Question

If CE is parallel to DB in the given figure, then the value of ‘x’ will be:

Answer 1

In $\Delta ABD$
$\angle ADB+{30}^{\circ }+{110}^{\circ }={180}^{\circ }$ [Angle sum property of a triangle]
$\Rightarrow \angle ADB={40}^{\circ }$
So, $\angle BDC={75}^{\circ }-{40}^{\circ }={35}^{\circ }$
$\angle ECF=\angle BDC={35}^{\circ }$
($\because$ Given:CE || DB, using corresponding angles axiom)
Now, DF is a straight line
So, ${60}^{\circ }+x+{35}^{\circ }={180}^{\circ }$
$\Rightarrow x={180}^{\circ }-{95}^{\circ }$
$\therefore x={85}^{\circ }$