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.

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.

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

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..

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.

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 = $180^o$

## How to do Angle Relationships?

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,

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

### Transversal Angles

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