Question

In each of the figures given below, AB || CD. Find the value of x in each case.

Answer 1

Draw $EF\parallel AB\parallel CD$

Now, $AB\parallel EFandBEisthetransversal.$
Then,
$\angle ABE=\angle BEFAlternateInteriorAngles\Rightarrow \angle BEF={35}^0$
Again, $EF\parallel CDandDEisthetransversal.$
Then,
\angleDEF=\angleFED⇒\angleFED=65°

$\therefore x^0=\angle BEF+\angle FED={35}^0+{65}^0$
$={100}^{0}orx={100}^0$

(ii)
Draw $EO\parallel AB\parallel CD$
Then, ( \angleEOB+\angleEOD=x ^0 \)
Now, $EO\parallel ABandBOisthetransversal.$
$\therefore \angle EOB+\angle ABO={180}^{0}ConsecutiveInteriorAngles$
$\to \angle EOB+{55}^0={180}^0$

$\to \angle EOB={125}^0$
Again,$EO\parallel CDandDOisthetransversal$
$\therefore \angle EOD+\angle CDO={180}^{0}ConsecutiveInteriorAngles$
(\rightarrow \angle EOD+25^0=180^0\)
( \rightarrow \angleEOD=1556^0 \)
$\therefore x^0=\angle EOB+\angle EOD={125}^0+{155}^0={280}^0$

or, $x={280}^0$

(iii)Draw $EF\parallel AB\parallel CD$
Then, $\angle AEF+\angle CEF=x^0$
Now, $EF\parallel ABandAEisthetransversal$
$\therefore \angle AEF+\angle BAE={180}^{0}ConsecutiveInteriorAngles$
(\rightarrow \angle AEF+116^0 =180^0\)
$\to \angle AEF={64}^0$
$Again,EF\parallel CDandCEisthetransversal$
$\angle CEF+\angle ECD={180}^0.ConsecutiveInteriorAngles$
$\to \angle CEF+{124}^0={180}^0$
$\therefore \angle CEF=56°$
$\therefore ,x^0=\angle AEF+\angle CEF={64}^0+56°0=120°or,x={120}^0$