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Here is the diagram

enter image description here

If we only know distance PA to line l1, distance PC to line l3, and angle alpha between l1 - l2, angle beta between l2 - l3, how to calculate PB from P to l2 the fastest?

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  • $\begingroup$ In which form the lines are given? $\endgroup$ – Dr. Sonnhard Graubner Jun 30 '18 at 11:08
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Hint.

Calling $|OP| = 2r$ we have

$$ 2r\sin\phi = d_1\\ 2r\sin(\alpha+\phi) = d_2\\ 2r\sin(\beta+\alpha+\phi) = d_3 $$

three equations and three unknowns $\phi, r, d_2$

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Call $\gamma$ as the angle between $L_1$ and $OP$ $$\sin(\beta+\alpha+\gamma)=\frac{PC}{OP}\\ \sin\gamma=\frac{AP}{OP}\\ \gamma\implies\cot ^{-1}\left(\frac{\csc (\alpha +\beta ) (PC-AP \cos (\alpha +\beta ))}{AP}\right)\\ OP\implies\csc (\alpha +\beta ) (PC-AP \cos (\alpha +\beta )) \sqrt{\frac{AP^2 \sin ^2(\alpha +\beta )}{(PC-AP \cos (\alpha +\beta ))^2}+1}$$ Since $\sin (\gamma+\alpha)=\frac{BP}{OP}$, then: $$\bbox[10px,border: 1px black solid]{\therefore BP\implies OP\sin(\alpha+\gamma)\implies \csc (\alpha +\beta ) (AP \sin (\beta )+PC \sin (\alpha ))}$$

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