In the fiqure where PQRTU is a circle, ISTI = IRSI and angle TSR = 52o. Find the angle marked m
128o
52o
104o
64o
116o
Correct answer is E
< STR = \(\frac{180 - 52}{2}\) = \(\frac{128}{2}\) = 64o
< PTR = 180 - < STR(angle on a straight line)
= 180 - 64 = 116o
< PQR + < PTR = 180(Supplementary)
< PQR + 118 = 180
< PQR = 180 - 118
= 64
M = 180 - < PQR
= 180 - < PQR = 180 - 64
= 116o
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