A) \[{{10}^{o}}\]
B) \[{{20}^{\text{o}}}\]
C) \[{{30}^{o}}\]
D) \[{{40}^{o}}\]
Correct Answer: B
Solution :
\[PO=OQ=BO=OC=r\] \[{{140}^{o}}+26={{180}^{o}}\][linear pair] \[\therefore \] \[\theta ={{20}^{o}}\] Now, in \[\Delta PBO\] \[2\alpha +\theta ={{180}^{o}}\] \[\therefore \] \[\alpha ={{80}^{o}}\] \[\angle APO={{100}^{o}}\] (Exterior angle) Similarly, \[\angle AQO={{100}^{o}}\] In quadrilateral APOQ, \[{{100}^{o}}+{{100}^{o}}+{{140}^{o}}+\angle PAQ={{360}^{o}}\] \[\therefore \] \[\angle PAQ={{20}^{o}}\]You need to login to perform this action.
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