A) \[{{40}^{o}},{{140}^{o}}\]
B) \[{{50}^{o}},{{140}^{o}}\]
C) \[{{60}^{o}},{{120}^{o}}\]
D) \[{{70}^{o}},\text{ }{{40}^{o}}\]
Correct Answer: D
Solution :
Since, PQL is a tangent and OQ is a radius, so\[\angle OQL={{90}^{o}}\] \[\therefore \] \[\angle OQS=({{90}^{o}}-{{50}^{o}})={{40}^{o}}\] Now, \[OQ=OS\Rightarrow \angle OSQ=\angle OQS={{40}^{O}}\] Similarly, \[\angle ORS=({{90}^{o}}-{{60}^{o}})={{30}^{o}}\] And, \[OR=OS\Rightarrow \angle OSR=\angle ORS={{30}^{o}}\] \[\therefore \]\[\angle QSR=\angle OSQ+\angle OSR=({{40}^{o}}+{{30}^{o}})={{70}^{o}}\] Now, \[\angle ROQ=2\,\,\,\angle QSR={{140}^{o}}\] \[\angle ROQ+\,\,\angle ORP+\angle OQP+\angle RPQ={{360}^{o}}\] (Angle sum property of quadrilateral QORP) \[\Rightarrow \] \[{{140}^{o}}+{{90}^{o}}+{{90}^{o}}+\angle RPQ={{360}^{o}}\] \[\Rightarrow \] \[\angle RPQ={{40}^{o}}\]
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