question_answer

PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If\[\angle QPR={{67}^{o}}\]and \[\angle SPR={{72}^{o}},\]then \[\angle QRS=\_\_\_\_.\]

A) \[{{41}^{o}}\]

B)        \[{{23}^{o}}\]

C)        \[{{67}^{o}}\]

D)        \[{{18}^{o}}\]

Correct Answer: A

Solution :

\[\angle QPS=\angle QPR+\angle SPR\] \[={{67}^{o}}+{{72}^{o}}={{139}^{o}}\] PQRS is a cyclic quadrilateral. \[\therefore \]\[\angle QRS+\angle QPS={{180}^{o}}\] \[\Rightarrow \]\[\angle QRS={{180}^{o}}-{{139}^{o}}={{41}^{o}}\]