In the diagram, PR||SV||WY|, TX||QY|, < PQT = 48o and < TXW = 60o.Find < TQU.
120o
108o
72o
60o
Correct answer is C
In the diagram, < TUQ + 60o(corresp. angles)
< QTU = 48o (alternate angles)
< QU + 60o + 48o = 180o(sum of angles of a \(\Delta\))
< TQU = 180o - 108o
= 72o
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