A) \[{{100}^{o}}\]
B) \[{{50}^{o}}\]
C) \[{{80}^{o}}\]
D) None of these
Correct Answer: A
Solution :
In \[\Delta OAC\] \[\angle OAC+\angle OCA+\angle AOC={{180}^{o}}\] (angle sum property) \[\angle OAC+{{60}^{o}}+{{20}^{o}}={{180}^{o}}\] \[\angle OAC={{180}^{o}}-({{60}^{o}}+{{20}^{o}})={{180}^{o}}-{{80}^{o}}={{100}^{o}}\] As \[\Delta \,AOC\cong \Delta \,ODB\] \[\therefore \] \[\angle OAC=\angle OBD\] (By CPCT) So. \[\angle OBD={{100}^{o}}\]
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