Textbook - Lines and Angles Questions in English

(N/A) Since $PQ$ is a straight line,the sum of angles forming a linear pair is $180^o$.
$\angle POR + \angle ROQ = 180^o$ (Linear pair axiom)
Given that $\angle POR : \angle ROQ = 5 : 7$.
Let $\angle POR = 5x$ and $\angle ROQ = 7x$.
Substituting these into the equation:
$5x + 7x = 180^o$
$12x = 180^o$
$x = 15^o$
Therefore,$\angle POR = 5 \times 15^o = 75^o$ and $\angle ROQ = 7 \times 15^o = 105^o$.
Since $PQ$ and $RS$ are intersecting lines,vertically opposite angles are equal:
$\angle POS = \angle ROQ = 105^o$
$\angle SOQ = \angle POR = 75^o$
Thus,the angles are $75^o, 105^o, 75^o,$ and $105^o$.

(B) Ray $OS$ stands on the line $POQ$.
Therefore,$\angle POS + \angle SOQ = 180^o$ (Linear pair axiom).
Given that $\angle POS = x$.
Therefore,$x + \angle SOQ = 180^o$,which implies $\angle SOQ = 180^o - x$.
Now,ray $OR$ bisects $\angle POS$,therefore $\angle ROS = \frac{1}{2} \times \angle POS = \frac{x}{2}$.
Similarly,ray $OT$ bisects $\angle SOQ$,therefore $\angle SOT = \frac{1}{2} \times \angle SOQ = \frac{1}{2} \times (180^o - x) = 90^o - \frac{x}{2}$.
Now,$\angle ROT = \angle ROS + \angle SOT$.
Substituting the values,$\angle ROT = \frac{x}{2} + 90^o - \frac{x}{2} = 90^o$.

(N/A) To prove this,we need to produce any of the rays $OP$,$OQ$,$OR$,or $OS$ backwards to a point. Let us produce ray $OQ$ backwards to a point $T$ such that $TOQ$ is a straight line.
Now,ray $OP$ stands on line $TOQ$.
Therefore,$\angle TOP + \angle POQ = 180^o$ ........ $(1)$ (Linear pair axiom)
Similarly,ray $OS$ stands on line $TOQ$.
Therefore,$\angle TOS + \angle SOQ = 180^o$ ........ $(2)$
But,$\angle SOQ = \angle SOR + \angle QOR$.
Substituting this into $(2)$,we get:
$\angle TOS + \angle SOR + \angle QOR = 180^o$ ........ $(3)$
Now,adding $(1)$ and $(3)$,we get:
$\angle TOP + \angle POQ + \angle TOS + \angle SOR + \angle QOR = 360^o$ ........ $(4)$
Since $\angle TOP + \angle TOS = \angle POS$,equation $(4)$ becomes:
$\angle POQ + \angle QOR + \angle SOR + \angle POS = 360^o$.

(D) Since $AB$ is a straight line,the sum of angles on one side of the line at point $O$ is $180^o$.
$\therefore \angle AOC + \angle COE + \angle BOE = 180^o$
Rearranging the terms,we get:
$(\angle AOC + \angle BOE) + \angle COE = 180^o$
Given that $\angle AOC + \angle BOE = 70^o$,substituting this value:
$70^o + \angle COE = 180^o$
$\angle COE = 180^o - 70^o = 110^o$
Reflex $\angle COE = 360^o - \angle COE = 360^o - 110^o = 250^o$
Since lines $AB$ and $CD$ intersect at $O$,vertically opposite angles are equal:
$\angle AOC = \angle BOD$
Given $\angle BOD = 40^o$,therefore $\angle AOC = 40^o$.
Using the given equation $\angle AOC + \angle BOE = 70^o$:
$40^o + \angle BOE = 70^o$
$\angle BOE = 70^o - 40^o = 30^o$
Thus,$\angle BOE = 30^o$ and reflex $\angle COE = 250^o$.

(A) $XOY$ is a straight line.
Therefore,$\angle b + \angle a + \angle POY = 180^o$ (Linear pair axiom).
Given that $\angle POY = 90^o$,so $\angle b + \angle a + 90^o = 180^o$,which implies $\angle b + \angle a = 90^o$.
Given $a: b = 2: 3$,let $a = 2x$ and $b = 3x$.
Then $2x + 3x = 90^o$,so $5x = 90^o$,which gives $x = 18^o$.
Thus,$\angle a = 2 \times 18^o = 36^o$ and $\angle b = 3 \times 18^o = 54^o$.
Since lines $XY$ and $MN$ intersect at $O$,$\angle XON$ and $\angle MOY$ are vertically opposite angles,and $\angle XOM$ and $\angle NOY$ are vertically opposite angles.
Here,$\angle c = \angle XOM = \angle a + \angle POY$ is not correct based on the figure. Looking at the figure,$\angle c = \angle XON$.
Since $XOY$ is a straight line,$\angle XON + \angle NOY = 180^o$.
Also,$\angle NOY = \angle b = 54^o$ (vertically opposite angles).
Therefore,$\angle c = 180^o - 54^o = 126^o$.

(N/A) Since $ST$ is a straight line,
$\therefore \angle PQS + \angle PQR = 180^\circ$ ........... $(1)$
Similarly,$\angle PRT + \angle PRQ = 180^\circ$ ........... $(2)$
From $(1)$ and $(2)$,we have
$\angle PQS + \angle PQR = \angle PRT + \angle PRQ$
But $\angle PQR = \angle PRQ$ [Given]
$\therefore \angle PQS = \angle PRT$

(N/A) The sum of all the angles around a point is $360^\circ$.
Therefore,$x+y+z+w=360^\circ$.
We are given that $x+y=w+z$.
Substituting $(w+z)$ with $(x+y)$,we get:
$(x+y)+(x+y)=360^\circ$
$2(x+y)=360^\circ$
$x+y = \frac{360^\circ}{2} = 180^\circ$.
Since the sum of the adjacent angles $x$ and $y$ is $180^\circ$,they form a linear pair.
Therefore,$AOB$ is a straight line.

(N/A) $POQ$ is a straight line. [Given]
$\therefore \angle POS + \angle ROS + \angle ROQ = 180^o$
But $OR \perp PQ$,therefore $\angle ROQ = 90^o$.
Substituting $\angle ROQ = 90^o$ in the equation:
$\angle POS + \angle ROS + 90^o = 180^o$
$\Rightarrow \angle POS + \angle ROS = 90^o$ --- $(1)$
Now,we have $\angle QOS = \angle ROQ + \angle ROS$
Since $\angle ROQ = 90^o$,we have:
$\angle QOS = 90^o + \angle ROS$
$\Rightarrow 90^o = \angle QOS - \angle ROS$ --- $(2)$
From $(1)$ and $(2)$,we have:
$\angle POS + \angle ROS = \angle QOS - \angle ROS$
$\Rightarrow \angle ROS + \angle ROS = \angle QOS - \angle POS$
$\Rightarrow 2 \angle ROS = \angle QOS - \angle POS$
$\therefore \angle ROS = \frac{1}{2}(\angle QOS - \angle POS)$

(A) $XYP$ is a straight line.
Therefore,$\angle XYZ + \angle ZYQ + \angle QYP = 180^o$.
Since $64^o + \angle ZYQ + \angle QYP = 180^o$ and $YQ$ bisects $\angle ZYP$,we have $\angle ZYQ = \angle QYP$.
Thus,$64^o + 2 \angle QYP = 180^o$.
$2 \angle QYP = 180^o - 64^o = 116^o$.
$\angle QYP = 58^o$.
Reflex $\angle QYP = 360^o - 58^o = 302^o$.
Now,$\angle XYQ = \angle XYZ + \angle ZYQ = 64^o + 58^o = 122^o$.

(B) To solve this,we draw a line $AB$ passing through point $M$ such that $AB \parallel PQ$. Since $PQ \parallel RS$ and $AB \parallel PQ$,it follows that $AB \parallel RS$.
Now,consider the transversal $XM$ intersecting parallel lines $PQ$ and $AB$. The interior angles on the same side of the transversal are supplementary.
Therefore,$\angle QXM + \angle XMB = 180^o$.
Given $\angle QXM = 135^o$,we have $135^o + \angle XMB = 180^o$,which gives $\angle XMB = 180^o - 135^o = 45^o$ (Equation $1$).
Next,consider the transversal $YM$ intersecting parallel lines $RS$ and $AB$. The alternate interior angles are equal.
Therefore,$\angle BMY = \angle MYR$.
Given $\angle MYR = 40^o$,we have $\angle BMY = 40^o$ (Equation $2$).
Finally,$\angle XMY = \angle XMB + \angle BMY$.
Substituting the values from Equation $1$ and Equation $2$,we get $\angle XMY = 45^o + 40^o = 85^o$.

(N/A) In the figure,a transversal $AD$ intersects two lines $PQ$ and $RS$ at points $B$ and $C$ respectively. Ray $BE$ is the bisector of $\angle ABQ$ and ray $CG$ is the bisector of $\angle BCS$; and $BE \parallel CG$.
We are to prove that $PQ \parallel RS$.
It is given that ray $BE$ is the bisector of $\angle ABQ$.
Therefore,$\angle ABE = \frac{1}{2} \angle ABQ$ ...... $(1)$
Similarly,ray $CG$ is the bisector of $\angle BCS$.
Therefore,$\angle BCG = \frac{1}{2} \angle BCS$ ...... $(2)$
But $BE \parallel CG$ and $AD$ is the transversal.
Therefore,$\angle ABE = \angle BCG$ (Corresponding angles axiom) ...... $(3)$
Substituting $(1)$ and $(2)$ in $(3)$,you get
$\frac{1}{2} \angle ABQ = \frac{1}{2} \angle BCS$
That is,$\angle ABQ = \angle BCS$
But,they are the corresponding angles formed by transversal $AD$ with $PQ$ and $RS$; and are equal.
Therefore,$PQ \parallel RS$ (Converse of corresponding angles axiom).

(D) Given that $AB \parallel CD$ and $CD \parallel EF$,therefore $AB \parallel EF$.
Since $CD \parallel EF$ and $ED$ is a transversal,the interior angles on the same side of the transversal are supplementary.
Thus,$y + 55^o = 180^o$.
$y = 180^o - 55^o = 125^o$.
Since $AB \parallel CD$ and $BE$ is a transversal,the corresponding angles are equal.
Thus,$x = y = 125^o$.
Since $EA \perp AB$ and $AB \parallel EF$,then $EA \perp EF$. Therefore,$\angle AEF = 90^o$.
From the figure,$\angle AEF = z + 55^o$.
$z + 55^o = 90^o$.
$z = 90^o - 55^o = 35^o$.
Therefore,$x = 125^o, y = 125^o, z = 35^o$.

(N/A) From the figure,the line $PQ$ intersects the line $CD$ at point $F$.
Therefore,$y = 130^o$ (Vertically opposite angles) .......... $(1)$
Now,consider the line $AB$. The angle $x$ and the angle $50^o$ form a linear pair at the intersection point $E$.
Therefore,$x + 50^o = 180^o$ (Linear pair axiom)
$x = 180^o - 50^o = 130^o$ .......... $(2)$
From $(1)$ and $(2)$,we have $x = y = 130^o$.
Since these are alternate interior angles and they are equal,the lines $AB$ and $CD$ must be parallel.
Therefore,$AB \parallel CD$.

(B) Given: $AB \parallel CD$ and $CD \parallel EF$.
Since lines parallel to the same line are parallel to each other,we have $AB \parallel EF$.
Let the transversal intersect the lines $AB$,$CD$,and $EF$ at points $P$,$Q$,and $R$ respectively.
Since $AB \parallel EF$,the interior alternate angles are equal. Therefore,$\angle x = \angle z$ .......... $(1)$
Again,$AB \parallel CD$,so the sum of interior angles on the same side of the transversal is $180^o$.
Thus,$y + z = 180^o$.
Given $y: z = 3: 7$,let $y = 3k$ and $z = 7k$.
Then,$3k + 7k = 180^o \Rightarrow 10k = 180^o \Rightarrow k = 18^o$.
So,$z = 7 \times 18^o = 126^o$.
From $(1)$,$x = z = 126^o$.

(N/A) Given that $AB \parallel CD$,$EF \perp CD$,and $\angle GED = 126^o$.
$1$. Since $AB \parallel CD$ and $GE$ is a transversal,the alternate interior angles are equal.
Therefore,$\angle AGE = \angle GED$.
Given $\angle GED = 126^o$,so $\angle AGE = 126^o$.
$2$. We know that $\angle GED = \angle GEF + \angle FED$.
Since $EF \perp CD$,$\angle FED = 90^o$.
Therefore,$126^o = \angle GEF + 90^o$.
$\angle GEF = 126^o - 90^o = 36^o$.
$3$. Since $AB \parallel CD$ and $GE$ is a transversal,the sum of interior angles on the same side of the transversal is $180^o$.
Therefore,$\angle FGE + \angle GED = 180^o$.
$\angle FGE + 126^o = 180^o$.
$\angle FGE = 180^o - 126^o = 54^o$.
Thus,$\angle AGE = 126^o$,$\angle GEF = 36^o$,and $\angle FGE = 54^o$.

(D) Given: $PQ \parallel ST$,$\angle PQR = 110^o$,and $\angle RST = 130^o$.
Construction: Draw a line $EF$ parallel to $ST$ passing through point $R$.
Since $PQ \parallel ST$ and $EF \parallel ST$,it follows that $PQ \parallel EF$.
Since $PQ \parallel EF$ and $QR$ is a transversal,the sum of interior angles on the same side of the transversal is $180^o$ (or by alternate interior angles if extended). Alternatively,using alternate interior angles: $\angle PQR + \angle QRF = 180^o$ is incorrect here; rather,extend $PQ$ to meet $EF$ or use the property that $\angle PQR + \angle QRF = 180^o$ is not applicable directly. Let's use: $\angle PQR + \angle QRF = 180^o$ (consecutive interior angles) is wrong. Correct approach: $\angle PQR + \angle QRE = 180^o$ (consecutive interior angles) $\Rightarrow 110^o + \angle QRE = 180^o \Rightarrow \angle QRE = 70^o$.
Similarly,for $ST \parallel EF$ and transversal $RS$,$\angle RST + \angle SRE = 180^o$ (consecutive interior angles) $\Rightarrow 130^o + \angle SRE = 180^o \Rightarrow \angle SRE = 50^o$.
Now,$\angle QRS = \angle QRE + \angle SRE = 70^o + 50^o = 120^o$ is incorrect based on the figure. Let's re-evaluate: $\angle PQR = 110^o$. Draw $EF \parallel PQ$. Then $\angle PQR + \angle QRE = 180^o \Rightarrow \angle QRE = 70^o$. Since $PQ \parallel ST$ and $EF \parallel PQ$,then $EF \parallel ST$. Thus $\angle RST + \angle SRE = 180^o \Rightarrow 130^o + \angle SRE = 180^o \Rightarrow \angle SRE = 50^o$. From the figure,$\angle QRS = 360^o - (\angle QRE + \angle SRE)$ is not correct. Actually,$\angle QRS = 180^o - (70^o + 50^o) = 60^o$.

(A) Given: $AB \parallel CD$,$\angle APQ = 50^o$,and $\angle PRD = 127^o$.
$1$. Since $AB \parallel CD$ and $PQ$ is a transversal,the alternate interior angles are equal.
Therefore,$\angle PQR = \angle APQ$.
Given $\angle APQ = 50^o$,so $x = 50^o$.
$2$. Since $AB \parallel CD$ and $PR$ is a transversal,the alternate interior angles are equal.
Therefore,$\angle APR = \angle PRD$.
Given $\angle PRD = 127^o$,so $\angle APR = 127^o$.
We know that $\angle APR = \angle APQ + \angle QPR$.
Substituting the values,$127^o = 50^o + y$.
$y = 127^o - 50^o = 77^o$.
Thus,$x = 50^o$ and $y = 77^o$.

(N/A) $(i)$ Perpendiculars to parallel lines are parallel.
$(ii)$ According to the laws of reflection,the angle of incidence $=$ the angle of reflection.
Draw ray $BL \perp PQ$ and $CM \perp RS$.
$\because PQ \parallel RS$ [Given]
$\therefore BL \parallel CM$ [$\because BL \perp PQ$ and $CM \perp RS$]
And $BC$ is a transversal.
$\therefore \angle LBC = \angle MCB$ [Interior alternate angles] ......... $(1)$
Since,(Angle of incidence) $=$ (Angle of reflection)
$\therefore \angle ABL = \angle LBC$ and $\angle MCB = \angle MCD$
$\Rightarrow \angle ABL = \angle MCD$
$\therefore$ From $(1)$,we have
$\angle LBC + \angle ABL = \angle MCB + \angle MCD$ [Equals are added to equals]
$\Rightarrow \angle ABC = \angle BCD$
i.e.,the angles of a pair of interior alternate angles are equal.
$\therefore AB \parallel CD$.

(C) In $\Delta TQR$,since $QT \perp PR$,$\angle QTR = 90^o$.
By the angle sum property of a triangle,the sum of angles in $\Delta TQR$ is $180^o$:
$\angle QTR + \angle TQR + \angle TRQ = 180^o$
$90^o + 40^o + x = 180^o$
$130^o + x = 180^o$
$x = 180^o - 130^o = 50^o$
Now,consider $\Delta P S R$. The exterior angle $y$ of $\Delta P S R$ is equal to the sum of the two interior opposite angles:
$y = \angle SPR + \angle PRS$
$y = 30^o + x$
$y = 30^o + 50^o = 80^o$
Thus,$x = 50^o$ and $y = 80^o$.

(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 $(4)$:
$\angle BOC = \frac{1}{2}(180^o - x) = 90^o - \frac{x}{2} = 90^o - \frac{1}{2} \angle BAC$.

(A) Since $TQR$ is a straight line,
$\therefore \angle TQP + \angle PQR = 180^o$ [Linear pair axiom]
$\Rightarrow 110^o + \angle PQR = 180^o$
$\Rightarrow \angle PQR = 180^o - 110^o = 70^o$
Now,for $\Delta PQR$,the side $QP$ is produced to $S$.
Therefore,$\angle SPR$ is an exterior angle of $\Delta PQR$.
By the exterior angle property,the exterior angle is equal to the sum of the two interior opposite angles.
$\therefore \angle SPR = \angle PQR + \angle PRQ$
$135^o = 70^o + \angle PRQ$
$\Rightarrow \angle PRQ = 135^o - 70^o$
$\Rightarrow \angle PRQ = 65^o$

(B) In $\Delta XYZ$,the sum of angles is $180^o$.
$\angle XYZ + \angle YZX + \angle ZXY = 180^o$
Given $\angle XYZ = 54^o$ and $\angle ZXY = 62^o$.
$54^o + \angle YZX + 62^o = 180^o$
$\angle YZX = 180^o - 116^o = 64^o$.
Since $YO$ and $ZO$ are bisectors of $\angle XYZ$ and $\angle XZY$ respectively:
$\angle OZY = \frac{1}{2} \angle XZY = \frac{1}{2} (64^o) = 32^o$.
$\angle OYZ = \frac{1}{2} \angle XYZ = \frac{1}{2} (54^o) = 27^o$.
In $\Delta OYZ$,the sum of angles is $180^o$:
$\angle YOZ + \angle OYZ + \angle OZY = 180^o$
$\angle YOZ + 27^o + 32^o = 180^o$
$\angle YOZ = 180^o - 59^o = 121^o$.
Thus,$\angle OZY = 32^o$ and $\angle YOZ = 121^o$.

(C) Given that $AB \parallel DE$ and $AE$ acts as a transversal.
Since $AB \parallel DE$,the alternate interior angles are equal.
Therefore,$\angle AED = \angle BAC = 35^o$.
Now,consider the triangle $\Delta CDE$.
The sum of the angles in a triangle is $180^o$.
So,$\angle CDE + \angle DEC + \angle DCE = 180^o$.
Substituting the known values,we get $53^o + 35^o + \angle DCE = 180^o$.
$88^o + \angle DCE = 180^o$.
$\angle DCE = 180^o - 88^o = 92^o$.

(D) In $\Delta PRT$,
$\angle P + \angle R + \angle PTR = 180^\circ$ [By the angle sum property of a triangle]
$\Rightarrow 95^\circ + 40^\circ + \angle PTR = 180^\circ$
$\Rightarrow \angle PTR = 180^\circ - 135^\circ = 45^\circ$
Since lines $PQ$ and $RS$ intersect at $T$,the vertically opposite angles are equal.
$\therefore \angle QTS = \angle PTR = 45^\circ$
Now,in $\Delta TQS$,
$\angle TSQ + \angle STQ + \angle SQT = 180^\circ$ [By the angle sum property of a triangle]
$\Rightarrow 75^\circ + 45^\circ + \angle SQT = 180^\circ$
$\Rightarrow 120^\circ + \angle SQT = 180^\circ$
$\Rightarrow \angle SQT = 180^\circ - 120^\circ = 60^\circ$
Thus,$\angle SQT = 60^\circ$.

(A) In $\Delta QRS$,the side $SR$ is produced to $T$.
Therefore,the exterior angle $\angle QRT = \angle RQS + \angle RSQ$ (Exterior angle property).
Given $\angle RQS = 28^o$ and $\angle QRT = 65^o$.
So,$28^o + \angle RSQ = 65^o$.
$\angle RSQ = 65^o - 28^o = 37^o$.
Since $PQ \parallel SR$ and $QS$ is a transversal,the alternate interior angles are equal.
Therefore,$\angle PQS = \angle RSQ$.
$x = 37^o$.
Given $PQ \perp PS$,so $\angle SPQ = 90^o$.
In $\Delta PQS$,by the angle sum property,$\angle SPQ + \angle PQS + \angle PSQ = 180^o$.
$90^o + x + y = 180^o$.
$90^o + 37^o + y = 180^o$.
$127^o + y = 180^o$.
$y = 180^o - 127^o = 53^o$.
Thus,$x = 37^o$ and $y = 53^o$.

(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.,$\angle QTR = \frac{1}{2} \angle QPR$.