Question

Find the measure of angle CBE.

Answer 1

So... what would that angle be? namely, ABD + ABC

CBE = ABD  <--- vertical angles
thus
CBE = x + 56

and CBE + ABC = 180
or
$\bf (x+56)+(x^2+2x)=180\implies x^2+3x+56-180=0 \\\\ x^2+3x-124=0$

Answer 2

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, $x\neq \frac{-3- \sqrt{505}}{2}$

,so $x= \frac{-3+\sqrt{505}}{2}$

$\angle CBE= \frac{-3+\sqrt{505}}{2}$

Also, as $\angle ABD \,,\,\angle CBE$ are vertically opposite angles,

$\angle CBE=\angle ABD=\frac{-3+\sqrt{505}}{2}$