<div>
change the degrees in 1 to radians
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<div>
I'm looking for an algebraic solution. You can work it in either degrees or radians that shoudl not make any difference.
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<div>
A first step could be use (1) to eliminate beta from (2)<br />
<br />
q * cos (85 - alpha) = 6<br />
<br />
...
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<div>
85 degrees = 1.48 radians<br />
<br />
solve 4 for q<br />
solve 1 for alpha<br />
<br />
put results in 2 and 3<br />
you now have two equations in two unknowns ( still messy but at least some progress)
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<div>
First 3 equations have only 3 unknowns, so don't consider equation 4 at first.<br />
I would start by putting (3) in the same form as (2); i.e., multiply both sides by q.<br />
You can then divide (2) by (3) to eliminate the q --&gt; equation (5).<br />
<br />
(1) and (5) are 2 equations in 2 unknowns (alpha, beta).<br />
beta = 85 degrees - alpha<br />
Plug beta into (5) and you have one equation in one unknown, i.e., alpha.
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<div>
&gt;&gt; You can work it in either degrees or radians that shoudl not make any difference. <br />
You are right. It doesn't make a difference:<br />
<a href="http://www.wolframalpha.com/input/?i=cos+(+85+degrees+-+alpha)%2F+cos+(+alpha+)+%3D+2" rel="ugc">http://www.wolframalpha.com/input/?i=cos+(+85+degrees+-+alpha)%2F+cos+(+alpha+)+%3D+2</a><br />
<a href="http://www.wolframalpha.com/input/?i=cos+(+85+*+pi%2F180+-+alpha)%2F+cos+(+alpha+)+%3D+2" rel="ugc">http://www.wolframalpha.com/input/?i=cos+(+85+*+pi%2F180+-+alpha)%2F+cos+(+alpha+)+%3D+2</a>
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<div>
I can get it down to one equation in one unknown without much trouble, but then I'm stuck on solving it because of the trigonometry...
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<div>
Once you solve for alpha, then (1) gives you beta easily.<br />
Either (2) or (3) gives you q.<br />
Once you solve for q, then (4) gives you b easily.
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<div>
...looks like wolfram may help...
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<div>
&gt;&gt; I can get it down to one equation in one unknown without much trouble, but then I'm stuck on solving it because of the trigonometry... <br />
<br />
Then why didn't you just ask that question, &quot;one equation in one unknown&quot; instead of giving us 4 equations?
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<div>
It doesn't matter what i start solving for.. i get stuck
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<div>
Don't you just need an answer? Why do you need a step-by-step solution (which wolfram says they can provide by going &quot;Pro&quot;)?<br />
<br />
Once getting one equation in one unknown, you can use plot utilities (or even by hand depending upon the number of precision required) to determine intersections of graphs.
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<div>
well the one equation looks very ugly. but some maybe less ugly than others depending how you proceed. 4 equations looks neat at least.
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<div>
I'm assuming the equations are solvable algebraically.. but i could be wrong. Some info here would be useful, my maths is rusty..
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<div>
example one equation, find b:<br />
<br />
sqrt ( b^2 + 9 ) * cos(85 - acos(3 / sqrt ( b^2 + 9 ) )) = 6
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<div>
Does anyone have wolfram pro to check this? I'm getting &quot;Standard computation time exceeded&quot;
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<div>
It cannot be done. You are one equation short.<br />
<br />
divide 2 by 3. That eliminates q.<br />
b and q now appear only in 4. &nbsp;(1 , 2, and 3 will allow you to find alpha and beta,leaving only 4 for both b and q.)
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<div>
aburr, if it can't be done then my one equation in one would be wrong.. but apart from being messy it seemed like a straightforward sequence of substitution...<br />
<br />
...unless solving my one equation just gives b=b &nbsp;<br />
<br />
hmmm
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<div>
1, 2, and 3 are completely independent of b and q. As far as they are concerned, b and q can be anything. They tell you nothing about b and q, so you have one equation in two unknowns.
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<div>
2) q * cos ( beta &nbsp;) = 6<br />
&nbsp;3) q * cos ( alpha ) = 3<br />
<br />
if you solved for alpha, then from (3) you know q.<br />
But, if you solved for alpha, then you now know beta, and then you can solve for q using (2):<br />
<br />
&lt;&lt;EDIT&gt;&gt;<br />
We solved for this equation using Wolfram:<br />
<b>cos ( beta &nbsp;) / cos ( alpha ) = 2</b><br />
alpha1 = 1.09065<br />
alpha2 = -2.05094<br />
which gives us beta:<br />
beta1 = &nbsp;0.3929<br />
beta2 = &nbsp;3.5345
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<div>
Thanks for your inputs. I think I need to sleep on this one...
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<div>
(Note: I edited my previous post since I had to also sleep on it in a power nap.)<br />
We have:<br />
alpha1 = 1.09065<br />
&nbsp;alpha2 = -2.05094<br />
which gives us beta:<br />
&nbsp;beta1 = &nbsp;0.3929<br />
&nbsp;beta2 = &nbsp;3.5345 <br />
<br />
We can now compute q for these four values of alpha and beta:<br />
Equation (2):<br />
&gt;&gt;&nbsp;q11 = 6/cos ( beta1 )<br />
q11 =<br />
&nbsp; &nbsp; 6.4948<br />
&gt;&gt;&nbsp;q12 = 6/cos ( beta2 )<br />
q12 =<br />
&nbsp; &nbsp;-6.4948<br />
<br />
Equation (3):<br />
&gt;&gt;&nbsp;q21 = 3/cos(alpha1)<br />
q21 =<br />
&nbsp; &nbsp; 6.4948<br />
&gt;&gt;&nbsp;q22 = 3/cos(alpha2)<br />
q22 =<br />
&nbsp; &nbsp;-6.4948
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<div>
equation (4)<br />
&gt;&gt;&nbsp;b = sqrt(q11^2 - 9)<br />
b =<br />
&nbsp; &nbsp; 5.7605<br />
<br />
(Another solution is: b = -5.7605)
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<div>
As for solving this equation for alpha:<br />

<pre><code class="code-1 nolinks" id="code-20-41483979-1"><span>cos ( 85 * pi/180 - alpha)/ cos ( alpha ) = 2</span>
</code></pre>
I am tempted to avoid initially trigonometry manipulations, and instead would start with Euler's formula and see where that leads.<br />
<a href="https://en.wikipedia.org/wiki/Euler%27s_formula#Relationship_to_trigonometry" rel="ugc">https://en.wikipedia.org/wiki/Euler%27s_formula#Relationship_to_trigonometry</a><br />
<br />
BTW - as you noticed, the Wolfram solution indicates that there are an infinite number of solutions when solving for alpha. Depending upon what the four equations are modeling, you have to decide which solution is pertinent for your needs.
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Problem set up to use with Solver add-in for Excel

4EquationsQ28929349.xlsx

YvsDegZoomIn.JPG

cos(85*pi/180 - alpha)/ cos ( alpha ) from -80 to +80 degrees

YvsDegFullScale.JPG

<div>
That's excellent thank you both for your help.
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<div>
I'd give 5000 points but there's a limit. Thank you all for shedding light on this tricky problem.<br />
<br />
byundt, you make the subscription fee worth it on your own.<br />
<br />
&nbsp;I may post a follow-up as I will be looking for a generic solution on google sheets.<br />
<br />
Thanks again
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<div>
phoffric,<br />
Excel uses 64-bit floating point numbers to achieve 1 part in 10^15 precision. That's the same degree of precision and accuracy as you were using.<br />
<br />
Solver does quit when the solution converges to a certain degree. And that convergence would be measured against the sum of the squares of the residuals for the four equations. The default precision constraint is 1 part in a million, but you can set higher precision in the Solver...Options dialog. I assume that this is why there is a small difference between your graphical solution and the Solver results. <br />
<br />
I didn't think xenium needed more than the default Solver accuracy, because this problem was &quot;for some cardboard box design.&quot;<br />
<br />
Brad
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<div>
Follow-up <a href="https://www.experts-exchange.com/questions/28930098/Simultaneous-equations-part-II.html" rel="ugc">https://www.experts-exchange.com/questions/28930098/Simultaneous-equations-part-II.html</a><br />
Sorry i don't have Excel handy!!
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<div>
Finally, an analytic solution..<br />
Recall the equation to solve for alpha<br />
Let C = 85 degrees<br />
cos ( C- a ) / cos ( a ) = 2, where a ~ alpha<br />
{ cos(C)*cos(a) + sin(C)*sin(a) } / cos(a) = 2<br />
cos(C) + sin(C)tan(a) = 2<br />
<br />
Now it is easy to isolate tan(a).<br />
sin(C)tan(a) = 2 - cos(C) <br />
tan(a) = [ { 2 - cos(C) } / sin(C) ]<br />
<br />
Here is value of alpha:<br />
alpha = arctan[ { 2 - cos(85) } / sin(85) ] = arctan[ { 2 - 0.087155743 } / 0.996194698 ]<br />
alpha = arctan( 1.920151011 ) = 62.489843518 degrees<br />
<br />
&gt;&gt;&nbsp;breaking my head, i think the following is possible<br />
Your intuition was right; it is possible.<br />
More importantly, no more need to break your head. :)
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<div class="content wysiwyg-content">
hi, I'm breaking my head, i think the following is possible, 4 unknowns, 4 equations, can anyone help?<br />
<br />
1) alpha + beta = 85 degrees<br />
2) q * cos ( beta) = 6<br />
3) cos(alpha) =3/q<br />
4) b^2 = q^2 -9<br />
<br />
Or perhaps one of the equations is derived from the others...<br />
<br />
This is for some cardboard box design.<br />
<br />
Thanks!
</div>
</div>


hi, I'm breaking my head, i think the following is possible, 4 unknowns, 4 equations, can anyone help?

1) alpha + beta = 85 degrees
2) q * cos ( beta) = 6
3) cos(alpha) =3/q
4) b^2 = q^2 -9

Or perhaps one of the equations is derived from the others...

This is for some cardboard box design.

Thanks!

Solve the following simultaneous equations

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