In the figure POQ is the diameter of the circle PQR. If PSR = 145o, find xo
< PRQ = \(\frac{1}{2}\) < POQ = 90o < PSR + < PQR = 180o < PQR = 180o - 145o = 35o \(\bigtriangleup\)PQR is a right angled triangle x = 90 - < PQR = 90o - 35o = 55o
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