Find angles $A$ and $B$ in equation $\frac{F_1}{\cos A }=\frac{F_2}{\cos B}=\frac{F_3}{\sin(A+B)}$ I have the following relation for equilibrium of surface tension
$$\frac{F_1}{\cos(A)}=\frac{F_2}{\cos(B)}=\frac{F_3}{\sin(A+B)}$$
While trying to find the angles A and B, I end up having the same equations while solving the above 
$F_2\sin(A)+F_1\sin(B)=F_3$
$F_2\sin(A)+F_1\sin(B)=F_3$
Can anyone explain or comment on how to approach such question to find the unknown angles? 
 A: Let $\alpha = 90- A$,  $\beta = 90-B$ and $\alpha+ \beta +\gamma=180$ to write $\frac{F_1}{\cos A}=\frac{F_2}{\cos B}=\frac{F_3}{\sin(A+B)}$ as
$$ \frac{F_1}{\sin \alpha}=\frac{F_2}{\sin \beta} =\frac{F_3}{\sin \gamma}$$
which is the sine rule for the triangle of angles $\alpha$, $\beta$, $\gamma$ and sides $F_1$, $F_2$, $F_3$, respectively. Thus, via the cosine rule, the solutions for the unknown angles are 
$$\sin A = \frac{F_3^2+ F_2^2- F_1^2}{2F_2F_3},\>\>\>\>\>
\sin B = \frac{F_3^2+ F_1^2- F_2^2}{2F_1F_3}
$$
A: Suggestion: Let $x$ be the number such that
\begin{equation}
\cos A = x F_1,\quad \cos B = x F_2,\quad \sin(A+B) = x F_3 
\end{equation}
Then
\begin{equation}
F_3 = F_2\sin A + F_1 \sin B =\epsilon_2 F_2\sqrt{1 - x^2 F_1^2} + \epsilon_1 F_1 \sqrt{1 - x^2 F_2^2}
\end{equation}
where both $\epsilon_i$ are $\pm 1$. Now isolate the first square root, square the equation. Then isolate the second square root, square the equation and you obtain a polynomial equation for $x$, which you can solve.
