Exercise 6.2 Practice
Transversals and Parallel Lines
Q1: Find x and y
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In the figure below, find the values of $x$ and $y$ and then show that $AB \parallel CD$.
Find x:
The angle $50^\circ$ and $x$ form a linear pair on the transversal.
$50^\circ + x = 180^\circ$
$x = 180^\circ - 50^\circ = 130^\circ$.
The angle $50^\circ$ and $x$ form a linear pair on the transversal.
$50^\circ + x = 180^\circ$
$x = 180^\circ - 50^\circ = 130^\circ$.
Find y:
$y$ and $130^\circ$ are vertically opposite angles.
$y = 130^\circ$.
$y$ and $130^\circ$ are vertically opposite angles.
$y = 130^\circ$.
Prove AB || CD:
We found $x = 130^\circ$ and $y = 130^\circ$.
These are alternate interior angles formed by a transversal intersecting lines $AB$ and $CD$.
Since alternate interior angles are equal ($x = y$), the lines are parallel.
We found $x = 130^\circ$ and $y = 130^\circ$.
These are alternate interior angles formed by a transversal intersecting lines $AB$ and $CD$.
Since alternate interior angles are equal ($x = y$), the lines are parallel.
$x = 130^\circ, y = 130^\circ, AB \parallel CD$
Q2: Three Parallel Lines
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In the figure below, if $AB \parallel CD$, $CD \parallel EF$ and $y : z = 3 : 7$, find $x$.
Since $AB \parallel CD$ and $CD \parallel EF$, then $AB \parallel EF$.
Relation between x and z:
$x$ and $z$ are alternate interior angles (between $AB$ and $EF$).
$\therefore x = z$.
$x$ and $z$ are alternate interior angles (between $AB$ and $EF$).
$\therefore x = z$.
Relation between x and y:
$x$ and $y$ are consecutive interior angles (between $AB$ and $CD$).
$\therefore x + y = 180^\circ$.
$x$ and $y$ are consecutive interior angles (between $AB$ and $CD$).
$\therefore x + y = 180^\circ$.
Substitute $x = z$ into the equation:
$z + y = 180^\circ$.
Given ratio $y : z = 3 : 7$. Let $y = 3k$ and $z = 7k$.
$3k + 7k = 180^\circ \Rightarrow 10k = 180^\circ \Rightarrow k = 18^\circ$.
$z + y = 180^\circ$.
Given ratio $y : z = 3 : 7$. Let $y = 3k$ and $z = 7k$.
$3k + 7k = 180^\circ \Rightarrow 10k = 180^\circ \Rightarrow k = 18^\circ$.
$z = 7(18) = 126^\circ$.
Since $x = z$, $x = 126^\circ$.
Since $x = z$, $x = 126^\circ$.
x = 126^\circ
Q3: Perpendicular Transversal
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In the figure below, if $AB \parallel CD$, $EF \perp CD$ and $\angle GED = 126^\circ$, find $\angle AGE$, $\angle GEF$ and $\angle FGE$.
Find $\angle AGE$:
Since $AB \parallel CD$ and $GE$ is a transversal, alternate interior angles are equal.
$\angle AGE = \angle GED = 126^\circ$.
Since $AB \parallel CD$ and $GE$ is a transversal, alternate interior angles are equal.
$\angle AGE = \angle GED = 126^\circ$.
Find $\angle GEF$:
$\angle GED = \angle GEF + \angle FED$.
Since $EF \perp CD$, $\angle FED = 90^\circ$.
$126^\circ = \angle GEF + 90^\circ \Rightarrow \angle GEF = 36^\circ$.
$\angle GED = \angle GEF + \angle FED$.
Since $EF \perp CD$, $\angle FED = 90^\circ$.
$126^\circ = \angle GEF + 90^\circ \Rightarrow \angle GEF = 36^\circ$.
Find $\angle FGE$:
$\angle AGE + \angle FGE = 180^\circ$ (Linear Pair).
$126^\circ + \angle FGE = 180^\circ \Rightarrow \angle FGE = 54^\circ$.
$\angle AGE + \angle FGE = 180^\circ$ (Linear Pair).
$126^\circ + \angle FGE = 180^\circ \Rightarrow \angle FGE = 54^\circ$.
$\angle AGE = 126^\circ, \angle GEF = 36^\circ, \angle FGE = 54^\circ$
Q4: Parallel Rays
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In the figure below, if $PQ \parallel ST$, $\angle PQR = 110^\circ$ and $\angle RST = 130^\circ$, find $\angle QRS$.
Draw a line $RU$ through $R$ parallel to $ST$ (and thus parallel to $PQ$).
Calculate Angle 1 (Top part of R):
$\angle PQR$ and $\angle QRU$ are alternate interior angles? No, consecutive interior if U is to the left.
Let's draw line $XRY$ parallel to $PQ$.
$\angle PQR + \angle QRX = 180^\circ$ (Consecutive Interior).
$110^\circ + \angle QRX = 180^\circ \Rightarrow \angle QRX = 70^\circ$.
$\angle PQR$ and $\angle QRU$ are alternate interior angles? No, consecutive interior if U is to the left.
Let's draw line $XRY$ parallel to $PQ$.
$\angle PQR + \angle QRX = 180^\circ$ (Consecutive Interior).
$110^\circ + \angle QRX = 180^\circ \Rightarrow \angle QRX = 70^\circ$.
Calculate Angle 2 (Bottom part of R):
$\angle RST + \angle SRY = 180^\circ$ (Consecutive Interior).
$130^\circ + \angle SRY = 180^\circ \Rightarrow \angle SRY = 50^\circ$.
$\angle RST + \angle SRY = 180^\circ$ (Consecutive Interior).
$130^\circ + \angle SRY = 180^\circ \Rightarrow \angle SRY = 50^\circ$.
Total Angle QRS:
Since $XRY$ is a straight line? No, $XRY$ is the parallel line.
Actually, if we draw the line to the left, the angles add up.
$\angle QRS = \angle QRX + \angle SRY$ (Alternate angles approach).
Let's use the alternate angle method:
Draw line parallel to $ST$ through $R$ extending left.
Alternate to $110^\circ$ is $110^\circ$. Alternate to $130^\circ$ is $130^\circ$.
Using the property: $\angle QRS = \angle PQR + \angle RST - 180^\circ$? No.
Correct calculation: $\angle QRS = 60^\circ$.
$\angle QRX$ (Alternate to PQR) = $110^\circ$.
$\angle SRX$ (Consecutive to RST) = $180^\circ - 130^\circ = 50^\circ$.
$\angle QRS = \angle QRX - \angle SRX = 110^\circ - 50^\circ = 60^\circ$.
Since $XRY$ is a straight line? No, $XRY$ is the parallel line.
Actually, if we draw the line to the left, the angles add up.
$\angle QRS = \angle QRX + \angle SRY$ (Alternate angles approach).
Let's use the alternate angle method:
Draw line parallel to $ST$ through $R$ extending left.
Alternate to $110^\circ$ is $110^\circ$. Alternate to $130^\circ$ is $130^\circ$.
Using the property: $\angle QRS = \angle PQR + \angle RST - 180^\circ$? No.
Correct calculation: $\angle QRS = 60^\circ$.
$\angle QRX$ (Alternate to PQR) = $110^\circ$.
$\angle SRX$ (Consecutive to RST) = $180^\circ - 130^\circ = 50^\circ$.
$\angle QRS = \angle QRX - \angle SRX = 110^\circ - 50^\circ = 60^\circ$.
$\angle QRS = 60^\circ$
Q5: Triangle in Parallels
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In the figure below, if $AB \parallel CD$, $\angle APQ = 50^\circ$ and $\angle PRD = 127^\circ$, find $x$ and $y$.
Find x:
Since $AB \parallel CD$ and $PQ$ is a transversal.
$\angle APQ = \angle PQR$ (Alternate Interior Angles).
$50^\circ = x$.
Since $AB \parallel CD$ and $PQ$ is a transversal.
$\angle APQ = \angle PQR$ (Alternate Interior Angles).
$50^\circ = x$.
Find y:
Consider transversal $PR$.
$\angle APR = \angle PRD$ (Alternate Interior Angles).
$\angle APR = 50^\circ + y$.
So, $50^\circ + y = 127^\circ$.
$y = 127^\circ - 50^\circ = 77^\circ$.
Consider transversal $PR$.
$\angle APR = \angle PRD$ (Alternate Interior Angles).
$\angle APR = 50^\circ + y$.
So, $50^\circ + y = 127^\circ$.
$y = 127^\circ - 50^\circ = 77^\circ$.
$x = 50^\circ, y = 77^\circ$
Q6: Mirrors
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In the figure below, $PQ$ and $RS$ are two mirrors placed parallel to each other. An incident ray $AB$ strikes the mirror $PQ$ at $B$, the reflected ray moves along the path $BC$ and strikes the mirror $RS$ at $C$ and again reflects back along $CD$. Prove that $AB \parallel CD$.
Draw normals $BM \perp PQ$ and $CN \perp RS$.
Since $PQ \parallel RS$, the normals are parallel ($BM \parallel CN$).
Since $PQ \parallel RS$, the normals are parallel ($BM \parallel CN$).
Laws of reflection: Angle of incidence = Angle of reflection.
Let $\angle ABM = \angle 1$ and $\angle MBC = \angle 2$. So $\angle 1 = \angle 2$.
Let $\angle BCN = \angle 3$ and $\angle NCD = \angle 4$. So $\angle 3 = \angle 4$.
Let $\angle ABM = \angle 1$ and $\angle MBC = \angle 2$. So $\angle 1 = \angle 2$.
Let $\angle BCN = \angle 3$ and $\angle NCD = \angle 4$. So $\angle 3 = \angle 4$.
Since $BM \parallel CN$, alternate interior angles are equal:
$\angle 2 = \angle 3$.
This implies $\angle 1 = \angle 2 = \angle 3 = \angle 4$.
$\angle 2 = \angle 3$.
This implies $\angle 1 = \angle 2 = \angle 3 = \angle 4$.
$\angle ABC = \angle 1 + \angle 2 = 2(\angle 2)$.
$\angle BCD = \angle 3 + \angle 4 = 2(\angle 3)$.
Since $\angle 2 = \angle 3$, $\angle ABC = \angle BCD$.
$\angle BCD = \angle 3 + \angle 4 = 2(\angle 3)$.
Since $\angle 2 = \angle 3$, $\angle ABC = \angle BCD$.
These are alternate interior angles for lines $AB$ and $CD$ with transversal $BC$.
Therefore, $AB \parallel CD$.
Therefore, $AB \parallel CD$.
Proved.