question_answer

In the given figure, O is the centre and SAT is a tangent to the circle at A. If\[\angle BAT=30{}^\circ ,\] find \[\angle AOB\] and \[\angle AQB\].                        

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]                      \[\therefore \] \[\angle AOB={{60}^{o}},\text{ }\angle AQB={{150}^{o}}\]