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I have the following three equations: \begin{cases} v_{1f}\cos(37^\circ)+v_{2f}\cos(\theta) & = 3.5 \times 10^5 \\ v_{1f}\sin(37^\circ)-v_{2f}\sin(\theta) & = 0\\ v_{1f}^2+v_{2f}^2 & =(3.5 \times 10^5)^2 \end{cases} And I want to solve for $v_{1f}$, $v_{2f}$, and $\theta$. This is a system of three equations but it doesn't seem solvable and I've tried everything I know to solve it.

For example, nothing can cancel with each other like you would in an easy system, and I've tried using the 3rd equation to solve for $v_{1f}$ or $v_{2f}$ but it still does not come out correctly. I do know the answers, just not how to get them. Here they are: \begin{cases} v_{1f}=2.8 \times 10^5 \\ v_{2f}=2.11 \times 10^3 \\ \theta=53^\circ \\ \end{cases} Am I missing some information needed for solving this?

I really appreciate any help with this question. Sorry that I could not show more of my work but I'm stuck and showed what I know so far. Thank you.

Also, if someone sees that it isn't solvable that would help as well.

EDIT: I messed up typing the 1st equation, fixed now.

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    $\begingroup$ Does $37$ mean 37 degrees? $\endgroup$ – saulspatz Mar 11 '18 at 16:57
  • $\begingroup$ @saulspatz Yes, sorry. $\endgroup$ – JustHeavy Mar 11 '18 at 16:59
  • $\begingroup$ One solution is easy to see: $v_{1f}=0$, $v_{2f}=3.5\cdot10^5$, and $\theta=0$ (which makes $\cos\theta=1$ and $\sin\theta=0$). $\endgroup$ – Barry Cipra Mar 11 '18 at 17:06
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a) the geometric solution

Let's change the symbols so as to keep notation cleaner, and rewrite your input as $$ \bbox[lightyellow] { \left\{ \matrix{ a\cos \alpha + b\cos \theta = A \hfill \cr a\sin \alpha = b\sin \theta \hfill \cr a^{\,2} + b^{\,2} = A^{\,2} \hfill \cr} \right. } \tag{1}$$

This corresponds to the right triangle shown in this sketch

3_inc_trig_1

and the solution is quite simple noting that $\theta = 90^\circ -\alpha$, and applying the sine law.
Note that a (limit) solution is also given by $$ \theta=0, \; a=0, \; b=A$$

b) the algebraic solution

The last equation in (1) suggests that we may introduce an additional angle $\phi$ and write $$ \left\{ \matrix{ a = A\cos \varphi \hfill \cr b = A\sin \varphi \hfill \cr A\cos \varphi \cos \alpha + A\sin \varphi \cos \theta = A \hfill \cr A\cos \varphi \sin \alpha - A\sin \varphi \sin \theta = 0 \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{ a = A\cos \varphi \hfill \cr b = A\sin \varphi \hfill \cr \cos \varphi \cos \alpha + \sin \varphi \cos \theta = 1 \hfill \cr \cos \varphi \sin \alpha - \sin \varphi \sin \theta = 0 \hfill \cr} \right. $$

The last two equations written in matrix terms become $$ \left( {\matrix{ {\cos \alpha } & {\cos \theta } \cr {\sin \alpha } & { - \sin \theta } \cr } } \right)\left( {\matrix{ {\cos \varphi } \cr {\sin \varphi } \cr } } \right) = \left( {\matrix{ 1 \cr 0 \cr } } \right)\quad \Rightarrow \quad {\bf M}_{\,\theta } \,{\bf v}_{\,\varphi } = {\bf u} $$ where the matrix determinant is $$ \left| {{\bf M}_{\,\theta } } \right| = - \sin \left( {\alpha + \theta } \right) $$ so that for $\alpha +\theta \ne n \pi$ the matrix is invertible, giving $$ {\bf M}_{\,\theta } ^{\, - \,{\bf 1}} = {1 \over {\sin \left( {\alpha + \theta } \right)}}\left( {\matrix{ {\sin \theta } & {\cos \theta } \cr {\sin \alpha } & { - \cos \alpha } \cr } } \right) $$ and we get that $$ {\bf v}_{\,\varphi } = {\bf M}_{\,\theta } ^{\, - \,{\bf 1}} {\bf u}\quad \Rightarrow \quad \left( {\matrix{ {\cos \varphi } \cr {\sin \varphi } \cr } } \right) = {1 \over {\sin \left( {\alpha + \theta } \right)}}\left( {\matrix{ {\sin \theta } \cr {\sin \alpha } \cr } } \right) $$

However we shall ad the condition that the modulus of ${\bf v}_{\,\varphi }$ be unitary, i.e. $$ 1 = \left| {{\bf v}_{\,\varphi } } \right|^{\,2} = \overline {\bf u} \,\overline {{\bf M}_{\,\theta } } ^{\, - \,{\bf 1}} {\bf M}_{\,\theta } ^{\, - \,{\bf 1}} {\bf u}\quad \Rightarrow \quad \sin ^{\,2} \theta + \sin ^{\,2} \alpha = \sin ^{\,2} \left( {\alpha + \theta } \right) $$ which develops to provide $$ \eqalign{ & \sin ^{\,2} \theta + \sin ^{\,2} \alpha = \sin ^{\,2} \alpha \cos ^{\,2} \theta + \cos ^{\,2} \alpha \sin ^{\,2} \theta + 2\sin \alpha \cos \alpha \cos \theta \sin \theta \cr & \sin ^{\,2} \alpha \sin ^{\,2} \theta + \sin ^{\,2} \alpha \sin ^{\,2} \theta = 2\sin \alpha \cos \alpha \cos \theta \sin \theta \cr & \sin \alpha \sin ^{\,2} \theta = \cos \alpha \cos \theta \sin \theta \cr & \left( {\sin \alpha \sin \theta - \cos \alpha \cos \theta } \right)\sin \theta = 0 \cr & \sin \theta = 0\; \vee \;\cos \left( {\alpha + \theta } \right) = 0 \cr} $$

Finally, resuming the various steps we conclude that $$ \bbox[lightyellow] { \left\{ \matrix{ \theta \in \left\{ {\pi /2 + k\pi - \alpha ,\;k\pi } \right\} \hfill \cr \varphi = \arcsin \left( {\sin \alpha /\sin \left( {\alpha + \theta } \right)} \right) \hfill \cr a = A\cos \varphi \hfill \cr b = A\sin \varphi \hfill \cr} \right. } \tag{2}$$

We can see that the algebraic method add some solutions wrt the geometric method , which actually reduce to those implying negative values for $b$
e.g. $\theta=\pi, \; a=0, \; b=-A$

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    $\begingroup$ Very curious how you made that sketch? $\endgroup$ – Alex Reinking Mar 12 '18 at 6:15
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    $\begingroup$ Very nice! But what happened to the other solution, with $\theta=0$? $\endgroup$ – Barry Cipra Mar 12 '18 at 10:17
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    $\begingroup$ @AlexReinking: if by "how" you mean which tool, that's Geogebra. If you are asking "how" I arrived to that, well, I started as everybody algebraically but suspecting from the beginning that there was a simple vectorial/geometrical interpretation, till I got and write the 2nd equation as $a \sin\alpha=b \sin \theta$ which was the key. $\endgroup$ – G Cab Mar 12 '18 at 10:53
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    $\begingroup$ @BarryCipra: you are right, in the geometric approach the solution $\theta =0 \; \to \; a=0, b=A$ is a limit one and does not come immediately to mind. So I added the algebraic solution which more naturally suggests also the solutions with negative values for $b$ $\endgroup$ – G Cab Mar 12 '18 at 11:34
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    $\begingroup$ @AlexReinking: Geogebra is a very efficient tool for a certain class of geometric drawing (and computation). One of the tools allows you to insert text, Latex formulas, pictures. $\endgroup$ – G Cab Mar 12 '18 at 13:36
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For notational convenience, let's let $r=3.5\times10^5$, $x=v_{1f}$, and $y=v_{2f}$. The three equations are

$$x\cos(37)+y\cos\theta=r$$ $$x\sin(37)-y\sin\theta=0$$ and $$x^2+y^2=r^2$$

From the first two we have

$$y^2=y^2(\cos^2\theta+\sin^2\theta)=(r-x\cos(37))^2+x^2\sin^2(37)=r^2-2rx\cos(37)+x^2$$

from which the third equation tells us

$$r^2=x^2+y^2=2x^2-2rx\cos(37)+r^2=r^2+2x(x-r\cos(37))$$

which implies

$$x(x-r\cos(37))=0$$

so either $x=0$ or $x=r\cos(37)\approx2.795\times10^5$. If $x=0$, we must have $y=r$ and $\theta=0$. If $x=r\cos(37)$, we must have $y=r\sin(37)\approx2.106\times10^5$ (in order to satisfy $x^2+y^2=r^2$) and thus

$$x\sin(37)-y\sin\theta=0\implies r\cos(37)\sin(37)=r\sin(37)\sin\theta\implies\sin\theta=\cos(37)$$

From the general trig identity $\sin\theta=\cos(90-\theta)$, we get $\theta=90-37=53$.

In summary, there are two solutions:

$$(v_{if},v_{2f},\theta)=(0,3.5\times10^5,0)$$ and $$(v_{1f},v_{2f},\theta)\approx(2.795\times10^5,2.106\times10^5,53)$$

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  • $\begingroup$ Isn't $2.106\times{10^5}$ supposed to be $2.106\times{10^3}$? Thanks for your help. $\endgroup$ – JustHeavy Mar 12 '18 at 0:38
  • $\begingroup$ @JustHeavy, no, I'm pretty sure the exponent is $5$, not $3$. Why do you think it's $3$? (You can check which one is correct by plugging the values into the third equation, $v_{1f}^2+v_{2f}^2=(3.5\times10^5)^2$.) $\endgroup$ – Barry Cipra Mar 12 '18 at 1:43
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Let: $$A=V_{1F}$$ $$B=V_{2F}$$ $$a=3.5\times10^5$$ Then: $$A\cos37+B\cos\theta=a\tag{1}$$ $$A\sin37-B\sin\theta=0\tag{2}$$ $$A^2+B^2=a^2\tag{3}$$ We square $(1)$ and let it equal to $(3)$, thus: $$A^2\cos^237+B^2\cos^2\theta+2AB\cos37\cos\theta=A^2+B^2\tag{4}$$ Now we square $(2)$, thus: $$A^2\sin^237+B^2\sin^2\theta-2AB\sin37\sin\theta=0$$ By the pythagorean identity, we know that $\sin^2\theta+\cos^2\theta=1$, thus the equation above can be written as: $$A^2(1-\cos^237)+B^2(1-\cos^2\theta)-2AB\sin37\sin\theta=0\tag{5}$$ Now adding $(4)$ and $(5)$ yields to: $$2AB(\sin37\sin\theta-\cos37\cos\theta)=0\Rightarrow 2AB=0, \cos(37+\theta)=0$$ Now we can solve $\theta$ as: $$\cos(37+\theta)=0\Rightarrow 37+\theta=\cos^{-1}0$$ $$\theta=53^\circ$$

Now $V_{1F}$ and $V_{2F}$ can be solved accordingly. From $(1)$ we get: $$A=\frac{-B\cos53}{\cos37}$$ Thus substituting $A$ in $(2)$ we get: $$\frac{\sin37\cdot a-B\cos53\sin37}{\cos37}-B\sin53=0$$ $$\sin37\cdot a-B\cos53\sin37-B\sin53\cos37=0$$ $$B=\frac{\sin37\cdot a}{\cos53\sin37+\sin53\cos37}=\frac{\sin37 a}{\sin(53+37)}=\sin 37 a$$ $$A=\frac{a-\sin37 a\cos53}{\cos37}$$

Mathematica confirms the complete solution for $A, B$ and $\theta$: $$\{\{A\to 0.,B\to -350000.,\theta\to -180.\}$$ $$\{A\to 0.,B\to -350000.,\theta\to 180.\}$$ $$\{A\to 279522.,B\to -210635.,\theta\to -127.\}$$ $$\{A\to 279522.,B\to 210635.,\theta\to 53.\}$$ $$\{A\to 0.,B\to 350000.,\theta\to 0.\}\}$$

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JustHeavy, coming from a Physics teacher's perspective, this looks like you were doing a classic two dimensional glancing collision. The reason I believe this is because it looks like the first equation is conservation of momentum in the $x$-direction, the second looks like conservation of momentum in the $y$-direction, and the third is the conservation of kinetic energy which occurs in an elastic collision of this type. A couple of notes; the mass of the two objects are the same, this is why it is not in any of the equations. And since this is a glancing blow the two angles must add up to $90˚$. This could be the missing value that you need to solve this problem.

$$v_{1f}\cos(37) + v_{2f}\cos \theta = 3.5 \cdot 10^5$$

This formula shows us that the total momentum before the collision was only in the x-direction and each equal mass's velocity components added together was the same.

$$v_{1f}\sin(37) - v_{2f}\sin\theta = 0$$

This formula confirms that the object was only moving in the $x$-direction. There is no y component after so the two momenta in the $y$ need to cancel each other out. One negative, one positive.

The final equation is just conservation of kinetic energy. $\frac1{2}mv^2$. Since each term had $\frac1{2}$ in it, we can cancel that out. And once again seeing that the masses are the same they are not in the equation either.

So I feel this problem was just testing you on the fact if the masses are the same, the angles needs to add to $90$. Once you use that, the first two equations are very solvable, with the third serving as a check of your work.

Cudos to all the mathematicians that were able to solve this without the fact the angles need to be $90˚$.

The following would be a picture I would put on the board to show the students.

enter image description here

I hope this clears things up. I would love to know if you were really doing a momentum problem!

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  • $\begingroup$ You are completely correct. This is a momentum physics problem. Thanks for your help. $\endgroup$ – JustHeavy Mar 12 '18 at 19:07
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    $\begingroup$ [+1] for uncovering the context. Besides, thanks for having introduced me to the word "Kudos" (with a "K") : I didn't know this term in English... $\endgroup$ – Jean Marie Apr 21 '18 at 16:34
  • $\begingroup$ @JeanMarie, you're welcome. Although I did spell it incorrectly!!! :-\ $\endgroup$ – Douglas Apr 23 '18 at 11:50
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I will use alphabetical constants instead of numbers, to make things simpler. Then you can always get back your numbers.

The three equations are writable as

$$a v_1 + v_2\cos\theta = b$$

$$\sqrt{1 - a^2}v_1 - v_2\sin\theta = 0$$

$$v_1^2 + v_2^2 = c$$

Where indeed $a, b, c$ are your numbers.

Notice from the second one that $$\sin\theta = \sqrt{1 - \cos^2\theta}$$

Hence substituting, the second equation gives you

$$v_1 = \frac{v_2\sqrt{1 - \cos^2\theta}}{\sqrt{1 - a^2}}$$

From this one, substitute into the third one and you get

$$v_2 = \sqrt{\frac{c}{1 + \frac{1 - \cos^2\theta}{1-a^2}}}$$

Finally use the first one, substituting $v_1$ and you will get

$$v_2 = \frac{b}{\frac{a\sqrt{1-\cos^2\theta}}{\sqrt{1-a^2}} + \cos\theta}$$

a second equation in terms of $v_2$ and $\theta$.

The two latter ones will uniquely give you $\theta$ and $v_2$.

ADD

The two latter equations can be compared to get a unique equation for $\theta$. Compare them

$$\sqrt{\frac{c}{1 + \frac{1 - \cos^2\theta}{1-a^2}}} = \frac{b}{\frac{a\sqrt{1-\cos^2\theta}}{\sqrt{1-a^2}} + \cos\theta}$$

Arance them, square both and arrange them again.

With a bit of patience, at the end you will obtain this:

$$A\cos^2\theta + B\sin(2\theta) + C = 0$$

Where

$$A = -b^2 - c(1 - 2a^2)$$

$$B = -ac\sqrt{1-a^2}$$

$$C = -a^2b^2 + 2b^2 - c^2a^2$$

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  • $\begingroup$ I understand that you used letter constants instead of my numbers, but looking at what the equation you have for $v_1$ it doesn't look like I can get the answers I posted in my questions since $\theta$ is not known still. $\endgroup$ – JustHeavy Mar 11 '18 at 17:03
  • $\begingroup$ @JustHeavy I added some details $\endgroup$ – Von Neumann Mar 11 '18 at 17:16
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Hint...$$(v_{2f}\cos\theta)^2+(v_{2f}\sin\theta)^2=v_{2f}^2$$

From the first equation, $v_{2f}\cos\theta =\ldots$

From the second equation, $v_{2f}\sin\theta =\ldots$

Substitute these expressions into the identity above and get an equation you can combine with the third equation to find $v_{1f}$

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  • $\begingroup$ I'm not seeing where that comes from. Did you add two equations together? $\endgroup$ – JustHeavy Mar 11 '18 at 17:05
  • $\begingroup$ $\cos^2\theta+\sin^2\theta=1$ is a standard trigonometric identity $\endgroup$ – David Quinn Mar 11 '18 at 17:07
  • $\begingroup$ Oh I see now, but I've looked at the places that I can substitute that and it doesn't really help me. I can't see where that simplifies things. Would you be able to take it one more step? $\endgroup$ – JustHeavy Mar 11 '18 at 17:13
  • $\begingroup$ I have added some lines $\endgroup$ – David Quinn Mar 11 '18 at 17:20
  • $\begingroup$ @DavidQuinn : Notice these conspicuous typographical differences: $$ \begin{align} & v_{2f} \cos \theta =... \\ & v_{2f} \cos \theta = \ldots \\ & v_{2f} \cos\theta = \cdots \end{align} $$ You wrote the first one and I changed it to the second. In my own typing I prefer the third. But the first is an orthographic error, somewhat akin to an error of spelling punctuation, by all standards. When Donald Knuth created TeX, he did it with this sort of consideration in mind. $\endgroup$ – Michael Hardy Mar 11 '18 at 17:30
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I'm going to rewrite this a bit to make it easier to type. Let$$ \begin{align} x&=v_{1f}\\ y&=v_{2f}\\ \phi &= 37^{\circ}\\ a &= 3.5\cdot 10^5 \end{align} $$ Then we have $$ \begin{align} x\cos\phi +y\cos\theta &= a\tag 1\\ y\sin\phi+y\sin\theta &= 0\tag 2\\ x^2+y^2=a^2 \tag 3 \end{align} $$ Squaring (1) and (2) gives $$ \begin{align} x^2\cos^2\phi+2xy\cos\phi\cos\theta+y^2\cos^2\theta&=a^2\tag 4\\ x^2\sin^2\phi-2xy\sin\phi\sin\theta+y^2\sin^2\theta&=0\tag 5\\ \end{align} $$ Adding (4) and (5) gives $$ \begin{align} x^2+2xy(\cos\phi\cos\theta-\sin\phi\sin\theta)+y^2=a^2\\ x^2+2xy\cos(\phi+\theta)+y^2=a^2\tag 6 \end{align} $$ From (3) and (6), $$ 2xy\cos(\phi+\theta)=0 $$ Translating back to the original notation,$$ 2v_{1f}v_{2f}\cos(37^{\circ}+\theta)=0$$

Now, we need some more information than is included in the problem statement. If we know $v_{1f},v_{2f}\ne 0,$ we can conclude that $\cos(37^{\circ}+\theta)=0.$ There are infinitely many possibilities for $\theta,$ one of which is $\theta = 57^{\circ}.$ Assuming we know that this is correct, we can substitute the values of the trigonometric functions in (1) and (2) to get two linear equations in two unknowns, which you know how to solve.

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Let's assume you know that trigonometric functions have inverses which are available to you on a standard scientific calculator. Further, let's assume that you know all about addition, multiplication, exponentiation and all of their inverses.

Your first step is to count the number of equations and the number of unknowns. If there are more unknowns, you need more equations. If there are more equations, you may be over-defined, but it'll be quick to find out. Fortunately, this problem has 3 of each and it's homework, so we're probably in the clear.

Substitution Method

First, examine the equations to see if you can isolate any of the variables in an $a = f(b)$ form (as opposed to $a = f(b,c)$). Since the third equation ($a^2 + b^2 = r^2$) has only 2 variables, start there. The next step is to substitute $f(b)$ for $a$ in equation 2 and solve for $b$. Could you also put it into Eq. 1? Yes, but Eq. 2 is simpler, so you choose that which gives you $c = f(b)$. Finally, you substitute both of those into the first equation which gives you an equation that could be written $f(a) = 0$. Now use algebra to write this equation as $a = f(37^\circ, r)$ and then find the number corresponding to $a$. Is it the same as the given answer?

Next, invert the two equations you built so they become: $f(a) = b$, and $f(a) = c$ and confirm that you did it correctly by comparing your calculated numerical answers to the given answers. When I do homework, I always try two algebraic methods to make sure my answers are correct because algebra is hard, and I make mistakes. Once I get to the test, I'll have done twice the work, and I will be better and faster.

There are other methods - many outlined here in painstaking, dare I say that since this is a homework problem, brazenly explicit, detail, but I'd like to offer you one more with similar opacity. Notice that $a^2(\sin^2(\alpha) + \cos^2(\alpha) ) = a^2$.

Elimination Method

$$\left[ (\text{Eq. 1})^2 + (\text{Eq. 2})^2 - \text{Eq. 3}\right] \rightarrow 2ab(f(c)) = 0 = f(c)$$

Solve for $c$. If you're feeling frisky, you can repeat the procedure using Equations 2 and 3 and use acres of trigonometric algebra to find $a$ and $b$, but you can also just revert back to the Substitution Method knowing $c$.

Unsolvable Systems

Sometimes, we have an 'under-defined' system even though we have 3 equations. For example, if your third equation was replaced with $2 \cdot (\text{Eq.1}) - 3 \cdot (\text{Eq.2})$, you would still only have 2 equations. The way to show that would be to find an equation $F(\text{Eq. 1}) + G(\text{Eq. 2}) = \text{Eq. 3}$. Sometimes, a system of 2 independent equations and 3 unknowns has one variable that you can solve for leaving one equation with 2 unknowns.

If, instead, you had a fourth independent equation, it's possible that it would be incompatible and you'd have an unsolvable system - for example, $a^2 + b^3 = r^2$. If you assumed that the parameters have an uncertainty associated with them, you might massage those to find the best fit for the unknowns. If you were uncertain about the equations, instead, you might massage those. In any event, there are methods to deal with this, but probably not in this class.

Lastly, there are some functions you can't invert. This gets more common as math gets harder, but if you can get the equation into the form $f(a) = 0$, there are numerical methods for finding $a$ and they get easier as you get better at math. Functions of the form $f(a,b, ...) = 0$ are significantly harder to approximate, and if you get there, especially in this class, you've made a horrible mistake.

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There is a common way of solving this kind of system translating it into a polynomial system and solving it using Gröbner basis. You can learn about it here.

First of all, as the previous answers did, I will use a different notation. I will call

$a=\cos(37º), b=\sin(37º), c=3.5\times 10^5$

$x=v_{1f}, y=v_{2f}, z=\cos\theta, t=\sin\theta$

Then, the system becomes

$ \begin{cases} ax+yz-c=0\\ bx-yt=0\\ x^2+y^2-c^2=0 \end{cases} $

Since $\cos\theta$ and $\sin\theta$ are related by $\cos^2\theta+\sin^2\theta=1$ we have to add the equation $z^2+t^2-1=0$, so we get

$ \begin{cases} ax+yz-c=0\\ bx-yt=0\\ x^2+y^2-c^2=0\\ z^2+t^2-1=0 \end{cases} $

This equations define an ideal $I=\langle ax+yz-c, bx-yt, x^2+y^2-c^2, z^2+t^2-1\rangle\subset \mathbb{R}[x,y,z,t]$, and the solutions are the elements of the algebraic set (or variety) $V(I)$.

Let's compute a Gröbner basis $G$ for $I$ using a computer algebra system (I use SAGE). Here, the monomial ordering is important, in general any lex order works fine, I will use lex with $x>y>z>t$.

$G=\{x - \frac{318367197940197676250000}{847527367068540249}t^2, y - 300000z + \frac{234667789200000}{163394518424627}t, z^2 + t^2 - 1, zt - \frac{1518598087}{2015242866}t^2, t^3 - \frac{4061203808963893956}{6367343958803953525}t\}=\{g_1,g_2,g_3,g_4,g_5\}$

Then, $V(I)=V(G)$. At first sight, the equations given by $G$ seem awful, but the last polynomial is a polynomial only on $t$, with one trivial solution $t=0$ (corresponding to $\theta=k\pi$ with $k$ integer). The other solutions are $t=\pm \sqrt{\frac{4061203808963893956}{6367343958803953525}}$ (you can clear $\theta$ using a calculator).

Then, let's go to the polynomials on $z$ and $t$. Following the notation of the book above, let's call $I_3=\langle g_5\rangle$ and $I_2=\langle g_3,g_4,g_5\rangle$. By the Extension Theorem, a partial solution $d\in V(I_3)$ extends to $(c,d)\in V(I_2)$ whenever $d\notin V(c_3,c_4)$, being $c_3,c_4$ the coefficient of the highest power of $z$ in $g_3$ and $g_4$ respectively (actually we should take $g_5$ into account, but since it has no $z$ we won't worry about it).

In this case, $V(c_3, c_4)=V(1, t)$ corresponding to $z^2$ and $z$ respectively. Since $V(1,t)=\emptyset$, all the partial solutions for $t$ that we found earlier extend to partial solutions of $V(I_2)$. So, for each solution $t$, you just have to clear $z$ in $g_3=0$ and $g_4=0$ and find common (approximate) solutions. For example, for $t=0$, $g_4=0$ for all $z$, and $g_3=0$ implies $z=\pm 1$, so here you've got two partial solutions $(1,0)$ and $(-1,0)$. Note that these values of $z$ are coherent since, coming back to your original system, $\cos\theta=\pm 1$ when $\sin\theta=0$.

Then, again, we always can extend a partial solution in $V(I_2)$ to $V(I_1)=V(\{g_2,g_3,g_4,g_5\})$ because the coefficient of $y$ (this is the maximum power of $y$) in $g_2$ is constant, so for every partial solution $(z,t)$ you can easily clear $y$ from $g_2=0$, giving a partial solution $(y,z,t)\in V(I_1)$. For example, for the partial solutions above, we've got $y=300000$ for $(1,0)$ and $y=-300000$ for $(-1,0)$, so the resulting partial solutions are $(300000, 1, 0)$ and $(-300000,-1,0)$.

Finally, extend these solutions to $V(I)$ (you can do it because the coefficient of $x$ is constant in $g_1$, and again the solutions are easy to get from $g_1=0$). Following the example above, we have found two solutions: $(0,-300000,-1,0)$ and $(0,300000,1,0)$. You can check that these are actually solutions for the polynomial system and hence from your original system. The solutions you've been given must come from one of the other possible values of $t$. In particular, from the positive value, which gives you exactly the angle $\theta=53º$.

The code to finding the Gröbner basis in SAGE, which you can try here, is the following

Q.<x,y,z,t>=PolynomialRing(QQ,order='lex')

ar = cos(37 / 360 * (2*pi)).numerical_approx()
br=sin(37 / 360 * (2*pi)).numerical_approx()
c= 3*10^5
a=ar.nearby_rational(max_error=10^(-5))
b=br.nearby_rational(max_error=10^(-5))

I=ideal(a*x+y*z-c, b*x-y*t, x^2+y^2-c^2, z^2+t^2-1 )
G=I.groebner_basis()
G

Note that these are rational approximations, which is the best we can expect from a computer, but you can make it more accurate by changing the value of max_error.

$\endgroup$

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