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I can't figure out how to solve this problem. In fact, it looks more like a double reflection to me than a rotation. Can anyone help?

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Is the answer $50^\circ$? –  Joel Reyes Noche Nov 13 '13 at 4:51
    
Unfortunately, I don't have the answer (I'm actually posting on behalf of someone else). –  barrycarter Nov 13 '13 at 4:52
    
It turns out that a reflection over two non-parallel lines is a rotation. –  Cameron Buie Nov 13 '13 at 10:31
    
I still don't see how this is a rotation. Around which point do we rotate WXY to get CBA? –  barrycarter Nov 13 '13 at 17:09
    
It's a rotation $100^\circ$ clockwise about point $O$. –  Joel Reyes Noche Nov 14 '13 at 0:44

2 Answers 2

up vote 2 down vote accepted

Here is my take:
Let's take only 3 corresponding points $W$, $T$, and $C$ instead of the triangles.
Because $T$ is $W$'s reflection: $\angle WOQ=\angle QOT=\theta_1$ (say)
Because $C$ is $T$'s reflection: $\angle TON=\angle NOC=\theta_2$ (say)
So now $\angle WOC=\theta_1+\theta_1+\theta_2+\theta_2=2\theta_1+2\theta_2=2\angle QON$
Hence $\angle QON=50^\circ$.

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This is much better than my answer. –  Joel Reyes Noche Nov 13 '13 at 13:11
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Thanks! @JoelReyesNoche I think your answer was equivalent, but this answer is more detailed, and I understood it better, so giving the checkmark here. –  barrycarter Nov 13 '13 at 17:08

I think the question would be solvable if points $W$ and $T$ were (the same point) on the $\overline{OQ}$ axis, and points $S$ and $B$ were (the same point) on the $\overline{ON}$ axis. By similarity of triangles $\Delta OTS$ and $\Delta OBC$ we get $\angle QON=50^\circ$.

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