Basic Terms and Definitions 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 $\angle COE = 250^o$.

(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,$\angle XYQ = \angle XYZ + \angle ZYQ = 64^o + 58^o = 122^o$.

(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.
$\therefore \angle SPR = \angle PQR + \angle PRQ$
$135^o = 70^o + \angle PRQ$
$\Rightarrow \angle PRQ = 135^o - 70^o$
$\Rightarrow \angle PRQ = 65^o$

(NONE) line is formed if the sum of adjacent angles around a point is $180^{\circ}$.
For $AOC$ to be a line,the sum of angles $\angle AOB + \angle BOC$ must be $180^{\circ}$.
Given $\angle AOB = 100^{\circ}$ and $\angle BOC = 82^{\circ}$,their sum is $100^{\circ} + 82^{\circ} = 182^{\circ}$.
Since $182^{\circ} \neq 180^{\circ}$,$AOC$ is not a straight line.
Similarly,for $BOD$ to be a line,the sum of angles $\angle BOC + \angle COD$ must be $180^{\circ}$.
Given $\angle BOC = 82^{\circ}$ and $\angle COD = 100^{\circ}$,their sum is $82^{\circ} + 100^{\circ} = 182^{\circ}$.
Since $182^{\circ} \neq 180^{\circ}$,$BOD$ is also not a straight line.

(C) line is a collection of points that extends infinitely in both directions. $A$ part of a line that has two distinct end points is known as a line segment. Therefore,the correct answer is a line segment.

(B) line is a straight path that extends infinitely in both directions. $A$ line segment is a part of a line with two end points. $A$ ray is a part of a line that starts at one point (the end point) and extends infinitely in one direction. Therefore,a part of a line with one end point is called a ray.

(D) An angle whose measure is exactly $180^{\circ}$ is called a straight angle. This angle represents a straight line.

(B) An angle that measures exactly $90^{\circ}$ is defined as a right angle. Therefore,the correct answer is a right angle.

(C) An angle whose measure is greater than $180^{\circ}$ and less than $360^{\circ}$ is known as a reflex angle.