A) \[40{}^\circ \]
B) \[60{}^\circ \]
C) \[80{}^\circ \]
D) \[100{}^\circ \]
Correct Answer: B
Solution :
[b] Quadrilateral AQBC is cyclic So, \[\angle ABC+\angle AQB=180{}^\circ \] \[\Rightarrow \] \[\angle AQB=180{}^\circ -120{}^\circ =60{}^\circ \] So, \[\angle AOB=2\times \angle AQB\] \[=20\times 60=120{}^\circ \] Now, in quadrilateral PAOB \[\angle APB+\angle PAO+\angle PBO+\angle AOB\] \[=360{}^\circ \] \[120+90{}^\circ +90+\angle APB=360{}^\circ \] \[\angle APB=360-300=60{}^\circ \] (using \[OA\bot PA\]and \[OB\bot PB\]) |
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