Question

If CE is parallel to DB in the given figure, then the value of ‘x’ is ${30}^{\circ }$ .

• A. True
• B. False

Answer 1

The correct option is B False

In $\Delta ABD$,
$\angle ADB+{30}^{\circ }+{110}^{\circ }={180}^{\circ }$
$\Rightarrow \angle ADB={40}^{\circ }$
So, $\angle BDC={75}^{\circ }-{40}^{\circ }={35}^{\circ }$
$\angle ECF=\angle BDC={35}^{\circ }$ ($\because$ Given:CE || DB)
Now, DF is a straight line.
So, ${60}^{\circ }+x+{35}^{\circ }={180}^{\circ }$
$\Rightarrow x={180}^{\circ }-{95}^{\circ }$
$\therefore x={85}^{\circ }$
Hence, the statement is false.