We are given a triangle $ABC$ with an angle $A = 55^\circ$. We are also given that $DE$ is parallel to $AB$ and the angle $D = 40^\circ$. We need to find the value of $x$, which is the angle $E$.

1. Problem Description

We are given a triangle $ABC$ with an angle $A = 55^\circ$. We are also given that $DE$ is parallel to $AB$ and the angle $D = 40^\circ$. We need to find the value of $x$, which is the angle $E$.

2. Solution Steps

Since $DE$ is parallel to $AB$, we know that angle $DBA$ is equal to angle $BDE$ because they are alternate interior angles. Therefore, angle $BDE = 40^\circ$.

Also, since $DE$ is parallel to $AB$, angle $ABC$ is equal to angle $DEC$ because they are corresponding angles.

Let's find the angle $ABC$ using the fact that the sum of angles in a triangle is $180^\circ$. In triangle $ABC$, we have:

$A + B + C = 180^\circ$

$55^\circ + B + C = 180^\circ$

In triangle $ABD$, the sum of the angles is $180^\circ$. Since the lines are parallel, $DBE=DBA$

Since $AB \parallel DE$ then angle $DBA$ is equal to angle $BDE$, which is given as $40^\circ$.

Angle $ABC = 180^\circ - 55^\circ -40^\circ =85^\circ$. Then the angle $DEC$ is corresponding angle and equal to $ABC$, so $DEC = x= 85$.

Since the sum of the angles on the same side is equal to $180^{\circ}$.

$40 + A= 180$

Because $DE||AB$ then angle $DBA = 40$

Then the angle $ABC = angle B$ can be calculated by:

angle $B + 55 = 180$, giving $ABC = angle B = 85$

The angle ACB is $angle ACB = 180 -angle A- angle B$

The angleACB = 180-55-85=40

The angle $x = 180-40=140$

The angles $ADE$ and $DAB$ sum up to

1

8

0.

$ADB+ABC=180$

In triangle ACB

55+x+y=180

Since $DE || AB$, $\angle BDE= \angle DBA = 40$

Then $\angle DBA = 40$ and $\angle A=55$

3. Final Answer

x=40