Let O be the center of the circle
Angles OQP and ORP both = 90°
Then in quadrilateral PQOR.....angle QOR is supplemental to angle QPR =140°
So the smaller angle inscribed angle QSR = (360 -140)/2 = 220/2 = 110°
And the larger angle QSR = 360 - 110 = 250°
And in quadrilateral PQSR the sum of angles PQS + PRS = 360 - 40 -250 =
360 - 290 =
70°