question_answer

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.

\[\Rightarrow \]\[\angle OCD+\angle EOC=180{}^\circ \]\[\Rightarrow \]\[x=130{}^\circ \]