+826 Points $T$ and $U$ lie on a circle centered at $O$, and point $P$ is outside the circle such that $\overline{PT}$ and $\overline{PU}$ are tangent to the circle. If $\angle TOP = 45^{\circ}$, then what is the measure of minor arc $TU$, in degrees?
+129852 
PT = PU )tangents form a point outside a circle are equal_
PO = PO
TO = UO = radii of the same circle
So, by SSS triangle TOP is congruent to triangle UOP
So angleTOP = angle UOP = 45
So angle TOU = 90
This is a central angle intercepting minor arc TU ...so minor arc TU = 90
