Find angles $A$ and $B$ in equation $\frac{F_1}{\cos A }=\frac{F_2}{\cos B}=\frac{F_3}{\sin(A+B)}$

Question

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?

Answer 1

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

$$\sin A = \frac{F_3^2+ F_2^2- F_1^2}{2F_2F_3},\>\>\>\>\> \sin B = \frac{F_3^2+ F_1^2- F_2^2}{2F_1F_3} $$

Answer 2

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.