A) \[{{30}^{o}}\]
B) \[{{15}^{o}}\]
C) \[{{90}^{o}}\]
D) \[{{60}^{o}}\]
Correct Answer: B
Solution :
\[\angle PRS=({{90}^{o}}+{{60}^{o}})={{150}^{o}}\] \[RQ=PR\] ?(i) [Sides of equilateral\[\Delta \]] and \[RQ=RS\] ?(ii) [Sides of square] \[\therefore \]\[RP=RS\] [from (i) and (ii)] \[\Rightarrow \]\[\angle RPS=\angle RSP=x\] In \[\Delta PSR,\angle PRS+\angle SPR+\angle RSP={{180}^{o}}\] \[\Rightarrow \]\[{{150}^{o}}+x+x={{180}^{o}}\] \[{{150}^{o}}+x+x={{180}^{o}}\] \[2x={{30}^{o}}\Rightarrow x={{15}^{o}}\]
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