question_answer

In the adjacent figure, BA and BC are produced to meet CD and AD produced in E and F. Then \[\angle AED+\angle CFD\] is.

A)  \[70{}^\circ \]                         

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

C)  \[55{}^\circ \]   

D)        \[75{}^\circ \]                             

Correct Answer: C

Solution :

(c): Let \[\angle AED=m\] and \[\angle CFD=n\] \[\angle DAE=90{}^\circ -m\]; Also  (Exterior\[\angle \]) \[\Rightarrow \]\[90{}^\circ -m=35{}^\circ +n;\]    \[\Rightarrow \] \[55{}^\circ =m+n\]