A) \[80{}^\circ \]
B) \[100{}^\circ \]
C) \[120{}^\circ \]
D) \[90{}^\circ \]
Correct Answer: B
Solution :
[b] From figure, sum of angles of quadrilateral \[ADOE=360{}^\circ \] \[\Rightarrow \] \[80{}^\circ +90{}^\circ +90{}^\circ +\angle DOE=360{}^\circ \] \[\Rightarrow \] \[\angle DOE=360{}^\circ -(80{}^\circ +90{}^\circ +90{}^\circ )\] \[=100{}^\circ \] \[\therefore \] \[\angle BOC=\angle DOE=100{}^\circ \] \[\therefore \] \[\angle BOC=\angle DOE=100{}^\circ \] |
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