I have the following equation:
$$\alpha = \arctan\Big(\frac{a}{q}\Big) \ - \ \arcsin\Big(\frac{b}{q}\Big)$$
The values $\alpha$, $a$ and $b$ are known, the only missing value is $q$. So I need to solve the equation for $q$ but I have absolutely no idea how to do it or even where to start.
I also tried putting this equation into Wolfram Alpha, but even this didn't give a usable output.
I've tried something like this, but it doesn't seem that it would lead to the correct solution
$$\sin(\alpha) = \sin\bigg(\arctan\Big(\frac{a}{q}\Big)\bigg) - \frac{b}{q}$$
EDIT:
I've tried your hint @Kavi and I came up with the following:
$$ \begin{align*} \tan(\alpha) &= \tan \Bigg(\arctan\Big(\frac{a}{q} - \arcsin\Big(\frac{b} {q}\Big)\Bigg) \\ &= \frac{\tan\Big(\arctan\big(\frac{a}{q}\big) - \tan\big(\arcsin\big(\frac{b}{q}\big)\big)\Big)}{1+\tan\Big(\arctan\big(\frac{a}{q}\big)\Big) \ \tan\Big(\arcsin\big(\frac{b}{q}\big)\Big)} \\ &= \frac{\frac{a}{q} - \frac{\frac{b}{q}}{\sqrt{1-\frac{b^2}{q^2}}}}{1+\frac{a}{q} \frac{\frac{b}{q}}{\sqrt{1-\frac{b^2}{q^2}}}} \end{align*} $$
Is this correct so far?
Hint: $\tan (A-B)=\frac {\tan\, A-\tan \, B} {1+\tan \,A \tan \, B}$. It is very easy to find $q$ once you apply this formula for $\tan \, \alpha$.