question_answer

A circle with centre O touches the sides of a quadrilateral ABCD at P, Q, R, S respectively. Find\[\angle AOD+\angle BOC.\].

A) \[{{90}^{o}}\]                     

B)        \[{{180}^{o}}\]                   

C)        \[{{160}^{o}}\]                   

D)        \[{{100}^{o}}\]                   

Correct Answer: B

Solution :

A circle with centre O touches the sides AB, BC, CD and DA of a quadrilateral ABCD at the point P, Q, R and S respectively.             Join OP, OQ, OR, OS. Since, the two tangents drawn from an external point to a circle subtend equal angles at the centre.                         \[\therefore \]\[\angle 1=\angle 2,\,\angle 3=\angle 4,\,\angle 5=\angle 6,\]and \[\angle 7=\angle 8,\] Now, \[\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\] \[\angle 8={{360}^{o}}\][Sum of all the angles subtended at a point is\[{{360}^{o}}\]]                           \[\Rightarrow \] \[2[\angle 1+\angle 4+\angle 5=\angle 8]={{360}^{o}}\] \[\Rightarrow \] \[(\angle 1+\angle 8)+(\angle 4+\angle 5]={{180}^{o}}\] \[\Rightarrow \]  \[\angle AOD+\angle BOC={{180}^{o}}\]