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Said best with a picture. Given angles a and b, solve for angle x. (Note that the top right vertex is also the center of the circle)

solve for X

What I've tried

Unable to find a simple method to get to x, I decided to draw all chords, and extend all segments to the edges of the triangle:

added chords and extended lines

I've managed to get just about every angle except the few I need to find x, so I'm not sure if this method of extending all lines has helped. Here's how far I got (excuse the rearrangement, I needed room to write):

everything but what I need

I think I'm overlooking something simple here. I'm not asking for the solution necessarily, just how I should get there.

Also if someone could help me with a more technically descriptive title, I would love an edit.

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  • $\begingroup$ What is a? It's created by a bisector? $\endgroup$ – Vladimir Fomenko Jan 4 '15 at 9:04
  • $\begingroup$ Not necessarily. It's just a known initial value. $\endgroup$ – uber5001 Jan 4 '15 at 9:05
  • $\begingroup$ Some info to check solutions against (possibly obvious to all already): If A is 0 then x must be 0. If A is (90-B), x = B. $\endgroup$ – turkeyhundt Jan 4 '15 at 9:21
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Scale the figure such that the circle is the unit circle.

Put a coordinate system such that $A$ is the center, and $C$ is on the $x$-axis

Now the point $D$ has coordinate

$$D=(\cos a,\,\sin a)$$

The sine of $b$ is $\frac{|AB|}{|AC|}$ so

$$\sin b=\frac1{|AC|}\implies|AC|=\csc b$$

Comparing $C$ and $D$ we find

$$\Delta x=\csc b-\cos a\qquad\qquad\Delta y=\sin a$$

Taking the arctangent of the slope of that, we get the angle

$$ \begin{align} x&=\arctan\frac{\sin a}{\csc b-\cos a}\\ &=\arctan\frac{\sin a\sin b}{1-\sin b\cos a} \end{align} $$

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  • $\begingroup$ For the uninformed: $$\csc x = \frac1{\sin x}$$ $\endgroup$ – Alice Ryhl Jan 4 '15 at 9:55
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    $\begingroup$ +1. You can tighten the argument slightly by avoiding coordinates, while taking advantage of your horizontal hypotenuse. I also might not scale, but would take $|AC|$ as my fundamental length. Then: Drop a perpendicular from $D$ to $F$ on $AC$. Since $|AD| = |AB| = |AC|\sin b$, we have $$|DF| = |AD|\sin a= |AC|\sin a\sin b \qquad\text{and}\qquad |AF| = |AD|\cos a = |AC|\sin a\cos a$$ Then, $$\tan x = \frac{|DF|}{|CF|} = \frac{|AC|\sin a\sin b}{|AC|-|AF|} = \frac{|AC|\sin a\sin b}{|AC|(1-\sin a\cos a)} = \frac{\sin a\sin b}{1 - \sin a \cos a}$$ This also avoids having to "explain" $\csc$. ;) $\endgroup$ – Blue Jan 4 '15 at 10:54
  • $\begingroup$ @Blue yeah, about csc, I have simplified it away now. $\endgroup$ – Alice Ryhl Jan 4 '15 at 11:03
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    $\begingroup$ @Blue also your formula is slightly wrong, you said $$|AD|\sin a=|AC|\sin a\sin b$$ and then $$|AD|\cos a=|AC|\sin a\cos a$$ where the last should be $$|AD|\cos a=|AC|\sin b\cos a$$ $\endgroup$ – Alice Ryhl Jan 4 '15 at 11:07
  • $\begingroup$ Ah, I should be more careful TeXing in comments, since there's a time limit on fixing typos. Oh, well ... $\endgroup$ – Blue Jan 4 '15 at 11:31

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