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Many angle relationships involve exactly a pair of angles. These relationships can be classified under the following categories: Relationships based on the measurement of anglesOn 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 AnglesOn 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 = $180^o$ |
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
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:
| 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 |  |