# Thread: Solve for alpha and beta

1. ## 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

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

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:

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

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 . . .

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

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

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