Answer:\(\angle CBE=\frac{-3+\sqrt{505}}{2}\)
Step-by-step explanation:
Given :
\(\angle ABD=x+56\\\angle ABC=x^2+2x\)
To find : \(\angle CBE\)
Solution:
As \(\angle ABD\,,\,\angle ABC\) lie on a straight line ,
\(\angle ABD+\angle ABC=180^{\circ}\)
\(x+56+x^2+2x=180\\x^2+3x+56-180=0\\x^2+3x-124=0\)
We will solve this equation using quadratic formula:
For equation \(ax^2+bx+c=0\), roots are given by \(x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)
Solving equation: \(x^2+3x-124=0\)
\(x=\frac{-3\pm \sqrt{9+496}}{2}=\frac{-3\pm \sqrt{505}}{2}\)
As value of angle can not be negative, \(x
eq \frac{-3- \sqrt{505}}{2}\)
,so \(x= \frac{-3+\sqrt{505}}{2}\)
\(\angle CBE= \frac{-3+\sqrt{505}}{2}\)
Also, as \(\angle ABD \,,\,\angle CBE\) are vertically opposite angles,
\(\angle CBE=\angle ABD=\frac{-3+\sqrt{505}}{2}\)
