A) \[{{110}^{o}}\]
B) \[{{125}^{o}}\]
C) \[{{152}^{o}}\]
D) \[{{98}^{o}}\]
Correct Answer: A
Solution :
Since, \[PQ||ST\] \[\therefore \]\[\angle PQS=\angle QST\] (Alternate angles) \[\Rightarrow \] \[\angle QST={{98}^{o}}\] \[\Rightarrow \] \[\angle QSA+\angle AST={{98}^{o}}\] \[\Rightarrow \] \[\angle AST={{98}^{o}}-{{28}^{o}}={{70}^{o}}\] Now, AB is a straight line. \[\therefore \] \[~\angle AST+~\angle TSB={{180}^{o}}\] (linear pair) \[\Rightarrow \] \[\angle TSB={{180}^{o}}-{{70}^{o}}={{110}^{o}}\]
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