question_answer

In the given figure, line RT is drawn parallel to \[SQ.\text{ }\,\text{If}~\,\angle QPS=100{}^\circ ,~\,\angle PQS=40{}^\circ ,\,\,\angle PSR=85{}^\circ \text{ }and\,\,\angle QRS~=70{}^\circ ,\text{ }then~\,\angle QRT\] is

A) \[45{}^\circ \]

B)        \[65{}^\circ \]  

C) \[85{}^\circ \]

D)        \[90{}^\circ \]

Correct Answer: B

Solution :

From \[\Delta PSQ\] \[\angle PSQ={{180}^{o}}-({{100}^{o}}+{{40}^{o}})={{40}^{o}}\] \[\therefore \]  \[\Delta QSR={{85}^{o}}-\angle PSQ\]                 \[={{85}^{o}}-{{40}^{o}}={{45}^{o}}\] Since \[RT||SQ,\] therefore                 \[\angle TRX={{45}^{o}}\] Now, \[~{{70}^{o}}+{{x}^{o}}+{{45}^{o}}={{180}^{o}}\] \[\therefore \]  \[{{x}^{o}}={{180}^{o}}-{{115}^{o}}={{65}^{o}}\]