When two parallel lines are intersected by another line, the sum of the measures of the interior angles on the same side of the line is 180°. Therefore, the sum of the angle that is labeled as 100° and angle y is 180°. As a result, angle y is 80°.

Another property of two parallel lines that are intersected by a third line is that the corresponding angles are congruent. So, the measurement of angle x is equal to the measurement of angle y, which is 80°.

Let A represent the measure, in degrees, of angle A. By definition, the sum of the measures of A and its complement is 90 degrees. We can write the following equation to determine an expression for the measure of the complement of angle A.

A + measure of complement of A = 90

Subtract A from both sides.

measure of complement of A = 90 – A

Similarly, because the sum of the measures of angle A and its supplement is 180 degrees, we can represent the measure of the supplement of A as 180 – A.

The problem states that the measure of the supplement of A is 40 degrees larger than twice the measure of the complement of A. We can write this as 2(90-A) + 40.

Next, we must set the two expressions 180 – A and 2(90 – A) + 40 equal to one another and solve for A:

180 – A = 2(90 – A) + 40

Distribute the 2:

180 - A = 180 – 2A + 40

Add 2A to both sides:

180 + A = 180 + 40

Subtract 180 from both sides:

A = 40

Therefore the measure of angle A is 40 degrees. 

The question asks us to find the sum of the measures of the supplement and complement of A. The measure of the supplement of A is 180 – A = 180 – 40 = 140 degrees. Similarly, the measure of the complement of A is 90 – 40 = 50 degrees.

The sum of these two is 140 + 50 = 190 degrees.

\(\displaystyle \dpi{100} \small \angle AEC\) & \(\displaystyle \dpi{100} \small \angle BEC\) must add up to 180 degrees. So, if \(\displaystyle \dpi{100} \small \angle AEC\) is 120, \(\displaystyle \dpi{100} \small \angle BEC\) (the supplementary angle) must equal 60, for a total of 180.

When a line intersects two parallel lines as a transversal, it always passes through both at identical angles (regardless of distance or length of arc).

Let's begin by observing the larger angle. \(\displaystyle \angle ABC\) is cut into two 10-degree angles by \(\displaystyle \overrightarrow{BD}\). This means that angles \(\displaystyle \angle ABD\) and \(\displaystyle \angle CBD\) equal 10 degrees. Next, we are told that \(\displaystyle \overrightarrow{BE}\) bisects \(\displaystyle \angle CBD\), which creates two 5-degree angles.  \(\displaystyle \angle ABE\) consists of \(\displaystyle \angle ABD\), which is 10 degrees, and \(\displaystyle \angle DBE\), which is 5 degrees. We need to add the two angles together to solve the problem.

\(\displaystyle \angle ABE=\angle ABD+\angle DBE\)

\(\displaystyle \angle ABE=10^{\circ}+5^{\circ}\)

\(\displaystyle \angle ABE=15^{\circ}\)

The question states that \(\displaystyle \overleftrightarrow{FH} \parallel \overleftrightarrow{AD}\). The alternate interior angle theorem states that if two parallel lines are cut by a transversal, then pairs of alternate interior angles are congruent; therefore, we know the following measure:

\(\displaystyle m\angle GCA = 58^\circ\)

The sum of angles of a triangle is equal to 180 degrees. The question states that \(\displaystyle \overline{BE}\perp \overleftrightarrow{GC}\); therefore we know the following measure:

\(\displaystyle m \angle BEC = 90^\circ\)

Use this information to solve for the missing angle: \(\displaystyle \angle EBC\)

\(\displaystyle 180^\circ=m\angle EBC+58^\circ+90^\circ\)

\(\displaystyle m\angle EBC=32^\circ\)

The degree measure of a straight line is 180 degrees; therefore, we can write the following equation:

\(\displaystyle 180^\circ=m\angle ABE+32^\circ\)

\(\displaystyle m\angle ABE=148^\circ\)

The measure of \(\displaystyle \angle ABE\) is 148 degrees. 

Since A and B are parallel, and the triangle is isosceles, we can use the supplementary rule for the two angles, \(\displaystyle 4x\) and \(\displaystyle 2x+30\) which will sum up to \(\displaystyle 180\). Setting up an algebraic equation for this, we get \(\displaystyle 4x+2x+30=180\). Solving for \(\displaystyle x\), we get \(\displaystyle x=25\). With this, we can get either \(\displaystyle 2(25)+30=80\) (for the smaller angle) or \(\displaystyle 4(25)=100\) (for the larger angle - must then use supplementary rule again for inner smaller angle). Either way, we find that the inner angles at the top are 80 degrees each. Since the sum of the angles within a triangle must equal 180, we can set up the equation as \(\displaystyle 80+80+y=180\)

\(\displaystyle y=20\) degrees.

A problem like this is very easy. A regular pentagon just looks like:

All of its sides are equal. Therefore, you know that \(\displaystyle 5s = 85\), where \(\displaystyle s\) is the length of one side. Solving for \(\displaystyle s\), you get \(\displaystyle s=17\).

Use the formula for perimeter to solve for the side length of the regular pentagon:

\(\displaystyle P=5s\)

Where \(\displaystyle P\) is the perimeter and \(\displaystyle s\) is the length of a side.

In this case:

\(\displaystyle 1,268=5s\)

\(\displaystyle s=\frac{1,268}{5}=253.6\)

Use the formula for perimeter to solve for the side length:

\(\displaystyle P=5s\)

\(\displaystyle 55=5s\)

\(\displaystyle s=11\)