If PQR is a straight line with QS = QR, calculate TPQ, If QT\\SR and TQS = 3y
°Exam Year# 1989
 

  • A 62°
  • B 56\(\frac{3}{2}\)°
  • C 20\(\frac{3}{2}\)°
  • D 18°
jamb 1989mathematics

The correct answer is C. 20\(\frac{3}{2}\)°

Since QS = QR
then, angle SQR = angle SRQ2 SQR = 180 - 56, SQR = \(\frac{124}{2}\) = 62° 

SQR = 180 - 56, SQR = \(\frac{124}{2}\) = 62

QTP = 62°

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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