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