A) \[60{}^\circ \]
B) \[90{}^\circ \]
C) \[120{}^\circ \]
D) \[150{}^\circ \]
Correct Answer: B
Solution :
(b): Since, tangents drawn from an external point to a circle are equal. \[\therefore \] AP = AC Thus, in \[\Delta \text{ }APO\]and \[\Delta \,ACO\] AP = AC AO = AO (Common side) OP = OC (Radius of circle) \[\therefore \] By SSS criterion of congruence, we have \[\Delta \,APO=\Delta \,ACO\] \[\angle PAO=\angle OAC\]\[\Rightarrow \]\[\angle PAC=2\text{ }\angle OAC\]. Similarly, we can prove that \[\angle CBO=\angle OBQ\] \[\Rightarrow \] \[\angle CBQ=2\angle CBO\] Since, \[XY\parallel X'Y'\text{ }\angle PAC+\angle QBC=180{}^\circ \] (Sum of interior angles on the same side of transversal is\[180{}^\circ \]) \[\therefore \] \[2\text{ }\angle CAO+2\text{ }\angle AOB\] \[=180{}^\circ \] .... (i) \[\Rightarrow \]\[\angle CAO+\angle CBO=90{}^\circ \] In \[\Delta AOB,\]\[\Rightarrow \]\[\angle CAO+\angle CBO+\angle AOB=180{}^\circ \] \[\Rightarrow \]\[\angle CAO+\angle CBO=180{}^\circ -\angle AOB\] ... (ii) \[\therefore \] From Eq. (i) and (ii), we get \[180{}^\circ -\angle AOB=90{}^\circ \] \[\therefore \] \[\angle AOB=90{}^\circ \]
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