First post...
Is it possible to rearrange this equation for the variable angle c in terms of the variable angles a and b?
$$\frac{\sin(\pi-a-b-c)}{\sin(a)}=\frac{\sin(c)}{\sin(b)}$$
(a and b variables in the range $0 < a,b < \pi$ and c is in the range $0 < c < \frac{\pi}{6}$)
I have arrived at this equation from a triangulation problem. If there is no solution to this equation, I can elaborate on the actual problem I am trying to solve.
Any ideas/solutions are much appreciated!
Notice that $\sin(\pi-x)=\sin(x)$, which allows us to simplify a bit. We then have the sum of angles formula, which states that $\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$. This simplifies things down a bit:
$$\sin(\pi-a-b-c)=\sin(a+b+c)=\sin(a+b)\cos(c)+\cos(a+b)\sin(c)$$
Now we divide both sides by $\sin(a)\sin(c)$ to get
$$\sin(a+b)\cot(c)+\cos(a+b)=\frac{\sin(a)}{\sin(b)}$$
Subtract $\cos(a+b)$ from both sides:
$$\sin(a+b)\cot(c)=\frac{\sin(a)}{\sin(b)}-\cos(a+b)$$
And divide by $\sin(a+b)$.
$$\cot(c)=\frac1{\sin(a+b)}\left(\frac{\sin(a)}{\sin(b)}-\cos(a+b)\right)$$
And then take inverse cotangent to find $c$.