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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!

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!

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solve 4 for q

solve 1 for alpha

put results in 2 and 3

you now have two equations in two unknowns ( still messy but at least some progress)

I would start by putting (3) in the same form as (2); i.e., multiply both sides by q.

You can then divide (2) by (3) to eliminate the q --> equation (5).

(1) and (5) are 2 equations in 2 unknowns (alpha, beta).

beta = 85 degrees - alpha

Plug beta into (5) and you have one equation in one unknown, i.e., alpha.

You are right. It doesn't make a difference:

http://www.wolframalpha.com/input/?i=cos+(+85+degrees+-+alpha)%2F+cos+(+alpha+)+%3D+2

http://www.wolframalpha.com/input/?i=cos+(+85+*+pi%2F180+-+alpha)%2F+cos+(+alpha+)+%3D+2

Either (2) or (3) gives you q.

Once you solve for q, then (4) gives you b easily.

Then why didn't you just ask that question, "one equation in one unknown" instead of giving us 4 equations?

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.

divide 2 by 3. That eliminates q.

b and q now appear only in 4. (1 , 2, and 3 will allow you to find alpha and beta,leaving only 4 for both b and q.)

3) q * cos ( alpha ) = 3

if you solved for alpha, then from (3) you know q.

But, if you solved for alpha, then you now know beta, and then you can solve for q using (2):

<<EDIT>>

We solved for this equation using Wolfram:

alpha1 = 1.09065

alpha2 = -2.05094

which gives us beta:

beta1 = 0.3929

beta2 = 3.5345

We have:

alpha1 = 1.09065

alpha2 = -2.05094

which gives us beta:

beta1 = 0.3929

beta2 = 3.5345

We can now compute q for these four values of alpha and beta:

Equation (2):

>> q11 = 6/cos ( beta1 )

q11 =

6.4948

>> q12 = 6/cos ( beta2 )

q12 =

-6.4948

Equation (3):

>> q21 = 3/cos(alpha1)

q21 =

6.4948

>> q22 = 3/cos(alpha2)

q22 =

-6.4948

```
cos ( 85 * pi/180 - alpha)/ cos ( alpha ) = 2
```

I am tempted to avoid initially trigonometry manipulations, and instead would start with Euler's formula and see where that leads.https://en.wikipedia.org/wiki/Euler%27s_formula#Relationship_to_trigonometry

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.

Constraints:

alpha between 0 and 85 degrees

beta between 0 and 85 degrees

Results:

alpha 62.5167201

beta 22.48327471

b 5.757738712

q 6.492426432

4EquationsQ28929349.xlsx

```
cos ( 85 * pi/180 - alpha)/ cos ( alpha ) = 2
```

Here is the question:http://www.experts-exchange.com/questions/28929373/How-to-solve-a-trigonometry-equation.html

In summary, here are the numerical results of your four equations:

```
alpha = +1.09065 or -2.05094
beta = +0.3929 or +3.5345
q = +6.4948 or -6.4948
b = +5.7605 or -5.7605
```

Here is what I had in mind (which is consistent with the Wolfram solution) showing the expression from -80 to +80 degrees.

Zooming in to where y = 2, we see that alpha = 62.4898 degrees * pi/180 = 1.09065 radians which is what was shown above.

Not sure why byundt's solution differs a little. My utilities were using 64-bit precision. Maybe byundt's solution used 32-bit, or possibly stopped short when a reasonable threshold was reached.

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.

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.

I didn't think xenium needed more than the default Solver accuracy, because this problem was "for some cardboard box design."

Brad

Recall the equation to solve for alpha

Let C = 85 degrees

cos ( C- a ) / cos ( a ) = 2, where a ~ alpha

{ cos(C)*cos(a) + sin(C)*sin(a) } / cos(a) = 2

cos(C) + sin(C)tan(a) = 2

Now it is easy to isolate tan(a).

sin(C)tan(a) = 2 - cos(C)

tan(a) = [ { 2 - cos(C) } / sin(C) ]

Here is value of alpha:

alpha = arctan[ { 2 - cos(85) } / sin(85) ] = arctan[ { 2 - 0.087155743 } / 0.996194698 ]

alpha = arctan( 1.920151011 ) = 62.489843518 degrees

>> breaking my head, i think the following is possible

Your intuition was right; it is possible.

More importantly, no more need to break your head. :)

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