Help please. Also, here is a problem for you people

Points T and U lie on a circle centered at O, and point P is outside the circle such that PT and PU are tangent to the circle. If \(\angle TPO =33^\circ\), then what is the measure of minor arc TU, in degrees?

AUnVerifiedTaxPayer Sep 7, 2024

#1**+2 **

Since Angle TPO is 33 degrees, then Angle OPU also is 33 degrees, since PO is the angle bisector of Angle TPU.

Therefore, Angle TPU is 66 degrees.

Since tangent lines intercect the circle at 90 degrees, we can form quadrilateral TOUP with Angles T and U being 90 degrees and angle P 66 degrees.

The 4 angles of a quadrilateral add up to 360 degrees, so 360 - 2(90) - 66 = 180 - 66 = 114 degrees.

Central angle TOU is 114 degrees, so the intercepted minor arc TU is 114 degrees.

mathrokus Sep 7, 2024