Question

In the given figure, ABCD is a cyclic quadrilateral. What is the value of $\frac{x^{\circ }}{{50}^{\circ }}$?

Answer 1

We have, $\angle CDA={180}^{\circ }-{80}^{\circ }={100}^{\circ }$
and, $\angle ABC+\angle x={180}^{\circ }$
$\angle ABC={180}^{\circ }-x^{\circ }$
Now, $\angle CDA+\angle ABC={180}^{\circ }$ [sum of opposite angles in a quadrilateral is ${180}^{\circ }$]
$\Rightarrow {100}^{\circ }+({180}^{\circ }-x^{\circ })={180}^{\circ }\Rightarrow {280}^{\circ }-x^{\circ }={180}^{\circ }$
$\Rightarrow x^{\circ }={100}^{\circ }$
$\therefore \frac{x^{\circ }}{{50}^{\circ }}=\frac{{100}^{\circ }}{{50}^{\circ }}=2^{\circ }$