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There are three planes, A, B, and C, all of which intersect at a single point, P. The angles between the planes are given: $$\angle\mathbf{AB}=\alpha$$ $$\angle\mathbf{BC}=\beta$$ $$\angle\mathbf{CA}=\gamma$$ $$0\lt\alpha,\beta,\gamma\le\frac{\pi}{2}$$

The intersection of any two of these planes form lines. The intersection of AB is $\mathbf{\overline{AB}}$, the intersection of BC is $\mathbf{\overline{BC}}$, and the intersection of CA is $\mathbf{\overline{CA}}$. It is given that none of these lines are parallel to each other and that they all intersect at the same single point, P.

Please express the lesser of the two angles formed by the intersection of these lines in terms of $\alpha$, $\beta$, and $\gamma$: $$\angle\mathbf{\overline{AB}\;\overline{BC}}=?$$ $$\angle\mathbf{\overline{AB}\;\overline{CA}}=?$$ $$\angle\mathbf{\overline{BC}\;\overline{CA}}=?$$

Thanks!

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closed as off-topic by Mario Carneiro, Shailesh, Vidyanshu Mishra, mechanodroid, Henrik Oct 14 '17 at 10:14

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  • $\begingroup$ It's hard to help you without knowing what you've tried and where you're stuck. $\endgroup$ – Qudit Oct 14 '17 at 0:13
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Let $(QPR)$, $(SPR)$ and $(QPS)$ be our planes, where $\measuredangle QPR=\alpha$, $\measuredangle SPR=\beta$, $\measuredangle QPS=\gamma$.

Let $\{\alpha,\beta,\gamma\}\subset\left(0^{\circ},90^{\circ}\right)$.

We can assume that $\measuredangle QRP=\measuredangle SRP=90^{\circ}$. Also, let $PR=a$.

Thus, $$QR=a\tan\alpha,$$ $$SR=a\tan\beta,$$ $$PQ=\frac{a}{\cos\alpha},$$ $$PS=\frac{a}{\cos\beta}$$ and $$QS^2=\left(\frac{a}{\cos\alpha}\right)^2+\left(\frac{a}{\cos\beta}\right)^2-2\cdot\frac{a}{\cos\alpha}\cdot\frac{a}{\cos\beta}\cdot\cos\gamma.$$ Let $\measuredangle QRS=\phi$.

Thus, $$\left(\frac{a}{\cos\alpha}\right)^2+\left(\frac{a}{\cos\beta}\right)^2-2\cdot\frac{a}{\cos\alpha}\cdot\frac{a}{\cos\beta}\cdot\cos\gamma=$$ $$=\left(\frac{a\sin^2\alpha}{\cos\alpha}\right)^2+\left(\frac{a\sin^2\beta}{\cos\beta}\right)^2-2\cdot\frac{a\sin\alpha}{\cos\alpha}\cdot\frac{a\sin\beta}{\cos\beta}\cdot\cos\phi$$ or $$\cos^2\alpha+\cos^2\beta-2\cos\alpha\cos\beta\cos\gamma=$$ $$=\sin^2\alpha\cos^2\beta+\sin^2\beta\cos^2\alpha-2\sin\alpha\sin\beta\cos\alpha\cos\beta\cos\phi$$ or $$2\cos^2\alpha\cos^2\beta-2\cos\alpha\cos\beta\cos\gamma=-2\sin\alpha\sin\beta\cos\alpha\cos\beta\cos\phi$$ or $$\cos\phi=\frac{\cos\gamma-\cos\alpha\cos\beta}{\sin\alpha\sin\beta}.$$

I changed your notations, but I am sure that you can get relevant formula with your notation.

By the same way you can get needed identities.

See also here: https://en.wikipedia.org/wiki/Spherical_trigonometry

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  • $\begingroup$ Do I need to know a new angle $\phi$, or is this one of the angles I am trying to solve for? $\endgroup$ – TheNewGuy Oct 14 '17 at 6:55
  • $\begingroup$ @TheNewGuy In our case $\phi$ it's $\gamma$ in your notations. I just showed how you can get laws from the link. $\endgroup$ – Michael Rozenberg Oct 14 '17 at 7:04

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