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] |
\[\Rightarrow \] \[65{}^\circ +x{}^\circ +80{}^\circ =180{}^\circ \] |
\[\therefore \] \[x{}^\circ =180{}^\circ -145{}^\circ =35{}^\circ \] |
You need to login to perform this action.
You will be redirected in
3 sec