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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?

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    $\begingroup$ $\sin\alpha = \frac{o \sqrt{1-\frac{b^2}{q^2}}-b}{q \sqrt{\frac{o^2}{q^2}+1}}$ $\endgroup$
    – Cesareo
    Jul 12, 2019 at 8:32
  • $\begingroup$ Thanks for your comment. But how did you made this up and how could I proceed with it? $\endgroup$
    – Codey
    Jul 12, 2019 at 8:36
  • $\begingroup$ $\sin(\arctan(t))=\frac{t}{\sqrt{t^2+1}}$. Now, squaring will lead to a montreous polynomial in $q$ $\endgroup$
    – Claude Leibovici
    Jul 12, 2019 at 8:42
  • $\begingroup$ @ClaudeLeibovici the monstreous polynomial will still contain roots of $q$. No? $\endgroup$
    – denklo
    Jul 12, 2019 at 8:46
  • $\begingroup$ @denklo. Yes, for sure ! One of the problems is that we shall need squaring which introduces extra roots. $\endgroup$
    – Claude Leibovici
    Jul 12, 2019 at 8:48

1 Answer 1

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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 $\tan \, \alpha$.

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  • $\begingroup$ Do you see any way to avoid a quartic polynomial in $q$ ? $\endgroup$
    – Claude Leibovici
    Jul 12, 2019 at 8:57
  • $\begingroup$ I don't think that is possible. $\endgroup$
    – Kavi Rama Murthy
    Jul 12, 2019 at 9:01

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