- #1

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Start with the Law of Sines:

$$ \frac{\sin \beta}{x} = \frac{\sin \varphi}{R} \implies \boxed{ x = R \frac{\sin \beta}{\sin \varphi} \tag{1} }$$

Relating angles:

$$ \theta + \alpha + \varphi = \frac{\pi}{2} \implies \boxed{ \alpha = \frac{\pi}{2} -( \theta+\varphi) \tag{2} } $$

Applying Law of Cosines ( for each triangle):

$$ \boxed{ R^2 = w^2 + x^2 - 2 w x \cos \varphi \tag{3}}$$

$$ \boxed{ l^2 = r^2 + x^2 - 2 r x \cos \alpha \tag{4}}$$

Sub ##(1) \to (3)## to find ##\sin \varphi## in terms of ##\sin \beta##:

$$ \left( \left( R^2-w^2\right)^2+ 4 w^2 R^2 \sin^2 \beta \right) \sin^4 \varphi - \left( 2R^2\left(R^2-w^2\right) + 4 w^2 R^2 \right) \sin^2 \beta \sin^2 \varphi + R^4 \sin^4 \beta = 0 \tag{5} $$

Eq ##(5)## can be solved for ##\sin \varphi## using the quadratic formula with the substitution ## u = \sin^2 \varphi ##:

## a = \left( \left( R^2-w^2\right)^2+ 4 w^2 R^2 \sin^2 \beta \right) ##

## b = - \left( 2R^2\left(R^2-w^2\right) + 4 w^2 R^2 \right) \sin^2 \beta ##

##c = R^4 \sin^4 \beta##

It follows that:

$$ \boxed{ \sin \varphi = \sqrt{ \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} } \tag{6} } $$

Now, if we return to ##(4)## substituting ##(1)## and the identity ##\cos \alpha = \sin ( \theta+ \varphi)##:

$$ l^2 = r^2 + \frac{R^2 \sin^2 \beta}{ \sin^2 \varphi}- 2 r \frac{R \sin \beta}{ \sin \varphi } \sin ( \theta + \varphi) $$

Then applying the sum-difference identity for ##\sin( \theta + \varphi )##:

$$\boxed{ l^2 = r^2 + \frac{R^2 \sin^2 \beta}{ \sin^2 \varphi}- 2 r \frac{R \sin \beta}{ \sin \varphi } \left( \sin \theta \cos \varphi + \cos \theta \sin \varphi \right) \tag{7} }$$

From this point I can use a similar trick as in ##5## to get a solution for either ##\sin \theta , \cos \theta ## in terms of ##\sin \beta ## and all the constants.

It's seeming to be a real mess. Is it really this complex of a function - Have I messed up?