A) \[{{60}^{o}}\]
B) \[{{30}^{o}}\]
C) \[{{150}^{o}}\]
D) \[{{100}^{o}}\]
Correct Answer: A
Solution :
Since, \[RT||UQ\]and QR is a transversal. \[\angle UQR=\angle QRT\] (Alternate angles) \[\Rightarrow \] \[y={{90}^{o}}\] Also, \[\angle PQU+{{150}^{o}}={{180}^{o}}\](Linear pair) \[\Rightarrow \] \[\angle PQU={{180}^{o}}-{{150}^{o}}={{30}^{o}}\] And \[PQ||RS\]and RQ is transversal. \[\angle PQR=\angle QRS~~\] (Alternate angles) \[\Rightarrow \] \[\angle PQU+\angle UQR=\angle SRT+\angle TRQ\] \[\Rightarrow \] \[{{30}^{o}}+{{90}^{o}}=x+{{90}^{o}}\Rightarrow x={{30}^{o}}\] Now, \[8x-2y\text{ =}8\times {{30}^{o}}-2\times {{90}^{o}}\] \[={{240}^{o}}-{{180}^{o}}={{60}^{o}}\]
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