In the figure, QS = 15, RT = 36, and RT is tangent to radius QR with the point of tangency at R. Find QT.

the radius going through the tangent point is perpendicular to the tangent line, so ΔQRT is a right triangle. 
QT²=QR²+RT²
QR and QS are both radii, so QR=QS=15
QT²=15²+36²
QT=39