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Math Help - Solve for alpha and beta

  1. #1
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    Solve for alpha and beta

    How do you solve alpha and beta for these two equations?
    (0.5)(4.00) = (0.500)(2.00)(\cos{\alpha}) + 3(4.47)(\cos(\beta))

    0 = 0.500(2.00)(\sin{\alpha}) - 0.300(4.47)(\sin{\beta})

    i tried but it gives me a math error
    the answers are \alpha = 36.9^{\circ},\beta = 26.6^{\circ}

    The hint is:
    solve the 1st equation for cos(beta) first
    then 2nd equation for sin(beta)
    then square each equation (cos(beta))^2 + (sin(beta))^2 = 1

    thats what i did but my answer at last part gives me
    1.798281 = 4 - 2\cos(\alpha)
    alpha gives a math error for this equation

    thank you very much
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    How do you solve alpha and beta for these two equations?
    (0.5)(4.00) = (0.500)(2.00)(\cos{\alpha}) + 3(4.47)(\cos(\beta))

    0 = 0.500(2.00)(\sin{\alpha}) - 0.300(4.47)(\sin{\beta})

    i tried but it gives me a math error
    the answers are \alpha = 36.9^{\circ},\beta = 26.6^{\circ}

    The hint is:
    solve the 1st equation for cos(beta) first
    then 2nd equation for sin(beta)
    then square each equation (cos(beta))^2 + (sin(beta))^2 = 1

    thats what i did but my answer at last part gives me
    1.798281 = 4 - 2\cos(\alpha)
    alpha gives a math error for this equation

    thank you very much
    (0.5)(4.00) = (0.500)(2.00)(\cos{\alpha}) + 3(4.47)(\cos(\beta))

    0 = 0.500(2.00)(\sin{\alpha}) - 0.300(4.47)(\sin{\beta})

    So you have the system:
    2 = cos (\alpha ) + 13.41 cos( \beta)

    0 = sin( \alpha ) - 1.341 sin( \beta )

    So
    cos(\beta) = \frac{2 - cos(\alpha)}{13.41}

    and
    sin(\beta) = \frac{sin(\alpha)}{1.341}

    Thus
    sin^2(\beta) + cos^2(\beta) = 1 \implies \left ( \frac{sin(\alpha)}{1.341} \right )^2 + \left ( \frac{2 - cos(\alpha)}{13.41} \right ) ^2 = 1

    \frac{sin^2(\alpha)}{1.341^2} + \left ( \frac{4 - 4cos(\alpha) + cos^2(\alpha)}{13.41^2} \right ) = 1 <-- Multiply both sides by 13.41^2

    100sin^2(\alpha) + 4 - 4cos(\alpha) + cos^2(\alpha) = 13.41^2<-- Add 99cos^2(\alpha) to both sides

    100sin^2(\alpha) + 4 - 4cos(\alpha) + 100cos^2(\alpha) = 13.41^2 + 99 cos^2(\alpha)

    100sin^2(\alpha) + 100cos^2(\alpha) + 4 - 4cos(\alpha) = 13.41^2 + 99 cos^2(\alpha)

    100 + 4 - 4cos(\alpha) = 13.41^2 + 99 cos^2(\alpha)

    99cos^2(\alpha) + 4 cos(\alpha) + 13.41^2 - 104 = 0

    99cos^2(\alpha) + 4 cos(\alpha) + 75.8281 = 0

    You would typically solve this using the quadratic formula, but note that the discriminant is negative. So there are no real solutions.

    Note that with your given solution the first equation is
    2 = 12.7903
    which is obviously untrue. (Though the given solution does satisfy the second equation.)

    -Dan
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  3. #3
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    Hello, ^_^Engineer_Adam^_^!

    Why weren't the equations simplified first?


    How do you solve \alpha and \beta for these two equations?

    (0.5)(4.00) = (0.500)(2.00)(\cos{\alpha}) + 3(4.47)(\cos(\beta))

    0 = 0.500(2.00)(\sin{\alpha}) - 0.300(4.47)(\sin{\beta})

    The answers are \alpha = 36.9^{\circ},\beta = 26.6^{\circ}
    There must be a typo . . . the answers don't check out.

    I think the equations look like this:

    . . \begin{array}{cccc}\cos\alpha + {\color{red}1.341}\cos\beta & = & 2 & {\color{blue}[1]} \\<br />
\sin\alpha - 1.341\sin\beta & = & 0 & {\color{blue}[2]}\end{array}


    The hint is:
    solve the 1st equation for \cos\beta first, .
    Really?
    . . then 2nd equation for \sin\beta. .
    Why introduce fractions?
    Then square each equation and add: . \cos^2\!\beta + \sin^2\!\beta\:= \:1
    This is the way I would do it . . .


    \begin{array}{ccccc}\text{From {\color{blue}[1]}:} & \cos\alpha & = & 2-1.341\cos\beta & {\color{blue}[3]}\\<br />
\text{From {\color{blue}[2]}:} & \sin\alpha & = & 1.341\sin\beta & {\color{blue}[4]}\end{array}

    \begin{array}{cccc}\text{Square {\color{blue}[3]}:} & \cos^2\!\alpha & = & 4 - 5.364\cos\beta + 1.798281\cos^2\!\beta \\<br />
\text{Square {\color{blue}[4]}:} & \sin^2\!\alpha & = & 1.798281\sin^2\!\beta \end{array}

    Add: . \cos^2\!\alpha + \sin^2\!\alpha \;=\;4 - 5.364\cos\beta + 1.798281\left(\cos^2\!\beta + \sin^2\!\beta\right)

    and we have: . 1 \;=\;4 - 5.364\cos\beta + 1.798281

    Hence: . 5.364\cos\beta \:=\:4.798281\quad\Rightarrow\quad \cos\beta \:=\:0.894534116\quad\Rightarrow\quad\boxed{\beta \:\approx\:26.6^o}

    Substitute into [4]: . \sin\alpha \;=\;1.341\sin26.6^o \:=\:0.600444937\quad\Rightarrow\quad\boxed{\alpha \:\approx\:36.9^o}

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