Options:
A) $ 30{}^\circ $
B) $ 35{}^\circ $
C) $ 40{}^\circ $
D) $ 45{}^\circ $
Show Answer
Answer:
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 $
BETA
