Angle Relationships
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 = $180^o$
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 = $(\frac{x}{3})^o$ = $(\frac{108}{3})^o$ = $36^o$
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
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
Theorem 1 (Right Angles Congruence Theorem)
All right angles are congruent.
Given: $\angle$1 and $\angle$2 are right angles.
To Prove: $\angle 1 \cong \angle 2$
Proof:
Theorem 2 (Congruent Supplements Theorem)
If two angles are supplementary to the same angle or to congruent angles, the angles are congruent.
Proof:
Given: $\angle$1 and $\angle$3 are supplementary angles to $\angle$2.
To Prove: $\angle$1 $\cong$ $\angle$3
Proof:
Theorem 3 (Congruent Complements Theorem)
If two angles are complement to the same angle or to congruent angles, the two angles are congruent.
Given: $\angle$1 and $\angle$3 are complement to $\angle$2.
To Prove: $\angle$1 $\cong$ $\angle$3
Proof:
Theorem 4 (Vertical Angles Congruence Theorem)
Vertical Angles are equal in measure.
Construction: Let $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ be two intersecting lines in a plane, intersecting at the point P, as shown in the figure. Here, the angles $\angle$APD and $\angle$BPC are vertical angles.
To Prove:
$\angle$APD $\cong$ $\angle$BPC
All right angles are congruent.
Given: $\angle$1 and $\angle$2 are right angles.
To Prove: $\angle 1 \cong \angle 2$
Proof:
| S. No. | Statement | Reason |
|---|---|---|
| 1 | ∠1 and ∠2 are right angles. | Given |
| 2 | m ∠1 = 90o | Definition of Right Angle |
| 3 | m ∠2 = 90o | Definition of Right Angle |
| 4 | 90o = m ∠2 | (3) and by Symmetric Property of Equality |
| 5 | m ∠1 = m ∠2 | (2), (4), and by Transitive Property of Equality |
| 6 | ∠1≅∠2 | Definition of Congruent Angles |
Theorem 2 (Congruent Supplements Theorem)
If two angles are supplementary to the same angle or to congruent angles, the angles are congruent.
Proof:
Given: $\angle$1 and $\angle$3 are supplementary angles to $\angle$2.
To Prove: $\angle$1 $\cong$ $\angle$3
Proof:
| S. No. | Statement | Reason |
|---|---|---|
| 1 | ∠1 and ∠2 are supplementary angles | Given |
| 2 | m ∠1 + m ∠2 = 180o | (1), Definition of Supplementary Angels |
| 3 | ∠2 and ∠3 are supplementary angles | Given |
| 4 | m ∠3 + m ∠2=180o | (3), Definition of Supplementary Angles |
| 5 | 180o = m ∠3 + m ∠2 | (4), Symmetric Property of Equality |
| 6 | m ∠1 + m ∠2 = m∠3+m ∠2 | (2), (5), Transitive Property of Equality |
| 7 | m ∠1 = m ∠3 | (6), Subtraction Property of Equality |
| 8 | ∠1≅∠3 | Definition of Congruent Angles |
Theorem 3 (Congruent Complements Theorem)
If two angles are complement to the same angle or to congruent angles, the two angles are congruent.
Given: $\angle$1 and $\angle$3 are complement to $\angle$2.
To Prove: $\angle$1 $\cong$ $\angle$3
Proof:
| S. No. | Statement | Reason |
|---|---|---|
| 1 | ∠1 and ∠2 are complementary | Given |
| 2 | m ∠1 + m ∠2 = 90o | Definition of complementary angles |
| 3 | ∠2 and ∠3 are complementary | Given |
| 4 | m ∠2 + m ∠3 = 90o | Definition of complementary angles |
| 5 | 90o = m ∠2 + m ∠3 | (4), Symmetric Property of Equality |
| 6 | m ∠1 + m ∠2 = m ∠2 + m ∠3 | (2), (5), Transitive Property of Equality |
| 7 | m ∠1 = m ∠3 | (6), Subtraction Property of Equality |
| 8 | ∠1≅∠3 | Definition of Congruent Angles |
Theorem 4 (Vertical Angles Congruence Theorem)
Vertical Angles are equal in measure.
Construction: Let $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ be two intersecting lines in a plane, intersecting at the point P, as shown in the figure. Here, the angles $\angle$APD and $\angle$BPC are vertical angles.
To Prove:
$\angle$APD $\cong$ $\angle$BPC
| S. No. | Statement | Reason |
|---|---|---|
| 1 | ∠APD and ∠APC are a linear pair of angles | Given |
| 2 | m ∠APD+m ∠APC = 180o | (1), Linear Pair Postulate |
| 3 | ∠BPC and ∠APC are a linear pair of angles | Given |
| 4 | m ∠BPC + m ∠APC = 180o | (3), Linear Pair Postulate |
| 5 | 180o = m ∠BPC + m ∠APC | (4), Symmetric Property of Equality |
| 6 | m ∠APD + m ∠APC = m∠BPC + m ∠APC | (2), (5), Transitive Property of Equality |
| 7 | m ∠APC = m ∠BPC | (6), Subtraction Property of Equality |
| 8 | ∠APC≅∠BPC | Definition of Congruent Angles |
The interior and exterior angles in a triangle are related to one another. The following is the description of angle relationships in Triangles.
Theorem 5 (Triangle Sum Theorem)
The sum of the measures of the interior angles of a triangle is equal to 180o.
Given: Let $\Delta$ABC be a triangle.
Construction: Draw an auxiliary line $\overleftrightarrow{DE}$ which passes through B and parallel to $\overline{AC}$, as shown in the figure.
To Prove: m $\angle$1+ m $\angle$2 + m $\angle$3 = 180o
Proof:
Corollary 1: The acute angles of a right triangle are complementary.
Given: $\Delta$ABC be a right triangle, right-angled at B.
To Prove: $\angle$A and $\angle$C are complementary
Proof
Similar to the previous corollary we can prove the following two statements:
1. The measure of each angle of an equiangular triangle is 60o.
2. The measure of an exterior angle of a triangle is equal to the sum of the measures of its remote interior angles.
Theorem 5 (Triangle Sum Theorem)
The sum of the measures of the interior angles of a triangle is equal to 180o.
Given: Let $\Delta$ABC be a triangle.
Construction: Draw an auxiliary line $\overleftrightarrow{DE}$ which passes through B and parallel to $\overline{AC}$, as shown in the figure.
To Prove: m $\angle$1+ m $\angle$2 + m $\angle$3 = 180o
Proof:
| S. No. | Statement | Reason |
|---|---|---|
| 1 | ∠1≅∠4 | Alternate Interior Angles Theorem |
| 2 | m ∠1= m ∠4 | (1), Definition of Congruent Angles |
| 3 | ∠3≅∠5 | Alternate Interior Angles Theorem |
| 4 | m ∠3=m ∠5 | (3), Definition of Congruent Angles |
| 5 | m ∠1+m ∠3=m ∠4+m ∠5 | (2), (4), Addition of Equalities |
| 6 | m ∠4+m ∠2+m ∠5 = 180o | Angle Addition Postulate and Definition of Straight Angle |
| 7 | m ∠2+m ∠1+m ∠3=180o | (5), (6), Substitution |
| 8 | m ∠1+m ∠2+m ∠3=180o | (7), Arrangement |
Corollary 1: The acute angles of a right triangle are complementary.
Given: $\Delta$ABC be a right triangle, right-angled at B.
To Prove: $\angle$A and $\angle$C are complementary
Proof
| S. No. | Statement | Reason |
|---|---|---|
| 1 | ∠B is a right angle | Given |
| 2 | m ∠B= 90o | Definition of Right Angle |
| 3 | m ∠A+m ∠B+m ∠C=180o | Triangle Sum Theorem |
| 4 | m ∠A+90o+m ∠C=180o | (2), (3), Substitution |
| 5 | m ∠A+m ∠C=90o | (4), Substitution Property of Equality |
| 6 | ∠A and ∠C are complementary | Definition of Complementary Angles |
Similar to the previous corollary we can prove the following two statements:
1. The measure of each angle of an equiangular triangle is 60o.
2. The measure of an exterior angle of a triangle is equal to the sum of the measures of its remote interior angles.
| Vertex of Angle | Measure of Angle | Diagram |
| On the Circle | Measure of Angle = Half of the measure of intercepted arc m$\angle$ = $\frac{1}{2}$m$\angle$ACB |  |