Answer:
(C)100°
Step-by-step explanation:
See attached diagram to understand the problem better.
[tex]|QT|=|PT|$ (radii of circle T)\\Therefore \triangle QPT $ is an isosceles triangle\\\angle PQT=\angle QPT=40^\circ$ (base angles of an isosceles triangle)[/tex]
[tex]\angle PQT+\angle QPT+\angle PTQ=180^\circ$ (sum of angles in a \triangle )\\40^\circ+40^\circ+ \angle PTQ=180^\circ\\\angle PTQ=180^\circ-80^\circ\\\angle PTQ=100^\circ[/tex]
The measure of arc SR=Central Angle STR
[tex]\angle PTQ=\angle STR $ (Vertically opposite angles)\\\angle STR=100^\circ\\mSR=100^\circ[/tex]
