question_answer

If 0 is the centre of the circle and PA and PB are tangents, what is the value of\[\angle APB\]?

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 \[OB\bot PB\])