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.

  • $\begingroup$ 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

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

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

  • $\begingroup$ For the uninformed: $$\csc x = \frac1{\sin x}$$ $\endgroup$ – Alice Ryhl Jan 4 '15 at 9:55
  • 1
    $\begingroup$ +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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.