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, \[~{{70}^{o}}+{{x}^{o}}+{{45}^{o}}={{180}^{o}}\] \[\therefore \] \[{{x}^{o}}={{180}^{o}}-{{115}^{o}}={{65}^{o}}\]You need to login to perform this action.
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