Geometry problem calculate arc, inner triangle.

Question

I learned how to solve this in geometry in middle school, but I can't remember. This isn't homework.

I want to calculate the distance between points A˜B and the arc A˜B as well. The know values are, the radius R and the S length. C is the center of the circle. The angle is not important, but it's probably involved on the arc calc.

R=6400

S=30

A˜B=?

this values are arbitrary, I only want to know how to solve.

Answer 1

Draw the altitude from $C$ to $AB$. It bisects $AB$ at a point (call it $H$) since $\triangle ABC$ is an isosceles triangle, and its length is $R - S$. Using the Pythagorean Theorem:

$$ (AH)^2 = (AC)^2 - (CH)^2 = R^2 - (R - S)^2 = 2RS - S^2 $$

Therefore:

$$ AB = 2(AH) = 2 \sqrt{2RS - S^2} $$

For the arc, use the lengths of $AH$ and $AC$ to calculate the sine of $\alpha/2$:

$$ \sin\left(\frac{\alpha}{2}\right) = \frac{AH}{AC} = \frac{R - S}{R} $$

This allows you to find the value of $\alpha$. The arc length is $R \times \alpha$ (where $\alpha$ is measured in radians).