Angles in Parallel Lines: Using angles in a triangle

Since this is an isosceles triangle, angle B and angle C are equal to each other.

$B=C=55$

Therefore we can calculate angle A since the sum of the angles in the triangle equals 180:

$A=180-55-55=180-110=70$

Since angle X is the vertex of angle A, they are equal, hence:

$A=X=70$

Our initial objective is to find the angle adjacent to the 94 angle.

Bearing in mind that adjacent angles are equal to 180, we can calculate the following:

$180-94=86$
Let's now observe the triangle.

Considering that the sum of the angles in a triangle is 180, we can determine the following:

$180=x+53+86$

$180=x+139$

$180-139=x$

$x=41$

Please note that according to the definition of corresponding angles, the angle $\alpha$corresponds to the angle located on line a and is also within the small triangle created in the drawing.

As we already have one angle in this triangle, we will try to find and calculate the remaining angles.

Furthermore the angle opposite to the angle 62 next to the vertex is also equal to 62 (vertex opposite angles are equal to one other)

Therefore, we can now calculate the missing angle in the small triangle created in the drawing, which is the angle

$\alpha$

$\alpha=180-54-62$

$\alpha=64$

Due to the fact that the lines are parallel, we will begin by drawing a further imaginary parallel line that crosses the 110 angle.

The angle adjacent to the angle 105 is equal to 75 (a straight angle is equal to 180 degrees) This angle is alternate with the angle that was divided using the imaginary line, therefore it is also equal to 75.

In the picture we are shown that the whole angle is equal to 110. Considering that we found only a part of it, we will indicate the second part of the angle as X since it alternates and is equal to the existing X angle.

Therefore we can say that:

$75+x=100$

$x=110-75=35$

First, let's draw another line parallel to the existing lines that will divide the given angle of 120 degrees in the following way:

Note that the line we drew creates two adjacent and straight angles, each equal to 90 degrees.

Now we can calculate the missing part of the angle known to us using the formula:

$120-90=30$

Let's write down the known data as follows:

Note that from the drawing we can see that angle alpha and the angle equal to 30 degrees are alternate angles, therefore they are equal to each other.

$\alpha=30$

Given that CE is parallel to AD, and AB equals CB

Observe angle C and notice that the alternate angles are equal to 2X

Observe angle A and notice that the alternate angles are equal to X-10

Proceed to mark this on the drawing as follows:

Notice that angle ACE which equals 2X is supplementary to angle DAC

Supplementary angles between parallel lines equal 180 degrees.

Therefore:

$2x+DAC=180$

Let's move 2X to one side whilst maintaining the sign:

$DAC=180-2x$

We can now create an equation in order to determine the value of angle CAB:

$CAB=180-2x-(x-10)$

$CAB=180-2x-x+10$

$CAB=190-3x$

Observe triangle CAB. We can calculate angle ACB according to the law that the sum of angles in a triangle equals 180 degrees:

$ACB=180-(3x-30)-(190-3x)$

$ACB=180-3x+30-190+3x$

Let's simplify 3X:

$ACB=180+30-190$

$ACB=210-190$

$ACB=20$

Proceed to write the values that we calculated on the drawing:

Note that from the given information we know that triangle ABC is isosceles, meaning AB equals BC

Therefore the base angles are also equal, meaning:

$190-3x=20$

Let's move terms accordingly whilst maintaining the sign:

$190-20=3x$

$170=3x$

Divide both sides by 3:

$\frac{170}{3}=\frac{3x}{3}$

$x=56.67$

Given that the three lines are parallel:

The 75 degree angle is an alternate angle with the one adjacent to angle X on the right side, and therefore is also equal to 75 degrees.

The 64 degree angle is an alternate angle with the one adjacent to angle X on the left side, and therefore is also equal to 64 degrees.

Now we can calculate:

$64+x+75=180$

$x=180-75-64=41$