Many angle relationships involve exactly a pair of angles. These relationships can be classified under the following categories:
1. Relationships based on the Measurement of Angles
2. Relationships based on the Position of Angles

Relationships based on the measurement of angles


On the basis of the measurement of the angles, the angles can be categorized into the following two categories:

Complementary Angles: Those angles for which the sum of the measures is $90^o$. That is, if $\angle$A + $\angle$B = $90^o$, each angle is the complement of the other.
Complementary Angles

Supplementary Angles: Those angles for which the sum of the measures is $180^o$. That is, if $\angle$A + $\angle$B = $180^o$, then each angle is the supplement of the other.Supplementary Angles

Congruent Angles: A pair of angles are said to be congruent angles if they have the same measure.Congruent Angles

Notation: If two angles $\angle$A and $\angle$B are congruent, we denote it by $\angle$A $\cong$ $\angle$B. It is customary to denote angles by small letters of the Greek Alphabet. That is, we use the Greek letters $\alpha$, $\beta$, $\gamma$ and so on to denote angles. Thus, if $\alpha$ and $\beta$ are congruent angles, we write $$\alpha \cong \beta$$

Relationships based on the Position of Angles


On the basis of the position of the angles, every pair of angles can be categorized into the following categories:

Adjacent Angles: Adjacent angles are a pair of angles which share a common vertex and side, but have no common interior points.Adjacent Angles.

Complementary angles and supplementary angles can be adjacent angles or non-adjacent angles.
Vertical Angles: Vertical angles are two non-adjacent angles formed by two intersecting lines.

In the figure, $\angle$1 and $\angle$2 are a pair of vertical angles. Another pair of vertical angles is $\angle$3 and $\angle$4.

Linear Pair of Angles: A linear pair is a pair of adjacent angles whose non-common sides are opposite rays.
Linear pair of Angles

In the figure,
$\angle$1 and $\angle$2 are vertical angles.
$\angle$3 and $\angle$4 are vertical angles.
$\angle$1 and $\angle$3 is a linear pair of angles.
$\angle$3 and $\angle$2 is a linear pair of angles.
$\angle$2 and $\angle$4 is a linear pair of angles.
$\angle$4 and $\angle$1 is a linear pair of angles.

Vertical angles are always congruent.
Thus, we have $\angle$1 $\cong$ $\angle$2 and $\angle$3 $\cong$ $\angle$4

Linear Pair Postulate: If two angles form a linear pair, then the angles are supplementary.

From the above diagram, it is clear that

m $\angle$1 + m $\angle$3 = $180^o$, m $\angle$2 + m$\angle$3 = $180^o$, m $\angle$2 + m $\angle$4 = $180^o$, and m $\angle$ 1 + m $\angle$4 = $180^o$

On the basis of the above discussion, we can find out the angles in a given diagram if the angles are expressed with unknowns.

Solved Examples

Question 1: Given that $\angle$A and $\angle$B are complementary angles, whose measures are given by

m $\angle$A = $(\frac{x}{2})^o$ and m $\angle$B = $(\frac{x}{3})^o$

Find the value of x and the given angles.
Solution:
 
Given that $\angle$A and $\angle$B are complementary angles.  Thus, we have
                 m $\angle$A + m $\angle$B = $90^o$

            $\rightarrow$ $(\frac{x}{2})^o$ + $(\frac{x}{3})^o$ = $90^o$

            $\rightarrow$ $\frac{x}{2}$ + $\frac{x}{3}$ = 90

            $\rightarrow$ $\frac{3x + 2x}{6}$ = 90

            $\rightarrow$ $\frac{5x}{6}$ = 90

            $\rightarrow$ x = 90 × $\frac{6}{5}$

            $\rightarrow$ x = 108
Thus,
    m $\angle$A = $(\frac{x}{2})^o$ = $(\frac{108}{2})^o$ = $54^o$ and m $\angle$B = $(\frac{x}{3})^o$ = $(\frac{108}{3})^o$ = $36^o$
 

Question 2: In the figure,
Solving Angle Relationships
        m $\angle$1 = (2x + 2y)$^o$
        m $\angle$2 = (4x - 2y)$^o$
        m $\angle$3 = (2x - y)$^o$

Find x and y.
Solution:
 
In the figure, $\angle$1 and $\angle$2 are vertical angles, and hence they are congruence.  Thus, we have
                 2x + 2y = 4x - 2y
            $\rightarrow$ 2x - 4y = 0
            $\rightarrow$ x - 2y = 0          (1)

Again, $\angle$1 and $\angle$3 are supplementary angles. Thus, we have
                  m $\angle$1 + m $\angle$3 = $180^o$
            $\rightarrow$(2x + 2y) + (2x - y) = 180
            $\rightarrow$4x + y = 180          (2)

From (1) and (2), we have
                 (x - 2y) + 2(4x + y) = 0 + 2(180)
            $\rightarrow$x - 2y + 8x + 2y = 360
            $\rightarrow$9x = 360
            $\rightarrow$x = $\frac{360}{9}$
            $\rightarrow$x = 40

From (2),
                 4x + y = 180
            $\rightarrow$4 (40) + y = 180
            $\rightarrow$160 + y = 180
            $\rightarrow$y = 180 - 160
            $\rightarrow$y = 20