Angles in a Circle 2

Draw line segment $\overline{AD}$ and name the resulting intersection point $O$ . We'll look at the formed triangle $\triangle OCD$ .

$\angle AOB$ is $90-x$ and is equal to $\angle COD$ because they are corresponding angles. $\angle CDO$ is $180-x^{3}$ because it is supplementary with $\angle CDE$ . Now we can add all interior angles of said triangle which forms a cubic equation .

$\small \begin{array}{ccc} \begin{aligned} 180-x^{3}+x^{2}+90-x = & \space 180 \\ - x^{3}+x^{2}-x+90 = & \space 0 \end{aligned} & \implies & x \approx 4.7608 \end{array}$

Rounded to the nearest hundredths, the answer is $\boxed{4.76}$ .