question_answer

AB and CD are two parallel lines. PQ cuts AB and CD at E and F, respectively. EL is the bisector of \[\angle FEB.\] If\[\angle LEB=35{}^\circ ,\] then \[\angle CFQ\] will be

A)  \[110{}^\circ \]

B)  \[85{}^\circ \]

C)  \[70{}^\circ \]

D)  \[95{}^\circ \]

Correct Answer: A

Solution :

\[\angle LEB=35{}^\circ \] \[\angle FEB=2\times \angle LEB=70{}^\circ \] \[\angle AEB=\angle AEF+\angle BEF=180{}^\circ \] \[\Rightarrow \]   \[\angle AEF=180{}^\circ -70{}^\circ =110{}^\circ \] \[\angle CFQ=\angle AEQ=110{}^\circ \](Alternate angles)