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.
Best Answer
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
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. $\theta=\pi, \; a=0, \; b=-A$