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