In the given figure, \[AB\parallel CD.\] If \[\angle AOC=30{}^\circ \] and \[\angle OAB=100{}^\circ ,\] then \[\angle OCD\] is equal to |
A) \[130{}^\circ \]
B) \[150{}^\circ \]
C) \[80{}^\circ \]
D) \[100{}^\circ \]
Correct Answer: A
Solution :
Let \[\angle OCD=x{}^\circ \] |
Draw \[OE\parallel AB\parallel CD\] |
Now, \[AB\parallel OE\] and OA is the transversal. |
\[\therefore \] \[\angle OAB+\angle AOE=180{}^\circ \] |
\[\Rightarrow \] \[\angle BAO+\angle AOC+\angle COE=180{}^\circ \] |
\[\Rightarrow \] \[100{}^\circ +30{}^\circ +\angle COE=180{}^\circ \]\[\Rightarrow \]\[\angle COE=50{}^\circ \] |
Again, \[CD||OE\] and OC is transversal. |
\[\Rightarrow \]\[\angle OCD+\angle EOC=180{}^\circ \]\[\Rightarrow \]\[x=130{}^\circ \] |
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