A square has four right angles which each have \(\displaystyle 90\) degrees.  

To get the total you just multiple the measure of one by \(\displaystyle 4\) to get 

\(\displaystyle 90*4=360\).

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. 

Use the slope fomula, setting \(\displaystyle x_{1} = -2, x_{2} = 2,y_{1} = 3, y_{2} = 5\)

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{5-3}{2-(-2)} = \frac{2}{4} =\frac{1}{2}\)

Any point with a positive \(\displaystyle x\)-coordinate and a negative \(\displaystyle y\)-coordinate is located in the lower right quadrant - Quadrant IV.

A line with slope 0 has equation \(\displaystyle y = b\) for some real value \(\displaystyle b\).

The slope of the line is 

\(\displaystyle m = \frac{y_{2}- y_{1}}{x_{2}- x_{1}} = \frac{6- (-1)}{0-(-4)} = \frac{7}{4}\)

The \(\displaystyle y\)-intercept of the line has \(\displaystyle y\)-coordinate \(\displaystyle b= 6\)

The slope-intercept form can be written:

\(\displaystyle y = mx+b\)

Replace:

\(\displaystyle y = \frac{7}{4}x+6\)

The slope of the blue line, being perpendicular to the green line, is the opposite of the reciprocal of the slope of the green line. The slope of the green line can be found using the slope formula:

\(\displaystyle m = \frac{y_{2}- y_{1}}{x_{2}- x_{1}} = \frac{10-0}{0-30} = -\frac{1}{3}\)

The opposite of the reciprocal of \(\displaystyle -\frac{1}{3}\) is 3, and this is the slope of the blue line.

The \(\displaystyle x\)-coordinate of the midpoint of a line segment is the mean of the \(\displaystyle x\)-coordinates of its endpoints. Therefore, the \(\displaystyle x\)-coordinate is 

\(\displaystyle x= \frac{(a-5)+(b+8)}{2} = \frac{a+b-5+8}{2} = \frac{a+b+3}{2}\).

The \(\displaystyle y\)-coordinate of the midpoint of a line segment is the mean of the \(\displaystyle y\)-coordinates of its endpoints. Therefore, the \(\displaystyle y\)-coordinate is 

\(\displaystyle y= \frac{(a-4)+(-b)}{2} = \frac{a-b-4}{2}\).