# Geometry problem calculate arc, inner triangle.

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 $\widehat {AB}$ 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 in the arc calculation.

$R=6400$

$S=30$

$\widehat {AB}=?$

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

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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).

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