question_answer

In the given figure, ABCD is a trapezium. EF is parallel to AD and BC. Then, \[\angle y\]is equal to [CDS 2012]

A) \[30{}^\circ \]                          

B) \[45{}^\circ \]

C) \[60{}^\circ \]  

D) \[65{}^\circ \]

Correct Answer: C

Solution :

From figure,

\[x{}^\circ =z{}^\circ =50{}^\circ \]

[alternate interior angles]

\[\theta +z{}^\circ =180{}^\circ \]                    [linear pair]

\[\theta =180{}^\circ -50{}^\circ =130{}^\circ \]

Now, in quadrilateral AQFD,

\[x{}^\circ +y{}^\circ +120{}^\circ +\theta =360{}^\circ \]

[\[\therefore \]the sum of all angles in a quadrilateral is equal to \[360{}^\circ \]]

\[\Rightarrow \]\[50{}^\circ +y{}^\circ +120{}^\circ +130{}^\circ =360{}^\circ \]

\[\therefore \]\[y=360{}^\circ -300{}^\circ =60{}^\circ \]