Pairs of Angles Questions in English

(N/A) Ray $BO$ is the bisector of $\angle CBE$.
Therefore,$\angle CBO = \frac{1}{2} \angle CBE = \frac{1}{2}(180^o - y) = 90^o - \frac{y}{2}$ ......... $(1)$
Similarly,ray $CO$ is the bisector of $\angle BCD$.
Therefore,$\angle BCO = \frac{1}{2} \angle BCD = \frac{1}{2}(180^o - z) = 90^o - \frac{z}{2}$ ......... $(2)$
In $\Delta BOC$,$\angle BOC + \angle BCO + \angle CBO = 180^o$ ......... $(3)$
Substituting $(1)$ and $(2)$ in $(3)$,we get:
$\angle BOC + (90^o - \frac{z}{2}) + (90^o - \frac{y}{2}) = 180^o$
$\angle BOC + 180^o - \frac{1}{2}(y + z) = 180^o$
$\angle BOC = \frac{1}{2}(y + z)$ ......... $(4)$
In $\Delta ABC$,$x + y + z = 180^o$ (Angle sum property of a triangle).
Therefore,$y + z = 180^o - x$.
Substituting this in $(4)$:
$\angle BOC = \frac{1}{2}(180^o - x) = 90^o - \frac{x}{2} = 90^o - \frac{1}{2} \angle BAC$.

(N/A) In $\Delta PQR$,the side $QR$ is produced to $S$.
$\therefore$ Exterior $\angle PRS = \text{Sum of the interior opposite angles}$
$\Rightarrow \angle PRS = \angle P + \angle Q$
Since $QT$ and $RT$ are bisectors of $\angle PQR$ and $\angle PRS$ respectively,
$\therefore \angle TQR = \frac{1}{2} \angle PQR$ and $\angle TRS = \frac{1}{2} \angle PRS$.
Now,in $\Delta QRT$,the exterior angle $\angle TRS = \angle TQR + \angle QTR$.
Substituting the values:
$\frac{1}{2} \angle PRS = \frac{1}{2} \angle PQR + \angle QTR$
$\angle QTR = \frac{1}{2} \angle PRS - \frac{1}{2} \angle PQR$
$\angle QTR = \frac{1}{2} (\angle PRS - \angle PQR)$
Since $\angle PRS = \angle P + \angle PQR$ (exterior angle property),
$\angle QTR = \frac{1}{2} (\angle P + \angle PQR - \angle PQR)$
$\angle QTR = \frac{1}{2} \angle P$
i.e.,$\angle QTR = \frac{1}{2} \angle QPR$.

(N/A) In the given figure,$x$ and $y$ are two adjacent angles formed at point $B$.
For $ABC$ to be a straight line,the sum of the adjacent angles $x$ and $y$ must be equal to $180^{\circ}$.
This is based on the linear pair axiom,which states that if a ray stands on a line,then the sum of two adjacent angles so formed is $180^{\circ}$.
Therefore,$x + y = 180^{\circ}$.

(B) No,it is not necessary that each of these angles will be a right angle.
Two adjacent angles are defined as angles that have a common vertex,a common arm,and non-common arms on opposite sides of the common arm.
If two adjacent angles are equal,let each angle be $x$. Their sum is $2x$.
For these angles to be right angles ($90^{\circ}$ each),their sum must be $180^{\circ}$ (i.e.,they must form a linear pair).
However,adjacent angles can have any measure as long as they satisfy the definition of adjacency. For example,two adjacent angles of $45^{\circ}$ each are equal but are not right angles.

(N/A) Given: In the figure,$OD \perp OE$. $OD$ and $OE$ are the bisectors of $\angle AOC$ and $\angle BOC$ respectively.
To prove: Points $A, O$,and $B$ are collinear,i.e.,$AOB$ is a straight line.
Proof: Since $OD$ and $OE$ bisect $\angle AOC$ and $\angle BOC$ respectively,
$\angle AOC = 2 \angle DOC$ ---$(1)$
$\angle BOC = 2 \angle COE$ ---$(2)$
Adding equations $(1)$ and $(2)$,we get:
$\angle AOC + \angle BOC = 2 \angle DOC + 2 \angle COE$
$\Rightarrow \angle AOC + \angle BOC = 2(\angle DOC + \angle COE)$
$\Rightarrow \angle AOC + \angle BOC = 2 \angle DOE$
Since $OD \perp OE$,$\angle DOE = 90^{\circ}$.
$\Rightarrow \angle AOC + \angle BOC = 2 \times 90^{\circ} = 180^{\circ}$.
Since the sum of adjacent angles $\angle AOC$ and $\angle BOC$ is $180^{\circ}$,they form a linear pair. Therefore,$AOB$ is a straight line,and points $A, O$,and $B$ are collinear.

(N/A) Given: Two lines $AB$ and $CD$ intersect at point $O$.
To prove:
$(i)$ $\angle AOC = \angle BOD$
$(ii)$ $\angle AOD = \angle BOC$
Proof:
$(i)$ Since ray $OA$ stands on line $CD$,by the linear pair axiom:
$\angle AOC + \angle AOD = 180^{\circ}$ ....$(1)$
Similarly,since ray $OD$ stands on line $AB$,by the linear pair axiom:
$\angle AOD + \angle BOD = 180^{\circ}$ ....$(2)$
From equations $(1)$ and $(2)$,we get:
$\angle AOC + \angle AOD = \angle AOD + \angle BOD$
Subtracting $\angle AOD$ from both sides,we get:
$\angle AOC = \angle BOD$
Hence,proved.
$(ii)$ Since ray $OD$ stands on line $AB$,by the linear pair axiom:
$\angle AOD + \angle BOD = 180^{\circ}$ ....$(3)$
Similarly,since ray $OB$ stands on line $CD$,by the linear pair axiom:
$\angle BOD + \angle BOC = 180^{\circ}$ ....$(4)$
From equations $(3)$ and $(4)$,we get:
$\angle AOD + \angle BOD = \angle BOD + \angle BOC$
Subtracting $\angle BOD$ from both sides,we get:
$\angle AOD = \angle BOC$
Hence,proved.

(N/A) Given that $\angle ABC = 70^{\circ}$ and $\angle ACB = 60^{\circ}$.
Since $DBC$ is a straight line,$\angle ABC$ and $\angle ABD$ form a linear pair.
Therefore,$\angle ABD + \angle ABC = 180^{\circ}$.
$\angle ABD + 70^{\circ} = 180^{\circ} \implies \angle ABD = 180^{\circ} - 70^{\circ} = 110^{\circ}$.
Similarly,$\angle ACB$ and $\angle ACE$ form a linear pair on the straight line $BCE$.
Therefore,$\angle ACE + \angle ACB = 180^{\circ}$.
$\angle ACE + 60^{\circ} = 180^{\circ} \implies \angle ACE = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Thus,$\angle ABD = 110^{\circ}$ and $\angle ACE = 120^{\circ}$.

(N/A) Given that $\angle DBC = 70^{\circ}$ and $ABC$ is a straight line.
Therefore,$\angle DBA + \angle DBC = 180^{\circ}$ (Linear pair axiom).
$\angle DBA + 70^{\circ} = 180^{\circ} \implies \angle DBA = 110^{\circ}$.
Since ray $BP$ is the bisector of $\angle DBA$,we have $\angle ABP = \angle PBD = \frac{1}{2} \times \angle DBA = \frac{1}{2} \times 110^{\circ} = 55^{\circ}$.
Now,$\angle PBC = \angle PBD + \angle DBC = 55^{\circ} + 70^{\circ} = 125^{\circ}$.
Reflex $\angle PBD = 360^{\circ} - \angle PBD = 360^{\circ} - 55^{\circ} = 305^{\circ}$.

(C) Two angles are said to be complementary if their sum is $90^{\circ}$.
Given that $\angle A + \angle B = 90^{\circ}$.
Also,the ratio $\angle A : \angle B = 1 : 5$. Let $\angle A = x$ and $\angle B = 5x$.
Substituting these into the sum equation: $x + 5x = 90^{\circ}$.
$6x = 90^{\circ}$.
$x = 90^{\circ} / 6 = 15^{\circ}$.
Therefore,$\angle A = 15^{\circ}$ and $\angle B = 5 \times 15^{\circ} = 75^{\circ}$.

(A) Two angles are supplementary if their sum is $180^{\circ}$.
Given $\angle X : \angle Y = 23 : 13$,let $\angle X = 23k$ and $\angle Y = 13k$.
Since they are supplementary,$23k + 13k = 180^{\circ}$.
$36k = 180^{\circ}$.
$k = \frac{180^{\circ}}{36} = 5^{\circ}$.
Therefore,$\angle X = 23 \times 5^{\circ} = 115^{\circ}$ and $\angle Y = 13 \times 5^{\circ} = 65^{\circ}$.

(A) Two angles are complementary if their sum is $90^{\circ}$.
Therefore,$\angle P + \angle Q = 90^{\circ}$.
Substituting the given expressions: $(3x + 15^{\circ}) + (x + 7^{\circ}) = 90^{\circ}$.
Combining like terms: $4x + 22^{\circ} = 90^{\circ}$.
Subtracting $22^{\circ}$ from both sides: $4x = 68^{\circ}$.
Dividing by $4$: $x = 17^{\circ}$.
Now,calculate $\angle P$: $\angle P = 3(17^{\circ}) + 15^{\circ} = 51^{\circ} + 15^{\circ} = 66^{\circ}$.
Calculate $\angle Q$: $\angle Q = 17^{\circ} + 7^{\circ} = 24^{\circ}$.
Thus,$\angle P = 66^{\circ}$ and $\angle Q = 24^{\circ}$.

(B) Two angles are supplementary if their sum is $180^{\circ}$.
Therefore,$\angle A + \angle B = 180^{\circ}$.
Substituting the given expressions: $(3x - 10^{\circ}) + (2x + 30^{\circ}) = 180^{\circ}$.
Combining like terms: $5x + 20^{\circ} = 180^{\circ}$.
Subtracting $20^{\circ}$ from both sides: $5x = 160^{\circ}$.
Dividing by $5$: $x = 32^{\circ}$.
Now,calculate $\angle A$: $\angle A = 3(32^{\circ}) - 10^{\circ} = 96^{\circ} - 10^{\circ} = 86^{\circ}$.
Calculate $\angle B$: $\angle B = 2(32^{\circ}) + 30^{\circ} = 64^{\circ} + 30^{\circ} = 94^{\circ}$.
Thus,$\angle A = 86^{\circ}$ and $\angle B = 94^{\circ}$.

(A) Two angles are complementary if their sum is $90^{\circ}$.
Given: $\angle X + \angle Y = 90^{\circ}$ and $\angle X = 4 \angle Y$.
Substitute the value of $\angle X$ in the first equation:
$4 \angle Y + \angle Y = 90^{\circ}$
$5 \angle Y = 90^{\circ}$
$\angle Y = 90^{\circ} / 5 = 18^{\circ}$.
Now,find $\angle X$:
$\angle X = 4 \times 18^{\circ} = 72^{\circ}$.
Thus,$\angle X = 72^{\circ}$ and $\angle Y = 18^{\circ}$.

(D) Two angles are supplementary if their sum is $180^{\circ}$.
Therefore,$\angle P + \angle Q = 180^{\circ}$.
Given that $\angle P = 11 \angle Q$.
Substituting the value of $\angle P$ in the first equation:
$11 \angle Q + \angle Q = 180^{\circ}$
$12 \angle Q = 180^{\circ}$
$\angle Q = \frac{180^{\circ}}{12} = 15^{\circ}$.
Now,find $\angle P$:
$\angle P = 11 \times 15^{\circ} = 165^{\circ}$.
Thus,$\angle P = 165^{\circ}$ and $\angle Q = 15^{\circ}$.

(A) Two angles are complementary if their sum is $90^{\circ}$.
Therefore,$\angle A + \angle B = 90^{\circ}$.
Given that $\angle A = \angle B - 20^{\circ}$.
Substituting the value of $\angle A$ in the first equation:
$(\angle B - 20^{\circ}) + \angle B = 90^{\circ}$
$2\angle B - 20^{\circ} = 90^{\circ}$
$2\angle B = 110^{\circ}$
$\angle B = 55^{\circ}$.
Now,find $\angle A$:
$\angle A = 55^{\circ} - 20^{\circ} = 35^{\circ}$.
Thus,$\angle A = 35^{\circ}$ and $\angle B = 55^{\circ}$.

(A) Two angles are supplementary if their sum is $180^{\circ}$.
Given: $\angle P + \angle Q = 180^{\circ}$ (Equation $1$)
Given: $\angle P = \angle Q + 35^{\circ}$ (Equation $2$)
Substitute Equation $2$ into Equation $1$:
$(\angle Q + 35^{\circ}) + \angle Q = 180^{\circ}$
$2\angle Q + 35^{\circ} = 180^{\circ}$
$2\angle Q = 180^{\circ} - 35^{\circ}$
$2\angle Q = 145^{\circ}$
$\angle Q = 145^{\circ} / 2 = 72.5^{\circ}$
Now,find $\angle P$ using Equation $2$:
$\angle P = 72.5^{\circ} + 35^{\circ} = 107.5^{\circ}$
Thus,$\angle P = 107.5^{\circ}$ and $\angle Q = 72.5^{\circ}$.

(A) Since $\angle APC$ and $\angle BPD$ are vertically opposite angles,they must be equal.
Therefore,$3x - 20^{\circ} = 2x + 30^{\circ}$.
Subtracting $2x$ from both sides,we get $x - 20^{\circ} = 30^{\circ}$.
Adding $20^{\circ}$ to both sides,we get $x = 50^{\circ}$.
Now,substitute $x = 50^{\circ}$ into the expressions for the angles:
$\angle APC = 3(50^{\circ}) - 20^{\circ} = 150^{\circ} - 20^{\circ} = 130^{\circ}$.
$\angle BPD = 2(50^{\circ}) + 30^{\circ} = 100^{\circ} + 30^{\circ} = 130^{\circ}$.
Thus,$x = 50^{\circ}$ and both angles measure $130^{\circ}$.

(A) Since $\angle ABD$ and $\angle DBC$ are adjacent angles,their sum is equal to $\angle ABC$.
Therefore,$\angle ABD + \angle DBC = \angle ABC$.
Substituting the given values: $1.5x + 2x = 70$.
$3.5x = 70$.
$x = \frac{70}{3.5} = 20$.
Now,calculate the angles:
$\angle ABD = 1.5 \times 20^{\circ} = 30^{\circ}$.
$\angle DBC = 2 \times 20^{\circ} = 40^{\circ}$.
Thus,$x = 20$,$\angle ABD = 30^{\circ}$,and $\angle DBC = 40^{\circ}$.

(D) Let $\angle XYZ = 2\theta$. Since $YM$ is the bisector of $\angle XYZ$,we have $\angle XYM = \angle MYZ = \theta$.
Since $YN$ is the bisector of $\angle MYZ$,we have $\angle MYN = \angle NYZ = \frac{\theta}{2}$.
Given that $\angle XYN = 45^{\circ}$,we can express this as $\angle XYM + \angle MYN = 45^{\circ}$.
Substituting the values,we get $\theta + \frac{\theta}{2} = 45^{\circ}$.
This simplifies to $\frac{3\theta}{2} = 45^{\circ}$,which gives $3\theta = 90^{\circ}$,so $\theta = 30^{\circ}$.
Therefore,$\angle XYZ = 2\theta = 2(30^{\circ}) = 60^{\circ}$.

(N/A) Given that ray $BE$ is the bisector of $\angle DBC$,we have $\angle EBC = \angle EBD = 19^{\circ}$.
Therefore,$\angle DBC = \angle EBC + \angle EBD = 19^{\circ} + 19^{\circ} = 38^{\circ}$.
Since ray $BD$ is the bisector of $\angle ABC$,we have $\angle ABD = \angle DBC = 38^{\circ}$.
Thus,$\angle ABC = \angle ABD + \angle DBC = 38^{\circ} + 38^{\circ} = 76^{\circ}$.
Finally,$\angle ABE = \angle ABD + \angle DBE = 38^{\circ} + 19^{\circ} = 57^{\circ}$.

(A) Since $\angle ABC$ and $\angle ABD$ form a linear pair,their sum is $180^{\circ}$.
Let $\angle ABC = 17x$ and $\angle ABD = 13x$.
Therefore,$17x + 13x = 180^{\circ}$.
$30x = 180^{\circ}$,which gives $x = 6^{\circ}$.
Now,$\angle ABC = 17 \times 6^{\circ} = 102^{\circ}$.
And $\angle ABD = 13 \times 6^{\circ} = 78^{\circ}$.

(C) Since $\angle PQR$ and $\angle PQS$ form a linear pair,their sum is $180^{\circ}$.
$\angle PQR + \angle PQS = 180^{\circ}$ (Equation $1$).
Given that $11 \angle PQR = 9 \angle PQS$,we can write $\angle PQS = \frac{11}{9} \angle PQR$.
Substitute this into Equation $1$:
$\angle PQR + \frac{11}{9} \angle PQR = 180^{\circ}$.
$\frac{9 \angle PQR + 11 \angle PQR}{9} = 180^{\circ}$.
$\frac{20 \angle PQR}{9} = 180^{\circ}$.
$20 \angle PQR = 1620^{\circ}$.
$\angle PQR = \frac{1620^{\circ}}{20} = 81^{\circ}$.
Now,find $\angle PQS$:
$\angle PQS = 180^{\circ} - 81^{\circ} = 99^{\circ}$.
Thus,$\angle PQR = 81^{\circ}$ and $\angle PQS = 99^{\circ}$.

(N/A) Let $\angle ABC = \angle B$,$\angle ACB = \angle C$,and $\angle BAC = \angle A$.
Since $AE$ is the bisector of $\angle BAC$,we have $\angle BAE = \angle CAE = \frac{1}{2} \angle A$.
In $\Delta ABC$,the exterior angle $\angle ACD = \angle ABC + \angle BAC = \angle B + \angle A$.
In $\Delta AEC$,the exterior angle $\angle AEC$ is not directly applicable,so we use the angle sum property: $\angle AEC = 180^\circ - (\angle C + \angle CAE) = 180^\circ - (\angle C + \frac{1}{2} \angle A)$.
Alternatively,in $\Delta AEC$,$\angle AEC = \angle B + \angle BAE = \angle B + \frac{1}{2} \angle A$.
Multiplying by $2$,we get $2 \angle AEC = 2 \angle B + \angle A$.
We know $\angle ABC + \angle ACD = \angle B + (\angle B + \angle A) = 2 \angle B + \angle A$.
Thus,$\angle ABC + \angle ACD = 2 \angle AEC$.

(N/A) $(1)$ False. The sum of all angles in a triangle is $180^{\circ}$. If a triangle had two right angles $(90^{\circ} + 90^{\circ} = 180^{\circ})$,the third angle would have to be $0^{\circ}$,which is impossible for a triangle.
$(2)$ True. The complementary angle of $33^{\circ}$ is $90^{\circ} - 33^{\circ} = 57^{\circ}$. The supplementary angle of $57^{\circ}$ is $180^{\circ} - 57^{\circ} = 123^{\circ}$.

(C) Supplementary angles are two angles whose sum is $180^o$.
Given that $\angle A = \angle B$ and they are supplementary,we have $\angle A + \angle B = 180^o$.
Substituting $\angle B$ with $\angle A$,we get $\angle A + \angle A = 180^o$.
$2 \angle A = 180^o$.
$\angle A = \frac{180^o}{2} = 90^o$.

(D) Let the measure of the angle be $x^{\circ}$.
Since the sum of complementary angles is $90^{\circ}$,the complementary angle of $x^{\circ}$ is $(90 - x)^{\circ}$.
According to the problem,the angle is $20^{\circ}$ more than its complement:
$x = (90 - x) + 20$
$x = 110 - x$
$2x = 110$
$x = 55^{\circ}$
Therefore,the measure of the angle is $55^{\circ}$.

(A) Let the measure of the angle be $x^{\circ}$.
Since the sum of supplementary angles is $180^{\circ}$,the supplementary angle is $(180 - x)^{\circ}$.
According to the problem,the angle is $40^{\circ}$ less than its supplementary angle:
$x = (180 - x) - 40$
$x = 140 - x$
$2x = 140$
$x = 70^{\circ}$.
Therefore,the measure of the angle is $70^{\circ}$.

(B) Two angles are supplementary if their sum is $180^o$.
Given $\angle A : \angle B = 3 : 7$,let $\angle A = 3x$ and $\angle B = 7x$.
Since they are supplementary,$3x + 7x = 180^o$.
$10x = 180^o$,which implies $x = 18^o$.
Therefore,$\angle B = 7x = 7 \times 18^o = 126^o$.

(C) linear pair is a pair of adjacent angles formed by intersecting lines.
The sum of the measures of the angles in a linear pair is always $180^{\circ}$.
Let the measure of the first angle be $x^{\circ}$.
Let the measure of the second angle be $y^{\circ}$.
According to the definition of a linear pair,$x^{\circ} + y^{\circ} = 180^{\circ}$.
Therefore,the measure of the other angle is $y^{\circ} = 180^{\circ} - x^{\circ}$.

(D) Vertically opposite angles are always equal to each other.
Given that the two angles are $(5x + 30^{\circ})$ and $(8x - 60^{\circ})$.
Therefore,we can set up the equation: $5x + 30 = 8x - 60$.
Subtract $5x$ from both sides: $30 = 3x - 60$.
Add $60$ to both sides: $90 = 3x$.
Divide by $3$: $x = 30$.
Thus,the value of $x$ is $30$.

(A) linear pair is a pair of adjacent angles formed by intersecting lines. The sum of the measures of the angles in a linear pair is always $180^{\circ}$.
Given that one angle is $75^{\circ}$.
Let the measure of the other angle be $x$.
Therefore,$75^{\circ} + x = 180^{\circ}$.
$x = 180^{\circ} - 75^{\circ}$.
$x = 105^{\circ}$.
Thus,the measure of the other angle is $105^{\circ}$.

(B) Step $1$: Find the complementary angle of $32^{\circ}$.
The sum of two complementary angles is $90^{\circ}$.
Complementary angle = $90^{\circ} - 32^{\circ} = 58^{\circ}$.
Step $2$: Find the supplementary angle of the result obtained in Step $1$.
The sum of two supplementary angles is $180^{\circ}$.
Supplementary angle = $180^{\circ} - 58^{\circ} = 122^{\circ}$.
Therefore,the correct option is $B$.

(C) Step $1$: Find the supplementary angle of $132^{\circ}$.
Two angles are supplementary if their sum is $180^{\circ}$.
Supplementary angle $= 180^{\circ} - 132^{\circ} = 48^{\circ}$.
Step $2$: Find the complementary angle of the result obtained in Step $1$.
Two angles are complementary if their sum is $90^{\circ}$.
Complementary angle $= 90^{\circ} - 48^{\circ} = 42^{\circ}$.
Therefore,the correct answer is $42^{\circ}$.

(D) Let the acute angle be $x$.
By definition,the complementary angle of $x$ is $(90^{\circ} - x)$.
The supplementary angle of $x$ is $(180^{\circ} - x)$.
The difference between the supplementary angle and the complementary angle is:
$(180^{\circ} - x) - (90^{\circ} - x)$
$= 180^{\circ} - x - 90^{\circ} + x$
$= 180^{\circ} - 90^{\circ}$
$= 90^{\circ}$
Therefore,the difference is always $90^{\circ}$.

(C) linear pair of angles is formed when two adjacent angles are created by the intersection of two lines,such that their non-common sides form a straight line.
By the linear pair axiom,if a ray stands on a line,then the sum of two adjacent angles so formed is $180^{\circ}$.
Therefore,the sum of a linear pair of angles is always $180^{\circ}$.

(A) When two lines intersect at a point,they form two pairs of vertically opposite angles. According to the Vertically Opposite Angles Theorem,these angles are always equal to each other.

(A) Let the measure of the angle be $x^o$.
Since the sum of complementary angles is $90^o$,the measure of its complementary angle is $(90 - x)^o$.
According to the problem,the angle is $\frac{5}{4}$ times its complementary angle:
$x = \frac{5}{4}(90 - x)$
Multiply both sides by $4$:
$4x = 5(90 - x)$
$4x = 450 - 5x$
Add $5x$ to both sides:
$9x = 450$
Divide by $9$:
$x = 50$
Therefore,the measure of the angle is $50^o$.

(A) Let the measure of the angle be $x^o$.
Two angles are supplementary if their sum is $180^o$.
Therefore,the supplementary angle of $x^o$ is $(180 - x)^o$.
According to the problem,the angle is one-eighth of its supplementary angle:
$x = \frac{1}{8}(180 - x)$
Multiply both sides by $8$:
$8x = 180 - x$
Add $x$ to both sides:
$9x = 180$
Divide by $9$:
$x = 20$
Thus,the measure of the angle is $20^o$.

(A) Two angles are said to be supplementary if the sum of their measures is $180^{\circ}$.
Let the supplementary angle of $52^{\circ}$ be $x$.
According to the definition,$52^{\circ} + x = 180^{\circ}$.
Therefore,$x = 180^{\circ} - 52^{\circ}$.
$x = 128^{\circ}$.
Thus,the measure of the supplementary angle is $128^{\circ}$.

(B) Two angles are said to be complementary if the sum of their measures is $90^{\circ}$.
Let the measure of the complementary angle be $x$.
According to the definition,$x + 44^{\circ} = 90^{\circ}$.
Subtracting $44^{\circ}$ from both sides,we get $x = 90^{\circ} - 44^{\circ}$.
Therefore,$x = 46^{\circ}$.

(C) Step $1$: Find the complementary angle of $62^{\circ}$.
The sum of two complementary angles is $90^{\circ}$.
Complementary angle = $90^{\circ} - 62^{\circ} = 28^{\circ}$.
Step $2$: Find the supplementary angle of the result obtained in Step $1$.
The sum of two supplementary angles is $180^{\circ}$.
Supplementary angle = $180^{\circ} - 28^{\circ} = 152^{\circ}$.

(D) Step $1$: Find the supplementary angle of $128^{\circ}$.
The sum of two supplementary angles is $180^{\circ}$.
Supplementary angle = $180^{\circ} - 128^{\circ} = 52^{\circ}$.
Step $2$: Find the complementary angle of the result obtained in Step $1$.
The sum of two complementary angles is $90^{\circ}$.
Complementary angle = $90^{\circ} - 52^{\circ} = 38^{\circ}$.

(A) Let the measure of the acute angle be $x^{\circ}$.
The supplementary angle of $x^{\circ}$ is $(180 - x)^{\circ}$.
The complementary angle of $x^{\circ}$ is $(90 - x)^{\circ}$.
The difference between the supplementary angle and the complementary angle is:
Difference $= (180 - x)^{\circ} - (90 - x)^{\circ}$
Difference $= 180 - x - 90 + x$
Difference $= 180 - 90 = 90^{\circ}$.