In the diagram $\ov{TU}$ is a tangent to the circle SPQR at P. If $∠$PTS = 44º, $∠$SQP = 35º, find $∠$PST
$∠$SPT = 35º (alternate segment)
From $△$TSP
44 + $∠$SPT + $∠$PST = 180º (sum of angles in a triangle)
$∠$PST = 180 - 44 - 35 = 101º
