Mix Examples - Lines and Angles Questions in English

(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 $\angle DBC = 40^{\circ}$.

(A) Draw a line $PQ \parallel AB$,passing through point $X$ as shown in the figure.
Now,$PQ \parallel AB$ and $AB \parallel CD$.
Therefore,$CD \parallel PQ$ (since $AB \parallel CD$ and $AB \parallel PQ$).
Consider the transversal $MX$ intersecting parallel lines $AB$ and $PQ$. The interior angles on the same side of the transversal are supplementary.
$\therefore \angle BMX + \angle MXQ = 180^{\circ}$
Given $\angle BMX = 125^{\circ}$,we have:
$125^{\circ} + \angle MXQ = 180^{\circ}$
$\therefore \angle MXQ = 180^{\circ} - 125^{\circ} = 55^{\circ}$ $(1)$
Now,consider the transversal $NX$ intersecting parallel lines $CD$ and $PQ$. The alternate interior angles are equal.
$\therefore \angle NXQ = \angle XNC = 55^{\circ}$ $(2)$
Finally,$\angle MXN = \angle MXQ + \angle NXQ$.
Substituting the values from $(1)$ and $(2)$:
$\angle MXN = 55^{\circ} + 55^{\circ} = 110^{\circ}$.

$(96^{\circ})$ Given: $PQ \parallel RS$ and $RS \parallel TU$.
Since lines parallel to the same line are parallel to each other,we have $PQ \parallel TU$.
Therefore,$x = z$ (Alternate interior angles).
Now,for $PQ \parallel RS$,the interior angles on the same side of the transversal are supplementary.
Therefore,$x + y = 180^{\circ}$.
Substituting $x = z$ in the equation,we get $z + y = 180^{\circ}$.
Given the ratio $y: z = 7: 8$,let $y = 7k$ and $z = 8k$.
Substituting these into the equation: $7k + 8k = 180^{\circ} \implies 15k = 180^{\circ} \implies k = 12^{\circ}$.
Thus,$y = 7 \times 12^{\circ} = 84^{\circ}$ and $z = 8 \times 12^{\circ} = 96^{\circ}$.
Since $x = z$,we have $x = 96^{\circ}$.

(C) $1$. From the figure,the angle $x$ and the angle $70^{\circ}$ are vertically opposite angles. Therefore,$x = 70^{\circ}$.
$2$. The angle $y$ and the angle $110^{\circ}$ form a linear pair on the line $PQ$. Therefore,$y + 110^{\circ} = 180^{\circ}$,which gives $y = 180^{\circ} - 110^{\circ} = 70^{\circ}$. Wait,looking at the figure,$y$ is vertically opposite to $110^{\circ}$,so $y = 110^{\circ}$.
$3$. Alternatively,the interior angles on the same side of the transversal are $y$ and $x$. Since $y = 110^{\circ}$ and $x = 70^{\circ}$,their sum is $110^{\circ} + 70^{\circ} = 180^{\circ}$.
$4$. Since the sum of the interior angles on the same side of the transversal is $180^{\circ}$,the lines $PQ$ and $XY$ must be parallel $(PQ \parallel XY)$.

(D) Given that $PQ || RS$ and $RS || MN$. Therefore,$PQ || MN$.
From the figure,the angle $x$ and $z$ are alternate interior angles if we consider the transversal line,but looking at the positions,$x$ and $z$ are interior angles on the same side of the transversal for lines $PQ$ and $MN$,so $x + z = 180^o$.
Also,$y$ and $z$ are consecutive interior angles between lines $RS$ and $MN$,so $y + z = 180^o$.
This implies $y = x$ (as both are supplementary to $z$).
Given $y: z = 7: 11$,let $y = 7k$ and $z = 11k$.
Since $y + z = 180^o$,we have $7k + 11k = 180^o$,which gives $18k = 180^o$,so $k = 10^o$.
Thus,$y = 7 \times 10^o = 70^o$ and $z = 11 \times 10^o = 110^o$.
Since $x + z = 180^o$ (consecutive interior angles for $PQ || MN$),we have $x + 110^o = 180^o$.
Therefore,$x = 180^o - 110^o = 70^o$.
Wait,checking the figure again: $x$ and $y$ are consecutive interior angles between $PQ$ and $RS$,so $x + y = 180^o$. Since $y = 70^o$,$x = 180^o - 70^o = 110^o$.

(N/A) $1$. Given: $PQ \parallel RS$,$YZ \perp RS$ (i.e.,$\angle YZS = 90^{\circ}$),and $\angle SZX = 143^{\circ}$.
$2$. Since $RS$ is a straight line,$\angle RZX + \angle SZX = 180^{\circ}$ (Linear pair).
$\angle RZX + 143^{\circ} = 180^{\circ} \implies \angle RZX = 37^{\circ}$.
$3$. Since $PQ \parallel RS$,the alternate interior angles are equal. Thus,$\angle PXZ = \angle RZX = 37^{\circ}$.
$4$. Since $YZ \perp RS$,$\angle YZS = 90^{\circ}$. Also,$\angle XZY + \angle YZS = \angle XZS$. Since $\angle XZS$ and $\angle RZX$ are vertically opposite angles,$\angle XZS = \angle RZX = 37^{\circ}$. However,looking at the geometry,$\angle XZY = \angle YZS - \angle XZS = 90^{\circ} - 37^{\circ} = 53^{\circ}$.
$5$. In $\triangle XYZ$,the sum of angles is $180^{\circ}$. $\angle YXZ + \angle XZY + \angle XYZ = 180^{\circ}$. Since $PQ \parallel RS$ and $YZ \perp RS$,$\angle XYZ = 90^{\circ}$.
$\angle YXZ + 53^{\circ} + 90^{\circ} = 180^{\circ} \implies \angle YXZ = 180^{\circ} - 143^{\circ} = 37^{\circ}$.

(B) To find $\angle BPC$,draw a line $EF$ passing through point $P$ such that $EF \parallel AB$. Since $AB \parallel CD$ and $EF \parallel AB$,it follows that $EF \parallel CD$.
$1$. Consider the parallel lines $AB$ and $EF$. Since $BP$ is a transversal,the interior angles on the same side sum to $180^{\circ}$.
$\angle PBA + \angle BPE = 180^{\circ}$
$140^{\circ} + \angle BPE = 180^{\circ}$
$\angle BPE = 180^{\circ} - 140^{\circ} = 40^{\circ}$.
$2$. Consider the parallel lines $CD$ and $EF$. Since $CP$ is a transversal,the interior angles on the same side sum to $180^{\circ}$.
$\angle PCD + \angle CPE = 180^{\circ}$
$130^{\circ} + \angle CPE = 180^{\circ}$
$\angle CPE = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
$3$. Now,$\angle BPC = \angle BPE + \angle CPE = 40^{\circ} + 50^{\circ} = 90^{\circ}$.

(C) Given $AB \parallel CD$.
Since $AB \parallel CD$ and $XY$ is a transversal,the alternate interior angles are equal. Therefore,$\angle C Y X = \angle AXY = 80^{\circ}$.
Since $CYD$ is a straight line,$\angle C Y X + \angle X Y Z = 180^{\circ}$ (linear pair).
$80^{\circ} + x = 180^{\circ} \implies x = 100^{\circ}$.
Wait,looking at the figure,$\angle X Y Z$ is an interior angle. Since $AB \parallel CD$,$\angle AXY + \angle X Y C = 180^{\circ}$ (consecutive interior angles).
$80^{\circ} + \angle X Y C = 180^{\circ} \implies \angle X Y C = 100^{\circ}$.
Since $CYD$ is a straight line,$\angle X Y C + \angle X Y Z = 180^{\circ} \implies 100^{\circ} + x = 180^{\circ} \implies x = 80^{\circ}$.
Now,for $\angle XZD = 140^{\circ}$,since $CYD$ is a straight line,$\angle X Z Y + \angle X Z D = 180^{\circ} \implies \angle X Z Y + 140^{\circ} = 180^{\circ} \implies \angle X Z Y = 40^{\circ}$.
In $\triangle XYZ$,the sum of angles is $180^{\circ}$.
$\angle X Y Z + \angle X Z Y + \angle Y X Z = 180^{\circ} \implies 80^{\circ} + 40^{\circ} + y = 180^{\circ} \implies 120^{\circ} + y = 180^{\circ} \implies y = 60^{\circ}$.
Thus,$x = 80^{\circ}$ and $y = 60^{\circ}$.

(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 $\angle PQS = 99^{\circ}$.

(A) When two lines intersect,the vertically opposite angles are equal.
Since lines $AB$ and $CD$ intersect at $P$,$\angle APC$ and $\angle BPD$ are vertically opposite angles.
Therefore,$\angle APC = \angle BPD$.
Substituting the given expressions: $2x + 30^{\circ} = 4x - 20^{\circ}$.
Rearranging the terms: $30^{\circ} + 20^{\circ} = 4x - 2x$.
$50^{\circ} = 2x$,which gives $x = 25^{\circ}$.
Now,calculate the angles:
$\angle APC = 2(25) + 30 = 50 + 30 = 80^{\circ}$.
$\angle BPD = 4(25) - 20 = 100 - 20 = 80^{\circ}$.
Thus,$x = 25$,$\angle APC = 80^{\circ}$,and $\angle BPD = 80^{\circ}$.

(N/A) Given: $AB \parallel DE$ and $BC \parallel EF$.
To prove: $\angle ABC = \angle DEF$.
Construction: Extend $DE$ to intersect $BC$ at point $G$.
Proof:
$1$. Since $AB \parallel DE$ and $BC$ is a transversal,$\angle ABC = \angle DGC$ (Corresponding angles).
$2$. Since $BC \parallel EF$ and $DE$ is a transversal,$\angle DGC = \angle DEF$ (Corresponding angles).
$3$. From equations $(1)$ and $(2)$,we get $\angle ABC = \angle DEF$.
Hence proved.

(N/A) Given: $AB \parallel DE$ and $BC \parallel EF$.
To prove: $\angle ABC + \angle DEF = 180^{\circ}$.
Construction: Extend $ED$ to meet $BC$ at point $G$.
Proof:
Since $AB \parallel DE$ and $ED$ is extended to $G$,we have $AB \parallel DG$.
Since $AB \parallel DG$ and $BG$ is a transversal,the interior angles on the same side are supplementary,but here $AB$ and $DG$ are parallel and $BC$ is a transversal,so $\angle ABC + \angle BGD = 180^{\circ}$ (Consecutive interior angles).
Now,consider $BC \parallel EF$. Since $BC \parallel EF$ and $EG$ is a transversal,$\angle BGD = \angle DEF$ (Corresponding angles).
Substituting $\angle BGD = \angle DEF$ in the first equation,we get $\angle ABC + \angle DEF = 180^{\circ}$.
Hence proved.

(A) In $\Delta ABC$,the sum of the interior angles is $\angle A + \angle B + \angle C = 180^{\circ}$.
Given the ratio $\angle A : \angle B : \angle C = 5 : 7 : 8$,let the angles be $5x, 7x,$ and $8x$.
Sum of the angles: $5x + 7x + 8x = 180^{\circ}$.
$20x = 180^{\circ}$.
$x = \frac{180^{\circ}}{20} = 9^{\circ}$.
Therefore,the angles are:
$\angle A = 5 \times 9^{\circ} = 45^{\circ}$.
$\angle B = 7 \times 9^{\circ} = 63^{\circ}$.
$\angle C = 8 \times 9^{\circ} = 72^{\circ}$.

(N/A) In $\Delta ABC$,the sum of angles is $180^{\circ}$,so $\angle A + \angle ABC + \angle ACB = 180^{\circ}$.
$\therefore \angle ABC + \angle ACB = 180^{\circ} - \angle A \quad \dots(1)$
The bisectors of $\angle B$ and $\angle C$ intersect at $I$.
$\therefore \angle IBC = \frac{1}{2} \angle ABC$ and $\angle ICB = \frac{1}{2} \angle ACB$.
In $\Delta IBC$,the sum of angles is $180^{\circ}$,so $\angle BIC + \angle IBC + \angle ICB = 180^{\circ}$.
$\therefore \angle BIC = 180^{\circ} - (\angle IBC + \angle ICB)$.
Substituting the values of $\angle IBC$ and $\angle ICB$:
$\therefore \angle BIC = 180^{\circ} - \left[ \frac{1}{2} \angle ABC + \frac{1}{2} \angle ACB \right]$.
$\therefore \angle BIC = 180^{\circ} - \frac{1}{2} (\angle ABC + \angle ACB)$.
Using equation $(1)$:
$\therefore \angle BIC = 180^{\circ} - \frac{1}{2} (180^{\circ} - \angle A)$.
$\therefore \angle BIC = 180^{\circ} - 90^{\circ} + \frac{1}{2} \angle A$.
$\therefore \angle BIC = 90^{\circ} + \frac{1}{2} \angle A$.

(A) Let the angles of $\Delta PQR$ be $5x$,$2x$,and $2x$ respectively.
According to the angle sum property of a triangle,the sum of all interior angles is $180^{\circ}$.
Therefore,$5x + 2x + 2x = 180^{\circ}$.
$9x = 180^{\circ}$.
$x = 20^{\circ}$.
Now,calculating each angle:
$\angle P = 5x = 5 \times 20^{\circ} = 100^{\circ}$.
$\angle Q = 2x = 2 \times 20^{\circ} = 40^{\circ}$.
$\angle R = 2x = 2 \times 20^{\circ} = 40^{\circ}$.
Thus,the angles are $100^{\circ}, 40^{\circ}, 40^{\circ}$.

(A) We know that the sum of all angles in a triangle is $180^{\circ}$,so $\angle A + \angle B + \angle C = 180^{\circ}$.
Given: $\angle A + \angle B = 80^{\circ}$ and $\angle B + \angle C = 150^{\circ}$.
Substituting $\angle A + \angle B = 80^{\circ}$ into the sum equation: $80^{\circ} + \angle C = 180^{\circ} \implies \angle C = 180^{\circ} - 80^{\circ} = 100^{\circ}$.
Substituting $\angle C = 100^{\circ}$ into $\angle B + \angle C = 150^{\circ}$: $\angle B + 100^{\circ} = 150^{\circ} \implies \angle B = 50^{\circ}$.
Substituting $\angle B = 50^{\circ}$ into $\angle A + \angle B = 80^{\circ}$: $\angle A + 50^{\circ} = 80^{\circ} \implies \angle A = 30^{\circ}$.
Thus,the angles are $\angle A = 30^{\circ}, \angle B = 50^{\circ}, \angle C = 100^{\circ}$.

(A) We know that the sum of angles in a triangle is $180^{\circ}$,so $\angle A + \angle B + \angle C = 180^{\circ}$.
Given $\angle B = \frac{\angle A + \angle C}{2}$,which implies $2\angle B = \angle A + \angle C$.
Substituting this into the sum equation: $2\angle B + \angle B = 180^{\circ} \implies 3\angle B = 180^{\circ} \implies \angle B = 60^{\circ}$.
Now,$\angle A + \angle C = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Given the ratio $\angle A : \angle C = 1 : 2$,let $\angle A = x$ and $\angle C = 2x$.
Then $x + 2x = 120^{\circ} \implies 3x = 120^{\circ} \implies x = 40^{\circ}$.
Therefore,$\angle A = 40^{\circ}$ and $\angle C = 2(40^{\circ}) = 80^{\circ}$.
The angles are $\angle A = 40^{\circ}, \angle B = 60^{\circ}, \angle C = 80^{\circ}$.

(A) Since $S-Q-R-T$ is a straight line,$\angle PQS$ and $\angle PQR$ form a linear pair. Therefore,$\angle PQR = 180^{\circ} - 100^{\circ} = 80^{\circ}$.
Similarly,$\angle PRT$ and $\angle PRQ$ form a linear pair. Therefore,$\angle PRQ = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
In $\Delta PQR$,the sum of all interior angles is $180^{\circ}$.
Thus,$\angle P + \angle Q + \angle R = 180^{\circ}$.
$\angle P + 80^{\circ} + 50^{\circ} = 180^{\circ}$.
$\angle P + 130^{\circ} = 180^{\circ}$.
$\angle P = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
Hence,the angles are $\angle P = 50^{\circ}, \angle Q = 80^{\circ}, \angle R = 50^{\circ}$.

(A) The sum of the angles in a triangle is $180^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
Substituting the given values: $(2x - 10^{\circ}) + (x + 10^{\circ}) + (2x - 20^{\circ}) = 180^{\circ}$.
Combining like terms: $5x - 20^{\circ} = 180^{\circ}$.
$5x = 200^{\circ}$.
$x = 40^{\circ}$.
Now,calculate each angle:
$\angle A = 2(40^{\circ}) - 10^{\circ} = 80^{\circ} - 10^{\circ} = 70^{\circ}$.
$\angle B = 40^{\circ} + 10^{\circ} = 50^{\circ}$.
$\angle C = 2(40^{\circ}) - 20^{\circ} = 80^{\circ} - 20^{\circ} = 60^{\circ}$.
Thus,the angles are $\angle A = 70^{\circ}, \angle B = 50^{\circ}, \angle C = 60^{\circ}$.

(N/A) Let $ABCD$ be a convex quadrilateral.
Draw a diagonal $AC$ which divides the quadrilateral into two triangles,$\triangle ABC$ and $\triangle ADC$.
We know that the sum of the angles of a triangle is $180^{\circ}$.
For $\triangle ABC$,$\angle BAC + \angle ABC + \angle BCA = 180^{\circ}$ (Equation $1$).
For $\triangle ADC$,$\angle DAC + \angle ADC + \angle DCA = 180^{\circ}$ (Equation $2$).
Adding Equation $1$ and Equation $2$,we get:
$(\angle BAC + \angle DAC) + \angle ABC + \angle ADC + (\angle BCA + \angle DCA) = 180^{\circ} + 180^{\circ}$.
From the figure,$\angle BAC + \angle DAC = \angle A$ and $\angle BCA + \angle DCA = \angle C$.
Therefore,$\angle A + \angle B + \angle C + \angle D = 360^{\circ}$.
Hence,the sum of the angles of a convex quadrilateral is $360^{\circ}$.

(N/A) Let $\angle ABC = y$ and $\angle ACB = z$. The exterior angles are $\angle CBD = 180^{\circ} - y$ and $\angle BCE = 180^{\circ} - z$.
Since $BO$ and $CO$ are bisectors of $\angle CBD$ and $\angle BCE$ respectively,we have:
$\angle CBO = \frac{1}{2} \angle CBD = \frac{1}{2} (180^{\circ} - y) = 90^{\circ} - \frac{y}{2}$
$\angle BCO = \frac{1}{2} \angle BCE = \frac{1}{2} (180^{\circ} - z) = 90^{\circ} - \frac{z}{2}$
In $\Delta BOC$,the sum of angles is $180^{\circ}$:
$\angle BOC + \angle CBO + \angle BCO = 180^{\circ}$
$\angle BOC + (90^{\circ} - \frac{y}{2}) + (90^{\circ} - \frac{z}{2}) = 180^{\circ}$
$\angle BOC + 180^{\circ} - \frac{1}{2}(y + z) = 180^{\circ}$
$\angle BOC = \frac{1}{2}(y + z)$
In $\Delta ABC$,$y + z + \angle A = 180^{\circ}$,so $y + z = 180^{\circ} - \angle A$.
Substituting this into the equation for $\angle BOC$:
$\angle BOC = \frac{1}{2}(180^{\circ} - \angle A) = 90^{\circ} - \frac{1}{2} \angle A$.

(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) In $\Delta ABC$,$\angle ACD$ is an exterior angle. By the exterior angle theorem,$\angle ACD = \angle BAC + \angle ABC$.
In $\Delta EBC$,$\angle ECD$ is an exterior angle. Therefore,$\angle ECD = \angle EBC + \angle BEC$.
Since $BE$ is the bisector of $\angle ABC$,$\angle EBC = \frac{1}{2} \angle ABC$.
Since $CE$ is the bisector of $\angle ACD$,$\angle ECD = \frac{1}{2} \angle ACD = \frac{1}{2} (\angle BAC + \angle ABC)$.
Substituting these into the equation for $\Delta EBC$: $\frac{1}{2} (\angle BAC + \angle ABC) = \frac{1}{2} \angle ABC + \angle BEC$.
Simplifying,$\frac{1}{2} \angle BAC + \frac{1}{2} \angle ABC = \frac{1}{2} \angle ABC + \angle BEC$.
Thus,$\angle BEC = \frac{1}{2} \angle BAC$.

(B) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
$\angle A + \angle B + \angle C = 180^{\circ}$
$\angle A + 70^{\circ} + 80^{\circ} = 180^{\circ}$
$\angle A = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
Since $AD$ is the bisector of $\angle A$,$\angle BAD = \angle CAD = \frac{30^{\circ}}{2} = 15^{\circ}$.
In $\Delta ABD$,$\angle ADB = 180^{\circ} - (\angle B + \angle BAD) = 180^{\circ} - (70^{\circ} + 15^{\circ}) = 180^{\circ} - 85^{\circ} = 95^{\circ}$.
In $\Delta ADC$,$\angle ADC = 180^{\circ} - (\angle C + \angle CAD) = 180^{\circ} - (80^{\circ} + 15^{\circ}) = 180^{\circ} - 95^{\circ} = 85^{\circ}$.

(N/A) In $\Delta PQR$,let $\angle QPR = 2\alpha$. Since $PA$ is the angle bisector,$\angle QPA = \angle RPA = \alpha$.
In $\Delta PQR$,$\angle Q + \angle R + \angle QPR = 180^{\circ}$,so $\angle Q + \angle R + 2\alpha = 180^{\circ}$,which implies $\alpha = 90^{\circ} - \frac{1}{2}(\angle Q + \angle R)$.
In $\Delta PMQ$,$\angle PMQ = 90^{\circ}$,so $\angle QPM = 90^{\circ} - \angle Q$.
Now,$\angle APM = \angle QPA - \angle QPM$.
Substituting the values: $\angle APM = \alpha - (90^{\circ} - \angle Q)$.
$\angle APM = [90^{\circ} - \frac{1}{2}(\angle Q + \angle R)] - 90^{\circ} + \angle Q$.
$\angle APM = \angle Q - \frac{1}{2}\angle Q - \frac{1}{2}\angle R$.
$\angle APM = \frac{1}{2}\angle Q - \frac{1}{2}\angle R = \frac{1}{2}(\angle Q - \angle R)$.
Hence,proved.

(B) $(1)$ False. If a transversal intersects two parallel lines,the interior angles on the same side of the transversal are supplementary (their sum is $180^{\circ}$),not equal.
$(2)$ False. In any triangle,the sum of all three angles is $180^{\circ}$. It is possible for all three angles to be acute (e.g.,in an equilateral triangle,all angles are $60^{\circ}$),so the maximum number of acute angles in a triangle is three.

(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 $180^{\circ} - 57^{\circ} = 123^{\circ}$.

(B) The sum of all interior angles in a triangle is always $180^{\circ}$.
Given that $\angle P = 95^{\circ}$ and $\angle Q = 95^{\circ}$.
Sum of these two angles $= 95^{\circ} + 95^{\circ} = 190^{\circ}$.
Since $190^{\circ} > 180^{\circ}$,it is impossible for a triangle to have two angles each equal to $95^{\circ}$ because the sum of the angles would exceed the angle sum property of a triangle.
Therefore,the statement is False.

(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 angles of $\Delta XYZ$ be $x$,$4x$,and $4x$ respectively.
We know that the sum of the angles in a triangle is $180^{\circ}$.
Therefore,$x + 4x + 4x = 180^{\circ}$.
$9x = 180^{\circ}$.
$x = 20^{\circ}$.
Now,$\angle Z = 4x = 4 \times 20^{\circ} = 80^{\circ}$.