In the diagram above, < SQR = 52º and < PRT = 16º. Find the value of the angle marked y
<RQS + <QRS + RSQ = 180º
52º + 16º + <RSQ = 180º
<RSQ = 180 - 52 - 16 = 112º
But <RSQ = y ( exterior angle of a cyclic quad.)
Therefore, y = 112º
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