alpha, \ \sin\beta from the last two equations and choose the sign such as the second equation is satisfied. Then find \alpha, \ \beta.
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  4. #4
    Senior Member pacman's Avatar
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    Solution by Mathematica





    .

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