Points $A$ and $B$ lie on a circle centered at $O$, and point $X$ is outside the circle such that $\overline{AX}$ and $\overline{BX}$ are tangent to the circle. If $\angle AXO = 26^{\circ}$, then what is the measure of minor arc $AB$, in degrees?
+23246 Points A and B divide the circle into two arcs: major arc(AB) and minor arc(AB).
The size of angle(AXO) = 26°.
This means that the size of angle(BXO) is also 26°.
The total size of angle(AXB) = 52°.
Since we want to find the size of minor arc(AB), call minor arc(AB) = x.
Since the sum of the major arc(AB) and minor arc(AB) = 360°, major arc(AB) = 360° - x.
The appropriate formula is: [ major arc(AB) - minor arc(AB) ] / 2 = angle(AXB).
[ (360° - x) - x ] / 2 = 52°
[ 360° - 2x] / 2 = 52°
360° - 2x = 104°
256° = 2x
x = 128°
