If PQR is a straight line with QS = QR, calculate TPQ, If QT\\SR and TQS = 3yo
62o
56\(\frac{3}{2}\)o
20\(\frac{3}{2}\)o
18o
Correct answer is C
Since QS = QR
then, angle SQR = angle SRQ
2 SQR = 180 - 56, SQR = \(\frac{124}{2}\) = 62o
QTP = 62o
QTP = 62o, corresponding angle
3y + 56 + 62 = 180 = 3y = 180 - 118
3y = 62 = 180
3y = 180 - 118
3y = 62
y = \(\frac{62}{3}\)
= 20\(\frac{3}{2}\)
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