Trigonometry: Value of constants

Question

I have this equation where $A\sin b=B\sin(a-b)$ I have to find the value of A and B. Is there any other way except for assuming $A=\sin(a-b)$ (if it's even correct in the first place) even though it is giving me the answer I want? Note the $\sin$ values are not zero neither the constants. The a and b are specific points, nos. radians,... edit:$sin(a)$ and $sin(a-b)$ is a constant value except zero.

Answer 1

$$A\sin b=B\sin(a-b)\iff \frac {A}{B}= \frac {\sin (a-b)}{\sin b}$$

$$A=\lambda \sin (a-b)$$ and$$ B=\lambda \sin b$$ where $\lambda $ is an arbitrary real number.

Answer 2

If you have specific values of $a$ and $b$, plug them into

$$A\sin(b) = B\sin(a-b) \tag{1}\label{eq1}$$

to get something like

$$Ac_1 = Bc_2 \tag{2}\label{eq2}$$

where $c_1 = \sin(b)$ and $c_2 = \sin(a-b)$. If $c_1$ and $c_2$ are not $0$, then there is no unique value for $A$ and $B$. Instead, you can choose one of the values and get the other one in terms of the first. In particular, you can assume a value for $B$ and get $A = \frac{Bc_2}{c_1}$, or assume a value for $A$ and get $B = \frac{Ac_1}{c_2}$.

Answer 3

Assuming that $a, b, A,$ and $B$ are parts of a triangle (which imposes some restrictions on their values) by Law of Sines we have

$$\frac{\sin a}{A} = \frac{\sin b}{B}.$$

If we un-cross-multiply your equation, we have

$$\frac{\sin (a-b)}{A} = \frac{\sin b}{B}.$$

We conclude that $\sin a = \sin(a-b).$

Under our restrictions, this forces $b=0$.