Exercise 6.1 Practice
Basic Terms and Definitions, Intersecting Lines
Q1: Intersecting Lines
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In the figure below, lines $AB$ and $CD$ intersect at $O$. If $\angle AOC + \angle BOE = 80^\circ$ and $\angle BOD = 35^\circ$, find $\angle BOE$ and reflex $\angle COE$.
Given: $\angle AOC + \angle BOE = 80^\circ$ and $\angle BOD = 35^\circ$.
Lines $AB$ and $CD$ intersect at $O$. Thus, vertically opposite angles are equal.
$\angle AOC = \angle BOD = 35^\circ$.
$\angle AOC = \angle BOD = 35^\circ$.
Substitute $\angle AOC$ in the given equation:
$35^\circ + \angle BOE = 80^\circ$
$\angle BOE = 80^\circ - 35^\circ = 45^\circ$.
$35^\circ + \angle BOE = 80^\circ$
$\angle BOE = 80^\circ - 35^\circ = 45^\circ$.
Now, angles on line $AB$ sum to $180^\circ$:
$\angle AOC + \angle COE + \angle BOE = 180^\circ$
$35^\circ + \angle COE + 45^\circ = 180^\circ$
$\angle COE = 180^\circ - 80^\circ = 100^\circ$.
$\angle AOC + \angle COE + \angle BOE = 180^\circ$
$35^\circ + \angle COE + 45^\circ = 180^\circ$
$\angle COE = 180^\circ - 80^\circ = 100^\circ$.
Reflex $\angle COE = 360^\circ - 100^\circ = 260^\circ$.
$\angle BOE = 45^\circ$, Reflex $\angle COE = 260^\circ$
Q2: Linear Pair Ratio
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In the figure below, lines $XY$ and $MN$ intersect at $O$. If $\angle POY = 90^\circ$ and $a : b = 2 : 3$, find $c$.
Given $\angle POY = 90^\circ$. Since $XY$ is a line, $\angle POX = 180^\circ - 90^\circ = 90^\circ$.
$\angle POX = a + b = 90^\circ$.
Given ratio $a : b = 2 : 3$. Let $a = 2x$ and $b = 3x$.
Given ratio $a : b = 2 : 3$. Let $a = 2x$ and $b = 3x$.
$2x + 3x = 90^\circ \Rightarrow 5x = 90^\circ \Rightarrow x = 18^\circ$.
$a = 2(18) = 36^\circ$, $b = 3(18) = 54^\circ$.
$a = 2(18) = 36^\circ$, $b = 3(18) = 54^\circ$.
$MN$ is a straight line. Thus, $b + c = 180^\circ$ (Linear Pair).
$54^\circ + c = 180^\circ \Rightarrow c = 180^\circ - 54^\circ = 126^\circ$.
$54^\circ + c = 180^\circ \Rightarrow c = 180^\circ - 54^\circ = 126^\circ$.
c = 126^\circ
Q3: Prove Angle Relation
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In the figure below, $POQ$ is a line. Ray $OR$ is perpendicular to line $PQ$. $OS$ is another ray lying between rays $OP$ and $OR$. Prove that $\angle ROS = \frac{1}{2}(\angle QOS - \angle POS)$.
$\angle ROS = \angle QOS - \angle QOR$
Since $OR \perp PQ$, $\angle QOR = 90^\circ$.
So, $\angle ROS = \angle QOS - 90^\circ$ ... (i)
Since $OR \perp PQ$, $\angle QOR = 90^\circ$.
So, $\angle ROS = \angle QOS - 90^\circ$ ... (i)
Also, $\angle ROS = \angle POR - \angle POS$
Since $OR \perp PQ$, $\angle POR = 90^\circ$.
So, $\angle ROS = 90^\circ - \angle POS$ ... (ii)
Since $OR \perp PQ$, $\angle POR = 90^\circ$.
So, $\angle ROS = 90^\circ - \angle POS$ ... (ii)
Adding (i) and (ii):
$2\angle ROS = (\angle QOS - 90^\circ) + (90^\circ - \angle POS)$
$2\angle ROS = \angle QOS - \angle POS$
$\angle ROS = \frac{1}{2}(\angle QOS - \angle POS)$.
$2\angle ROS = (\angle QOS - 90^\circ) + (90^\circ - \angle POS)$
$2\angle ROS = \angle QOS - \angle POS$
$\angle ROS = \frac{1}{2}(\angle QOS - \angle POS)$.
Proved.
Q4: Exterior Angles
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In the figure below, if $\angle PQR = \angle PRQ$, then prove that $\angle PQS = \angle PRT$.
Since $ST$ is a straight line, the sum of angles on it is $180^\circ$ (Linear Pair Axiom).
$\angle PQS + \angle PQR = 180^\circ$ ... (i)
$\angle PRT + \angle PRQ = 180^\circ$ ... (ii)
$\angle PRT + \angle PRQ = 180^\circ$ ... (ii)
From (i) and (ii):
$\angle PQS + \angle PQR = \angle PRT + \angle PRQ$
$\angle PQS + \angle PQR = \angle PRT + \angle PRQ$
Given that $\angle PQR = \angle PRQ$.
Subtracting these equal angles from both sides:
$\angle PQS = \angle PRT$.
Subtracting these equal angles from both sides:
$\angle PQS = \angle PRT$.
Proved.
Q5: Prove Line
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In the figure below, if $x + y = w + z$, then prove that $AOB$ is a line.
The sum of all angles around a point is $360^\circ$.
$\therefore x + y + w + z = 360^\circ$.
$\therefore x + y + w + z = 360^\circ$.
Given $x + y = w + z$. Substitute $(w + z)$ with $(x + y)$:
$(x + y) + (x + y) = 360^\circ$
$2(x + y) = 360^\circ$
$x + y = 180^\circ$.
$(x + y) + (x + y) = 360^\circ$
$2(x + y) = 360^\circ$
$x + y = 180^\circ$.
Since angles $x$ and $y$ form a linear pair (sum is $180^\circ$), the non-common arms $OA$ and $OB$ form a line.
Therefore, AOB is a line.
Q6: Angle Bisector
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It is given that $\angle XYZ = 64^\circ$ and $XY$ is produced to point $P$. Draw a figure from the given information. If ray $YQ$ bisects $\angle ZYP$, find $\angle XYQ$ and reflex $\angle QYP$.
Since $XY$ is produced to $P$, $PYX$ is a straight line.
$\angle XYZ + \angle ZYP = 180^\circ$ (Linear Pair)
$64^\circ + \angle ZYP = 180^\circ \Rightarrow \angle ZYP = 116^\circ$.
$\angle XYZ + \angle ZYP = 180^\circ$ (Linear Pair)
$64^\circ + \angle ZYP = 180^\circ \Rightarrow \angle ZYP = 116^\circ$.
Ray $YQ$ bisects $\angle ZYP$.
$\therefore \angle QYP = \angle ZYQ = \frac{116^\circ}{2} = 58^\circ$.
$\therefore \angle QYP = \angle ZYQ = \frac{116^\circ}{2} = 58^\circ$.
Find $\angle XYQ$:
$\angle XYQ = \angle XYZ + \angle ZYQ = 64^\circ + 58^\circ = 122^\circ$.
$\angle XYQ = \angle XYZ + \angle ZYQ = 64^\circ + 58^\circ = 122^\circ$.
Find Reflex $\angle QYP$:
Reflex $\angle QYP = 360^\circ - \angle QYP = 360^\circ - 58^\circ = 302^\circ$.
Reflex $\angle QYP = 360^\circ - \angle QYP = 360^\circ - 58^\circ = 302^\circ$.
$\angle XYQ = 122^\circ$, Reflex $\angle QYP = 302^\circ$