A) \[{{100}^{o}}\]
B) \[{{80}^{o}}\]
C) \[{{60}^{o}}\]
D) \[{{50}^{o}}\]
Correct Answer: D
Solution :
Since PA and PB are tangents. Also, tangent is \[\bot \] r to radius through point of contact, \[\therefore \] \[\angle PAO={{90}^{o}}\] and \[\angle PBO={{90}^{o}}\] in quadriiateral APBO; \[\angle APB+\angle PAO+\angle PBO+\angle AOB={{360}^{o}}\] \[{{80}^{o}}+{{90}^{o}}+{{90}^{o}}+\angle AOB={{360}^{o}}\] \[\Rightarrow \] \[\angle AOB={{100}^{o}}\] \[\Rightarrow \] \[\angle AQB=\frac{1}{2}\angle AQB=\frac{1}{2}\angle AOB={{50}^{o}}\]You need to login to perform this action.
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