GSEB Class 9 Maths Solutions Chapter 6 Lines and Angles Exercise 6.2

Question 1. In figure, find the values of x and y and then show that AB || CD.

Answer: We see that \( x + 50^\circ = 180^\circ \) because they form a linear pair.
\( \implies x = 180^\circ - 50^\circ \)
\( \implies x = 130^\circ \)
Also, \( y = 130^\circ \) because these are vertically opposite angles.
We have found that \( x = 130^\circ \) and \( y = 130^\circ \).
Therefore, \( x = y \).
Since \( x \) and \( y \) are alternate interior angles, and they are equal, the lines AB and CD must be parallel.
\( \implies AB || CD \)
In simple words: First, find x by subtracting 50 from 180 since they are on a straight line. Then, y is the same as the angle opposite to 130. Since x and y are alternate angles and are the same, the lines are parallel.

Exam Tip: Remember to clearly state the geometric reason for each step, such as "linear pair," "vertically opposite angles," or "alternate interior angles."

Question 2. In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Answer: Given that \( y : z = 3 : 7 \), we can let \( y = 3a \) and \( z = 7a \).
Since CD || EF, the angles \( y \) and \( z \) are interior angles on the same side of the transversal.
Therefore, their sum is \( 180^\circ \).
\( \implies y + z = 180^\circ \)
Substitute the values of \( y \) and \( z \):
\( \implies 3a + 7a = 180^\circ \)
\( \implies 10a = 180^\circ \)
\( \implies a = \frac {180^\circ}{10} \)
\( \implies a = 18^\circ \)
Now we find the values of \( y \) and \( z \):
\( \therefore y = 3a = 3 \times 18^\circ \)
\( \implies y = 54^\circ \)
\( z = 7a = 7 \times 18^\circ \)
\( \implies z = 126^\circ \)
We are given that AB || CD and CD || EF.
When lines are parallel to the same line, they are also parallel to each other.
Therefore, AB || EF.
Since AB || EF, angles \( x \) and \( z \) are alternate interior angles.
Thus, \( x = z \).
\( \implies x = 126^\circ \)
In simple words: Since y and z add up to 180 degrees (because lines CD and EF are parallel), and their ratio is 3:7, we can find their exact values. Once we know z, then x is also the same as z because lines AB and EF are parallel, and x and z are alternate interior angles.

Exam Tip: When dealing with multiple parallel lines, use the transitivity property of parallel lines (if a || b and b || c, then a || c). Also, clearly distinguish between alternate interior angles and interior angles on the same side of the transversal.

Question 3. In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Answer: We are given that \( AB || CD \), \( EF \perp CD \), and \( \angle GED = 126^\circ \).
Since \( AB || CD \), the alternate interior angles \( \angle AGE \) and \( \angle GED \) are equal.
\( \angle AGE = \angle GED \)
\( \implies \angle AGE = 126^\circ \)
Next, we know that \( \angle GED \) can be broken into two angles: \( \angle GEF \) and \( \angle FED \).
\( \angle GED = \angle GEF + \angle FED \)
Since \( EF \perp CD \), the angle \( \angle FED \) is \( 90^\circ \).
Substitute the known values into the equation:
\( \implies 126^\circ = \angle GEF + 90^\circ \)
\( \implies \angle GEF = 126^\circ - 90^\circ \)
\( \implies \angle GEF = 36^\circ \)
Finally, to find \( \angle FGE \), we can use the linear pair property with \( \angle AGE \).
\( \angle FGE + \angle AGE = 180^\circ \) (Angles on a straight line)
\( \implies \angle FGE + 126^\circ = 180^\circ \)
\( \implies \angle FGE = 180^\circ - 126^\circ \)
\( \implies \angle FGE = 54^\circ \)
Therefore, \( \angle AGE = 126^\circ \), \( \angle GEF = 36^\circ \), and \( \angle FGE = 54^\circ \).
In simple words: First, find angle AGE because it's the same as angle GED (alternate interior angles). Then, break down angle GED into two parts: angle GEF and angle FED. Since EF is straight up from CD, angle FED is 90 degrees, which helps find angle GEF. Lastly, angle FGE and angle AGE add up to 180 degrees because they make a straight line.

Exam Tip: Carefully identify the type of angle pair (alternate interior, corresponding, co-interior, linear pair, vertically opposite) for each calculation to ensure correct reasoning and values. Diagrams are crucial for visualization.

Question 4. In figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

Answer: First, we draw a line AB parallel to ST through point R.
Since AB || ST and RS is a transversal, the interior angles on the same side are supplementary.
\( \angle RST + \angle SRB = 180^\circ \)
\( \implies 130^\circ + \angle SRB = 180^\circ \)
\( \implies \angle SRB = 180^\circ - 130^\circ \)
\( \implies \angle SRB = 50^\circ \) ..........(1)
Next, since PQ || ST and we drew AB || ST, it follows that PQ || AB (lines parallel to the same line are parallel to each other).
With PQ || AB and QR as a transversal, the interior angles on the same side are supplementary.
\( \angle PQR + \angle QRA = 180^\circ \)
\( \implies 110^\circ + \angle QRA = 180^\circ \)
\( \implies \angle QRA = 180^\circ - 110^\circ \)
\( \implies \angle QRA = 70^\circ \) ..........(2)
Now, we look at the angles around point R on the line AB.
\( \angle QRA + \angle QRB = 180^\circ \) (Linear pair of angles on line AB)
From the figure, \( \angle QRB \) is made of \( \angle QRS \) and \( \angle SRB \).
So, \( \angle QRA + \angle QRS + \angle SRB = 180^\circ \)
Substitute the values from equations (1) and (2):
\( \implies 70^\circ + \angle QRS + 50^\circ = 180^\circ \)
\( \implies \angle QRS + 120^\circ = 180^\circ \)
\( \implies \angle QRS = 180^\circ - 120^\circ \)
\( \implies \angle QRS = 60^\circ \)
In simple words: Draw a helper line through R that's parallel to PQ and ST. Use the fact that same-side interior angles add up to 180 degrees to find the parts of angle QRS. Then add those parts together to get the full angle QRS.

Exam Tip: When auxiliary lines are helpful, remember to draw them and clearly state their properties. Breaking down a complex angle into simpler components is a common strategy in such problems.

Question 5. In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Answer: We are given that \( AB || CD \), \( \angle APQ = 50^\circ \), and \( \angle PRD = 127^\circ \).
Since \( AB || CD \) and PQ is a transversal, the alternate interior angles \( \angle APQ \) and \( \angle PQR \) are equal.
\( \angle APQ = \angle PQR \)
\( \implies 50^\circ = x \)
\( \implies x = 50^\circ \)
Next, consider transversal PR. Since \( AB || CD \), the alternate interior angles \( \angle APR \) and \( \angle PRD \) are equal.
So, \( \angle APR = \angle PRD = 127^\circ \).
We know that \( \angle APR \) is made up of \( \angle APQ \) and \( \angle QPR \).
\( \angle APQ + \angle QPR = \angle APR \)
\( \implies 50^\circ + y = 127^\circ \)
To find \( y \), subtract \( 50^\circ \) from \( 127^\circ \):
\( \implies y = 127^\circ - 50^\circ \)
\( \implies y = 77^\circ \)
Thus, the values are \( x = 50^\circ \) and \( y = 77^\circ \).
In simple words: Because lines AB and CD are parallel, angle x is the same as angle APQ (50 degrees). Also, angle APR is the same as angle PRD (127 degrees). Since angle APR is made of angle APQ and angle y, subtract angle APQ from angle APR to find y.

Exam Tip: Clearly identify the transversals and the parallel lines for each angle relationship. Alternate interior angles are equal when lines are parallel, and this is key to solving such problems efficiently.

Question 6. In figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB on 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 || CD.

Answer: Given: \( PQ || RS \), incident ray AB strikes mirror PQ at B, reflected ray BC strikes mirror RS at C, and again reflects back along CD.
To Prove: \( AB || CD \)
Construction: Draw BM perpendicular to PQ and CN perpendicular to RS.
Proof: We know that BM \( \perp \) PQ and CN \( \perp \) RS.
Since \( PQ || RS \), and BM and CN are both perpendicular to these parallel lines, then BM \( || \) CN.
Now, considering BC as a transversal for BM \( || \) CN, the alternate interior angles are equal.
\( \angle 2 = \angle 3 \) ..........(1)
According to the law of reflection, the angle of incidence is equal to the angle of reflection.
At mirror PQ, \( \angle 1 = \angle 2 \) ..........(2)
At mirror RS, \( \angle 3 = \angle 4 \) ..........(3)
From equations (1), (2), and (3), we can conclude that \( \angle 1 = \angle 4 \) ..........(4)
Now, add equation (1) and equation (4):
\( \angle 1 + \angle 2 = \angle 3 + \angle 4 \)
The sum \( \angle 1 + \angle 2 \) represents \( \angle ABC \).
The sum \( \angle 3 + \angle 4 \) represents \( \angle BCD \).
Therefore, \( \angle ABC = \angle BCD \).
Since \( \angle ABC \) and \( \angle BCD \) are alternate interior angles formed by transversals AB and CD with transversal BC, and they are equal, it implies that the lines AB and CD are parallel.
\( \implies AB || CD \) (If alternate interior angles are equal, then lines are parallel).
In simple words: We draw lines straight out from points B and C (normals). Since the mirrors are parallel, these normal lines are also parallel. Because of this, angle 2 and angle 3 are the same. Also, in reflections, the incoming angle is the same as the outgoing angle. So, angle 1 equals angle 2, and angle 3 equals angle 4. Putting all this together means angle 1 equals angle 4. If we add the angles, we find that the big angle ABC is the same as the big angle BCD. Since these are alternate interior angles, it proves that line AB is parallel to line CD.

Exam Tip: This question combines geometric properties of parallel lines with the physics principle of reflection. Remember to clearly state the law of reflection and how it applies to the angles at each mirror. The construction of normals is essential for solving this proof.