question_answer

In the given figure, O is the centre of the circle. If \[\angle ABC={{110}^{o}},\]find\[x.\]

A) \[70{}^\circ \]     

B) \[140{}^\circ \]

C) \[60{}^\circ \]       

D) \[120{}^\circ \]

Correct Answer: B

Solution :

Take any point D on the circle another than A, B and C. Join DA and DC. Now, ABCD is a cyclic quadrilateral. \[\therefore \]\[\angle B+\angle D={{180}^{o}}\] \[\Rightarrow \]\[\angle D={{180}^{o}}-\angle B={{70}^{o}}\] Now, \[x=\angle D=2\times {{70}^{o}}={{140}^{o}}\] Hence, \[x={{140}^{o}}\]