GSEB Class 9 Maths Solutions Chapter 6 Lines and Angles Exercise 6.1

Question 1. In the given figure AB and CD intersect at O. If \( \angle AOC + \angle BOE = 70^\circ \), and \( \angle BOD = 40^\circ \), find \( \angle BOE \) and reflex \( \angle COE \).
Answer: We are given that:
\( \angle AOC + \angle BOE = 70^\circ \) ...........(1)
Since AB and CD are intersecting lines, vertically opposite angles are equal.
\( \angle AOC = \angle BOD \)
We are given \( \angle BOD = 40^\circ \).
\( \implies \angle AOC = 40^\circ \)
Substituting this value into equation (1):
\( 40^\circ + \angle BOE = 70^\circ \)
\( \angle BOE = 70^\circ - 40^\circ \)
\( \implies \angle BOE = 30^\circ \)
Also, AOB is a straight line, so angles on this line sum to 180°. The angles \( \angle AOC \), \( \angle COE \), and \( \angle BOE \) lie on the line AB.
\( \angle AOC + \angle COE + \angle BOE = 180^\circ \)
\( \implies (\angle AOC + \angle BOE) + \angle COE = 180^\circ \)
From equation (1), we know \( \angle AOC + \angle BOE = 70^\circ \).
\( \implies 70^\circ + \angle COE = 180^\circ \)
\( \angle COE = 180^\circ - 70^\circ \)
\( \implies \angle COE = 110^\circ \)
To find the reflex angle of \( \angle COE \), we subtract it from 360°.
Reflex \( \angle COE = 360^\circ - 110^\circ = 250^\circ \)
Therefore, \( \angle BOE \) is \( 30^\circ \) and the reflex \( \angle COE \) is \( 250^\circ \).
In simple words: First, use the fact that opposite angles are equal to find \( \angle AOC \). Then, use the given sum to find \( \angle BOE \). Finally, remember that angles on a straight line add up to 180° to find \( \angle COE \), and then subtract that from 360° for the reflex angle.

Exam Tip: Clearly state reasons for each step, such as "vertically opposite angles" or "linear pair," as these justifications are essential for full marks.

 

Question 2. In figure lines XY and MN intersect at O. If \( \angle POY = 90^\circ \) and a : b = 2 : 3, find c.
Answer: We are given that lines XY and MN intersect at point O. Ray OP is perpendicular to XY, so \( \angle POY = 90^\circ \). The ratio of angles a and b is 2:3.
Let \( a = 2x \) and \( b = 3x \).
Since XOY is a straight line, \( \angle POX + \angle POY = 180^\circ \) (Linear pair).
We know \( \angle POY = 90^\circ \).
\( \angle POX + 90^\circ = 180^\circ \)
\( \implies \angle POX = 180^\circ - 90^\circ = 90^\circ \)
From the figure, \( a + b = \angle POX \).
\( \implies a + b = 90^\circ \)
Substitute the values of a and b:
\( 2x + 3x = 90^\circ \)
\( 5x = 90^\circ \)
\( \implies x = \frac {90^\circ}{5} = 18^\circ \)
Now, calculate the values of a and b:
\( a = 2x = 2 \times 18^\circ = 36^\circ \)
\( b = 3x = 3 \times 18^\circ = 54^\circ \)
Angles \( \angle XOM \) and \( \angle YON \) are vertically opposite angles, so they are equal.
Thus, \( \angle YON = \angle XOM = b = 54^\circ \).
Now, MN is a straight line. Angles \( \angle XON \) and \( \angle XOM \) (or \( \angle YOM \)) form a linear pair. Also angles \( \angle XON \) and \( \angle YON \) form a linear pair.
\( \angle XON + \angle YON = 180^\circ \) (Linear pair)
From the figure, \( c = \angle XON \).
\( c + 54^\circ = 180^\circ \)
\( \implies c = 180^\circ - 54^\circ \)
\( \implies c = 126^\circ \)
In simple words: First, find the value of x from the given ratio and the fact that angles a and b add up to 90°. Then, calculate 'a' and 'b'. Since 'c' and 'b' are vertically opposite angles, they are equal. Finally, use the linear pair property for angles on the line MN to find 'c'.

Exam Tip: Remember that vertically opposite angles are equal and angles on a straight line form a linear pair, summing to 180°. These are key principles for solving such problems.

 

Question 3. In figure \( \angle PQR = \angle PRQ \), then prove that \( \angle PQS = \angle PRT \).
Answer:

Given: \( \angle PQR = \angle PRQ \)
To Prove: \( \angle PQS = \angle PRT \)
Proof:
Angles \( \angle PQS \) and \( \angle PQR \) form a linear pair on line ST.
\( \angle PQS + \angle PQR = 180^\circ \) ...........(1) (Linear pair)
Similarly, angles \( \angle PRQ \) and \( \angle PRT \) form a linear pair on line ST.
\( \angle PRQ + \angle PRT = 180^\circ \) ...........(2) (Linear pair)
From equation (1), we can express \( \angle PQS \) as:
\( \angle PQS = 180^\circ - \angle PQR \) ...........(3)
From equation (2), we can express \( \angle PRT \) as:
\( \angle PRT = 180^\circ - \angle PRQ \) ...........(4)
We are given that \( \angle PQR = \angle PRQ \).
Since \( \angle PQR = \angle PRQ \), it follows that \( 180^\circ - \angle PQR = 180^\circ - \angle PRQ \).
Comparing equations (3) and (4), we can conclude:
\( \angle PQS = \angle PRT \)
Hence, proved.
In simple words: Since angles on a straight line add up to 180°, both \( \angle PQS \) and \( \angle PRT \) are found by subtracting an equal angle from 180°. Because they are both equal to \( 180^\circ - (\text{the same angle}) \), they must be equal to each other.

Exam Tip: When proving angle relationships, always clearly state the given information and what needs to be proven. Use established geometric properties like "linear pair" as justifications for your steps.

 

Question 4. In the given figure, if x + y = w + z, then prove that AOB is a line.
Answer:

Given: \( x + y = w + z \)
To Prove: AOB is a line.
Proof:
The sum of all angles around a point is 360°. Here, the angles around point O are x, y, w, and z.
So, \( x + y + w + z = 360^\circ \) ...........(1)
We are given that \( x + y = w + z \) ...........(2)
Substitute equation (2) into equation (1):
\( (w + z) + (w + z) = 360^\circ \)
\( \implies 2w + 2z = 360^\circ \)
Factor out 2:
\( \implies 2(w + z) = 360^\circ \)
Divide both sides by 2:
\( \implies w + z = \frac{360^\circ}{2} \)
\( \implies w + z = 180^\circ \)
Since the sum of angles w and z is 180°, they form a linear pair. In the figure, angles w and z are \( \angle AOD \) and \( \angle DOB \), respectively.
\( \implies \angle AOD + \angle DOB = 180^\circ \)
When two adjacent angles sum to 180°, their non-common arms form a straight line.
Therefore, AOB is a straight line.
In simple words: We know all angles around a point add to 360°. Since x + y equals w + z, we can say that two times (w + z) is 360°. This means w + z is 180°. If two angles add up to 180°, they form a straight line, so AOB is a line.

Exam Tip: For proofs involving angles around a point, remember the fundamental rule that their sum is 360°. This forms the basis for demonstrating linear pairs and straight lines.

 

Question 5. In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between ray OP and OR. Prove that \( \angle ROS = \frac{1}{2}(\angle QOS - \angle POS) \).
Answer:

Given: POQ is a line. Ray OR is perpendicular to line PQ. Ray OS lies between ray OP and ray OR.
To Prove: \( \angle ROS = \frac{1}{2}(\angle QOS - \angle POS) \)
Proof:
Since ray OR is perpendicular to line PQ,
\( \angle POR = 90^\circ \) and \( \angle QOR = 90^\circ \).
From the figure, we can write \( \angle POS \) as:
\( \angle POS = \angle POR - \angle ROS \)
\( \implies \angle POS = 90^\circ - \angle ROS \) ...........(1)
Also, from the figure, we can write \( \angle QOS \) as:
\( \angle QOS = \angle QOR + \angle ROS \)
\( \implies \angle QOS = 90^\circ + \angle ROS \) ...........(2)
Now, subtract equation (1) from equation (2):
\( \angle QOS - \angle POS = (90^\circ + \angle ROS) - (90^\circ - \angle ROS) \)
\( \implies \angle QOS - \angle POS = 90^\circ + \angle ROS - 90^\circ + \angle ROS \)
\( \implies \angle QOS - \angle POS = 2 \angle ROS \)
Divide both sides by 2:
\( \implies \angle ROS = \frac{1}{2}(\angle QOS - \angle POS) \)
Hence, proved.
In simple words: First, express \( \angle POS \) and \( \angle QOS \) using the 90° angle \( \angle POR \) and \( \angle QOR \), along with \( \angle ROS \). Then, subtract the expression for \( \angle POS \) from the expression for \( \angle QOS \). After simplifying, you will see that the difference equals twice \( \angle ROS \), proving the required relationship.

Exam Tip: Break down complex angles into sums or differences of simpler angles. Label your equations (1), (2), etc., to keep track of steps when performing algebraic manipulations like substitution or subtraction.

 

Question 6. 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 \( \angle QYP \).
Answer:

We are given that \( \angle XYZ = 64^\circ \). Line segment XY is extended to point P, forming a straight line PX.
Since PXY is a straight line, \( \angle XYZ \) and \( \angle PYZ \) form a linear pair.
\( \angle XYZ + \angle PYZ = 180^\circ \)
\( 64^\circ + \angle PYZ = 180^\circ \)
\( \implies \angle PYZ = 180^\circ - 64^\circ \)
\( \implies \angle PYZ = 116^\circ \)
Ray YQ bisects \( \angle ZYP \), which means it divides \( \angle ZYP \) into two equal angles.
So, \( \angle QYZ = \angle QYP = \frac{1}{2} \angle PYZ \)
\( \implies \angle QYZ = \frac{116^\circ}{2} = 58^\circ \)
Now, we need to find \( \angle XYQ \). From the figure, \( \angle XYQ \) is the sum of \( \angle XYZ \) and \( \angle QYZ \).
\( \angle XYQ = \angle XYZ + \angle QYZ \)
\( \angle XYQ = 64^\circ + 58^\circ \)
\( \implies \angle XYQ = 122^\circ \)
Next, we need to find reflex \( \angle QYP \). We already found \( \angle QYP = 58^\circ \).
To find the reflex angle, subtract the angle from 360°.
Reflex \( \angle QYP = 360^\circ - \angle QYP \)
Reflex \( \angle QYP = 360^\circ - 58^\circ \)
\( \implies \) Reflex \( \angle QYP = 302^\circ \)
Thus, \( \angle XYQ \) is \( 122^\circ \) and reflex \( \angle QYP \) is \( 302^\circ \).
In simple words: First, use the linear pair property to find \( \angle PYZ \). Since YQ cuts \( \angle PYZ \) in half, divide \( \angle PYZ \) by two to get \( \angle QYZ \) and \( \angle QYP \). Add \( \angle XYZ \) and \( \angle QYZ \) to find \( \angle XYQ \). Finally, subtract \( \angle QYP \) from 360° to get its reflex angle.

Exam Tip: Pay close attention to keywords like "produced" (meaning extending to form a straight line) and "bisects" (meaning divides into two equal parts). Drawing the figure accurately helps visualize the angles and their relationships.