If PQR is a straight line with QS = QR, calculate TPQ, If QT\SR and TQS = 3yo
-
A.
62o -
B.
56(frac{3}{2})o -
C.
20(frac{3}{2})o -
D.
18o
Correct Answer: Option C
Explanation
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})