question_answer

In the figure given below, AB is parallel to \[CD,\]\[\angle ABC=65{}^\circ ,\]\[\angle CDE=15{}^\circ \]and \[AB=AE.\] What is the value of \[\angle AEF?\]

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

B) \[35{}^\circ \]

C) \[40{}^\circ \]                          

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

Correct Answer: B

Solution :

Given that, \[\angle ABC=65{}^\circ \]and \[\angle CDE=15{}^\circ \]

Here, \[\angle ABC+\angle TCB=180{}^\circ \]    \[[\because AB||CD]\]

\[\therefore \]\[\angle TCB=180{}^\circ -65{}^\circ =115{}^\circ \]

\[\because \]       \[\angle TCB+\angle DCB=180{}^\circ \]            [linear pair]

\[\therefore \]                  \[\angle DCB=65{}^\circ \]

Now, in \[\Delta CDE,\]\[\angle CED=180{}^\circ -(\angle ECD+\angle EDC)\]

\[[\because \angle ECD=\angle BCD]\]

\[\because \]\[\angle DEC+\angle FEC=180{}^\circ \]        [linear pair]

\[\Rightarrow \]   \[\angle FEC=180{}^\circ -100{}^\circ =80{}^\circ \]

Given that,        \[AB=AE\]

i.e \[\Delta ABE\]an isosceles triangle.

\[\therefore \]      \[\angle ABE=\angle AEB=65{}^\circ \]

\[\because \]\[\angle AEB+\angle AEF+\angle FEC=180{}^\circ \][straight line]

\[\Rightarrow \]   \[65{}^\circ +x{}^\circ +80{}^\circ =180{}^\circ \]

\[\therefore \]                  \[x{}^\circ =180{}^\circ -145{}^\circ =35{}^\circ \]