A) \[{{60}^{o}},{{150}^{o}}\]
B) \[{{30}^{o}},{{150}^{o}}\]
C) \[{{60}^{o}},{{60}^{o}}\]
D) None of these
Correct Answer: A
Solution :
\[\angle BAT={{30}^{o}}\] (given) Since angle made by chord with a tangent is equal to angle made by it in alternate segment. \[\therefore \] \[\angle APB={{30}^{o}}\] Now.\[\angle APB+\angle AQB={{180}^{o}}\] [Opposite angles of cyclic quadrilateral APBQ] \[\Rightarrow \] \[{{30}^{o}}+\angle AQB={{180}^{o}}\]\[\Rightarrow \]\[\angle AQB={{150}^{o}}\] Also, \[\angle AOB={{60}^{o}}\] [Angle subtended by an arc at centre is double the angle subtended by it on remaining part of circle] \[\therefore \] \[\angle AOB={{60}^{o}},\text{ }\angle AQB={{150}^{o}}\]You need to login to perform this action.
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