# Geometry Question: Calculating the length of the side of a triangle inside a circle.

Given this picture:

The radius of the circle is $30$ inches. The angle between $A$ and $B$ is $22.5^o$.

How would I calculate the distance (not along the arc, but straight line) from $B$ to $A$, as depicted by the red line?

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@Eric Naslund Thank you for adding the image to the post. –  NW Tech Jun 3 '11 at 23:42
Thank Jonas Meyer, he was the one that did that. I merely added some latex for the letters, and changed the title to be more descriptive. (Usually we try to stay away from titles with "easy" or "hard" since these adjectives are very subjective.) –  Eric Naslund Jun 4 '11 at 0:32

Bisect the angle to get 2 right triangles with known hypotenuse and angles, then use $\sin$ to get the sides opposite the $22.5/2$ degree angles.

Or, use the triangle that's already there, having 2 angles equal to $(180-22.5)/2$ degrees, and apply the law of sines.

Or, apply the law of cosines immediately.

In case you want exact forms for the sines or cosines involved, you can use half/double angle formulas and the fact that $22.5=45/2$.

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Breaking that up into two right triangles, using the law of sine, and rounding to 1 decimal place, I came up with 11.7 inches. does that look right? –  NW Tech Jun 3 '11 at 23:49
@Chip: Yes, it's around 11.7054193. An exact form is $$30\sqrt{2-\sqrt{2+\sqrt{2}}}.$$ –  Jonas Meyer Jun 3 '11 at 23:55

Just for future reference, in case someone stumbles upon this problem hoping to learn how to solve a similar problem:

Let $O$ denote the point at the origin. We are given that $\angle AOB = 22.5^\circ$, and we are given that the radius of the circle is $30$ inches. That means the length of the line segments $OA$ and $OB$ are each $30$ inches, since they are both radii of the circle. As Jonas pointed out in his answer, there are a number of approaches to solving for the length of the line segment $AB$.

Note that $\triangle AOB$ is an isosceles triangle, and so the angles $\angle OAB, \angle OBA$ are equal. Let's call the measure of one of these two angles "$x$". Then, since the sum of the measures of the angles of any triangle is $180^{\circ}$, we know that $$22.5 + 2x = 180^{\circ}$$ Solving for $x$ gives us $\displaystyle x = \frac{180-22.5}{2} = 78.75^\circ$

Now, there are a few options: We have all the angles of $\triangle AOB$, and the length of two of its sides. We need only find the length of the segment $AB$.

We can use any of the following approaches find the length of $AB$:

1. Using the Law of sines: $$\frac{c}{\sin(22.5^\circ)} \,=\, \frac{30}{\sin(78.5^\circ)} \,=\, \frac{30}{\sin(78.75^\circ)}$$ where the numerators are the lengths of the sides of a triangle, $c$ the unknown length, and the denominator of each term is the sin of the angle opposite the side given in its numerator. From here, one can easily solve for $c$, the length of the segment $AB$.
2. Using the Law of cosines, in this case, $$(AB)^2 = (OA)^2 + (OB)^2 - 2(OA)(OB)\cos(\angle AOB) \rightarrow (AB)^2 = 2(30)^2 - 2(30)^2\cos(22.5^\circ)$$ One need only solve for segment $AB$.

3. Denote the midpoint of segment $AB$ by $M$ (which is also the point at which the bisector of $\angle AOB$ intersects segment $AB$), such that $\triangle AOM, \triangle BOM$ are both congruent right triangles (with $\angle OMA, \angle OMB\;\text{both}\; 90^\circ$). So, we have that $\cos(\angle OAB) = (AM)/(OA)$, so that $\cos(78.75^\circ) = (AM)/30$. Solving for $AM$ givens us $(AM) = 30\cdot \cos(78.75)^\circ$, and from there we need only compute $2(AM)$ to obtain the length of the segment $AB$.

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$\exp(+)^{\infty}$. Thanks for your support. I hope I will be a hand for others, Amy. :-) –  Babak S. Jul 22 '13 at 17:09
Thank you @Babak! –  amWhy Jul 22 '13 at 17:22