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?
$\endgroup$ 22 Answers
$\begingroup$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 $22.5=45/2$.
$\endgroup$ 2 $\begingroup$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 $AB$:
- 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 $AB$.
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 $AB$.
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 $AB$.
