Question

In the given figure, ABCD is a special quadrilateral in which the sum of its opposite internal angles is ${180}^{\circ }$.
What is the value of $\angle FDC$, given $\angle CBE={130}^{\circ }$?

• A. ${75}^{\circ }$
• B. ${30}^{\circ }$
• C. ${50}^{\circ }$
• D. ${130}^{\circ }$

Answer 1

The correct option is C ${50}^{\circ }$

Given, $\angle CBE={130}^{\circ }$
$\angle CBE+\angle CBA={180}^{\circ }$ [Linear pair]
$\Rightarrow \angle CBA={50}^{\circ }$

Sum of a pair of opposite interior angles in ABCD $={180}^{\circ }$
$\Rightarrow \angle ABC+\angle ADC={180}^{\circ }$
$\therefore \angle ADC={180}^{\circ }-{50}^{\circ }={130}^{\circ }$

$\angle ADC+\angle FDC={180}^{\circ }$ [Linear pair]
$\angle FDC={180}^{\circ }-{130}^{\circ }$
$\Rightarrow \angle FDC={50}^{\circ }$