In the diagram, \(\bar{PF}\), \(\bar{QT}\), \(\bar{RG}\) intersect at S and PG||RG. If < SPQ = 113º and < RSt = 22º, find < PSQ
In In the diagram given, \(\alpha\) = 22º (vertically opp. angles), \(\alpha\) = \(\beta\) (alternate angles)
< PSQ + 113º + \(\beta\) = 180º (sum of angles of a \(\Delta\))
< PSQ + 113º + 22º = 180º
< PSQ = 180º - (113 + 22)º
45º
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