Chapter 6: Lines and Angles
Exercise 6.1
Q1
In the given figure, lines AB and CD intersect at O. If \( \angle AOC + \angle BOE = 70^{\circ} \) and \( \angle BOD = 40^{\circ} \), find \( \angle BOE \) and reflex \( \angle COE \).
Solution:
Given that AB is a straight line, rays OC and OE stand on it.
\[ \angle AOC + \angle COE + \angle BOE = 180^{\circ} \]
\[ \Rightarrow (\angle AOC + \angle BOE) + \angle COE = 180^{\circ} \]
Substituting the given value \( \angle AOC + \angle BOE = 70^{\circ} \):
\[ 70^{\circ} + \angle COE = 180^{\circ} \]
\[ \Rightarrow \angle COE = 180^{\circ} - 70^{\circ} = 110^{\circ} \]
Reflex \( \angle COE = 360^{\circ} - 110^{\circ} = 250^{\circ} \).
Now, CD is a straight line; rays OE and OB stand on it.
\[ \angle COE + \angle BOE + \angle BOD = 180^{\circ} \]
Substituting the known values \( \angle COE = 110^{\circ} \) and \( \angle BOD = 40^{\circ} \):
\[ 110^{\circ} + \angle BOE + 40^{\circ} = 180^{\circ} \]
\[ \Rightarrow 150^{\circ} + \angle BOE = 180^{\circ} \]
\[ \Rightarrow \angle BOE = 180^{\circ} - 150^{\circ} = 30^{\circ} \]
Answer: \( \angle BOE = 30^{\circ} \) and Reflex \( \angle COE = 250^{\circ} \).
Q2
In the given figure, lines XY and MN intersect at O. If \( \angle POY = 90^{\circ} \) and \( a : b = 2 : 3 \), find \( c \).
Solution:
Let the common ratio between \( a \) and \( b \) be \( x \).
Therefore, \( a = 2x \) and \( b = 3x \).
Since XY is a straight line, rays OM and OP stand on it:
\[ \angle XOM + \angle MOP + \angle POY = 180^{\circ} \]
\[ b + a + \angle POY = 180^{\circ} \]
\[ 3x + 2x + 90^{\circ} = 180^{\circ} \]
\[ 5x = 90^{\circ} \]
\[ x = 18^{\circ} \]
Now, we can find the values of \( a \) and \( b \):
\[ a = 2x = 2 \times 18^{\circ} = 36^{\circ} \]
\[ b = 3x = 3 \times 18^{\circ} = 54^{\circ} \]
Since MN is a straight line and ray OX stands on it, \( b \) and \( c \) form a linear pair:
\[ b + c = 180^{\circ} \]
\[ 54^{\circ} + c = 180^{\circ} \]
\[ c = 180^{\circ} - 54^{\circ} = 126^{\circ} \]
Answer: \( c = 126^{\circ} \).
Q3
In the given figure, \( \angle PQR = \angle PRQ \), then prove that \( \angle PQS = \angle PRT \).
Solution:
In the given figure, ST is a straight line and ray QP stands on it.
\[ \angle PQS + \angle PQR = 180^{\circ} \quad \text{(Linear Pair)} \]
\[ \Rightarrow \angle PQR = 180^{\circ} - \angle PQS \quad \dots(1) \]
Similarly, ray RP stands on line ST.
\[ \angle PRT + \angle PRQ = 180^{\circ} \quad \text{(Linear Pair)} \]
\[ \Rightarrow \angle PRQ = 180^{\circ} - \angle PRT \quad \dots(2) \]
It is given that \( \angle PQR = \angle PRQ \).
Equating equations (1) and (2):
\[ 180^{\circ} - \angle PQS = 180^{\circ} - \angle PRT \]
\[ -\angle PQS = -\angle PRT \]
\[ \Rightarrow \angle PQS = \angle PRT \]
Hence proved.
Q4
In the given figure, if \( x + y = z + w \), then prove that AOB is a line.
Solution:
The sum of all angles around a point is \( 360^{\circ} \).
\[ x + y + z + w = 360^{\circ} \]
It is given that \( x + y = z + w \).
Substitute \( z + w \) with \( x + y \) in the equation:
\[ (x + y) + (x + y) = 360^{\circ} \]
\[ 2(x + y) = 360^{\circ} \]
\[ x + y = 180^{\circ} \]
Since \( x \) and \( y \) form a linear pair whose sum is \( 180^{\circ} \), AOB is a straight line.
Q5
In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that \( \angle ROS = \frac{1}{2}(\angle QOS - \angle POS) \).
Solution:
It is given that \( OR \perp PQ \).
Therefore, \( \angle POR = 90^{\circ} \).
\[ \Rightarrow \angle POS + \angle ROS = 90^{\circ} \]
\[ \Rightarrow \angle ROS = 90^{\circ} - \angle POS \quad \dots(1) \]
Also, \( \angle QOR = 90^{\circ} \) (As \( OR \perp PQ \)).
From the figure, we can express \( \angle QOR \) involving \( \angle QOS \):
\[ \angle QOS - \angle ROS = 90^{\circ} \]
\[ \Rightarrow \angle ROS = \angle QOS - 90^{\circ} \quad \dots(2) \]
Adding equations (1) and (2):
\[ 2\angle ROS = (90^{\circ} - \angle POS) + (\angle QOS - 90^{\circ}) \]
\[ 2\angle ROS = \angle QOS - \angle POS \]
\[ \Rightarrow \angle ROS = \frac{1}{2}(\angle QOS - \angle POS) \]
Hence proved.
Q6
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 \( \angle QYP \).
Solution:
Since line YQ bisects \( \angle ZYP \):
\[ \angle QYP = \angle ZYQ \]
It can be observed that PX is a straight line. Rays YQ and YZ stand on it.
\[ \angle XYZ + \angle ZYQ + \angle QYP = 180^{\circ} \]
Substituting \( \angle XYZ = 64^{\circ} \) and \( \angle ZYQ = \angle QYP \):
\[ 64^{\circ} + 2\angle QYP = 180^{\circ} \]
\[ 2\angle QYP = 180^{\circ} - 64^{\circ} = 116^{\circ} \]
\[ \Rightarrow \angle QYP = 58^{\circ} \]
Also, \( \angle ZYQ = \angle QYP = 58^{\circ} \).
Finding Reflex \( \angle QYP \):
\[ \text{Reflex } \angle QYP = 360^{\circ} - 58^{\circ} = 302^{\circ} \]
Finding \( \angle XYQ \):
\[ \angle XYQ = \angle XYZ + \angle ZYQ \]
\[ \angle XYQ = 64^{\circ} + 58^{\circ} = 122^{\circ} \]
Answer: \( \angle XYQ = 122^{\circ} \) and Reflex \( \angle QYP = 302^{\circ} \).
Exercise 6.2
Q1
In the given figure, find the values of \( x \) and \( y \) and then show that \( AB \parallel CD \).
Solution:
Using the property of a linear pair on the transversal line intersecting AB:
\[ 50^{\circ} + x = 180^{\circ} \]
\[ \Rightarrow x = 180^{\circ} - 50^{\circ} = 130^{\circ} \]
For angle \( y \), it is vertically opposite to the angle measuring \( 130^{\circ} \).
\[ \Rightarrow y = 130^{\circ} \]
Now, we see that \( x = 130^{\circ} \) and \( y = 130^{\circ} \). Therefore, \( x = y \).
Since \( x \) and \( y \) are alternate interior angles for lines AB and CD and their measures are equal, the lines must be parallel.
\[ \Rightarrow AB \parallel CD \]
Q2
In the given figure, if \( AB \parallel CD \), \( CD \parallel EF \) and \( y : z = 3 : 7 \), find \( x \).
Solution:
It is given that \( AB \parallel CD \) and \( CD \parallel EF \).
\[ \Rightarrow AB \parallel CD \parallel EF \] (Lines parallel to the same line are parallel to each other).
Since \( AB \parallel EF \), alternate interior angles are equal:
\[ \Rightarrow x = z \]
It is given that \( y : z = 3 : 7 \). Let the common ratio be \( a \).
Then \( y = 3a \) and \( z = 7a \).
Since \( AB \parallel CD \), interior angles on the same side of the transversal are supplementary (Co-interior angles):
\[ x + y = 180^{\circ} \]
Substituting \( x = z \):
\[ z + y = 180^{\circ} \]
\[ 7a + 3a = 180^{\circ} \]
\[ 10a = 180^{\circ} \Rightarrow a = 18^{\circ} \]
Now, finding \( x \):
\[ x = z = 7a = 7 \times 18^{\circ} = 126^{\circ} \]
Answer: \( x = 126^{\circ} \).
Q3
In the given figure, if \( AB \parallel CD \), \( EF \perp CD \) and \( \angle GED = 126^{\circ} \), find \( \angle AGE \), \( \angle GEF \) and \( \angle FGE \).
Solution:
Given: \( AB \parallel CD \), \( EF \perp CD \) (so \( \angle FED = 90^{\circ} \)), and \( \angle GED = 126^{\circ} \).
1. Finding \( \angle GEF \):
\[ \angle GED = \angle GEF + \angle FED \]
\[ 126^{\circ} = \angle GEF + 90^{\circ} \]
\[ \angle GEF = 126^{\circ} - 90^{\circ} = 36^{\circ} \]
2. Finding \( \angle AGE \):
Since \( AB \parallel CD \), alternate interior angles are equal.
\[ \angle AGE = \angle GED \]
\[ \Rightarrow \angle AGE = 126^{\circ} \]
3. Finding \( \angle FGE \):
\( \angle AGE \) and \( \angle FGE \) form a linear pair.
\[ \angle AGE + \angle FGE = 180^{\circ} \]
\[ 126^{\circ} + \angle FGE = 180^{\circ} \]
\[ \angle FGE = 180^{\circ} - 126^{\circ} = 54^{\circ} \]
Answer: \( \angle AGE = 126^{\circ} \), \( \angle GEF = 36^{\circ} \), \( \angle FGE = 54^{\circ} \).
Q4
In the given figure, if \( PQ \parallel ST \), \( \angle PQR = 110^{\circ} \) and \( \angle RST = 130^{\circ} \), find \( \angle QRS \).
Solution:
Let us draw a line XY parallel to ST passing through point R (and since \( ST \parallel PQ \), \( XY \parallel PQ \)).
Consider the side with PQ. Angles on the same side of the transversal are supplementary:
\[ \angle PQR + \angle QRX = 180^{\circ} \]
\[ 110^{\circ} + \angle QRX = 180^{\circ} \]
\[ \Rightarrow \angle QRX = 70^{\circ} \]
Consider the side with ST:
\[ \angle RST + \angle SRY = 180^{\circ} \]
\[ 130^{\circ} + \angle SRY = 180^{\circ} \]
\[ \Rightarrow \angle SRY = 50^{\circ} \]
Since XY is a straight line:
\[ \angle QRX + \angle QRS + \angle SRY = 180^{\circ} \]
\[ 70^{\circ} + \angle QRS + 50^{\circ} = 180^{\circ} \]
\[ \angle QRS = 180^{\circ} - 120^{\circ} = 60^{\circ} \]
Answer: \( \angle QRS = 60^{\circ} \).
Q5
In the given figure, if \( AB \parallel CD \), \( \angle APQ = 50^{\circ} \) and \( \angle PRD = 127^{\circ} \), find \( x \) and \( y \).
Solution:
Since \( AB \parallel CD \), alternate interior angles are equal.
1. \( \angle APQ = \angle PQR \)
\[ 50^{\circ} = x \]
\[ \Rightarrow x = 50^{\circ} \]
2. \( \angle APR = \angle PRD \)
\[ \angle APQ + \angle QPR = \angle PRD \]
\[ 50^{\circ} + y = 127^{\circ} \]
\[ y = 127^{\circ} - 50^{\circ} \]
\[ y = 77^{\circ} \]
Answer: \( x = 50^{\circ} \) and \( y = 77^{\circ} \).
Q6
In the given figure, 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 \( AB \parallel CD \).
Solution:
Let us draw normals \( BM \perp PQ \) and \( CN \perp RS \).
Since \( PQ \parallel RS \), their normals are also parallel: \( BM \parallel CN \).
Thus, BM and CN are two parallel lines and transversal BC cuts them.
\[ \angle 2 = \angle 3 \quad \text{(Alternate interior angles)} \]
By the laws of reflection, angle of incidence equals angle of reflection:
\[ \angle 1 = \angle 2 \quad \text{and} \quad \angle 3 = \angle 4 \]
Combining these:
\[ \angle 1 = \angle 2 = \angle 3 = \angle 4 \]
Also,
\[ \angle 1 + \angle 2 = \angle 3 + \angle 4 \]
\[ \Rightarrow \angle ABC = \angle DCB \]
These are alternate interior angles formed by transversal BC intersecting lines AB and CD.
Therefore, \( AB \parallel CD \). Hence proved.
Exercise 6.3
Q1
In the given figure, sides QP and RQ of \( \Delta PQR \) are produced to points S and T respectively. If \( \angle SPR = 135^{\circ} \) and \( \angle PQT = 110^{\circ} \), find \( \angle PRQ \).
Solution:
Using the linear pair property:
\[ \angle SPR + \angle QPR = 180^{\circ} \]
\[ 135^{\circ} + \angle QPR = 180^{\circ} \Rightarrow \angle QPR = 45^{\circ} \].
Also,
\[ \angle PQT + \angle PQR = 180^{\circ} \]
\[ 110^{\circ} + \angle PQR = 180^{\circ} \Rightarrow \angle PQR = 70^{\circ} \].
The sum of interior angles of a triangle is \( 180^{\circ} \). For \( \Delta PQR \):
\[ \angle QPR + \angle PQR + \angle PRQ = 180^{\circ} \]
\[ 45^{\circ} + 70^{\circ} + \angle PRQ = 180^{\circ} \]
\[ 115^{\circ} + \angle PRQ = 180^{\circ} \]
\[ \angle PRQ = 180^{\circ} - 115^{\circ} = 65^{\circ} \]
Answer: \( \angle PRQ = 65^{\circ} \).
Q2
In the given figure, \( \angle X = 62^{\circ} \), \( \angle XYZ = 54^{\circ} \). If YO and ZO are the bisectors of \( \angle XYZ \) and \( \angle XZY \) respectively of \( \Delta XYZ \), find \( \angle OZY \) and \( \angle YOZ \).
Solution:
In \( \Delta XYZ \), the sum of angles is \( 180^{\circ} \):
\[ \angle X + \angle XYZ + \angle XZY = 180^{\circ} \]
\[ 62^{\circ} + 54^{\circ} + \angle XZY = 180^{\circ} \]
\[ 116^{\circ} + \angle XZY = 180^{\circ} \]
\[ \angle XZY = 64^{\circ} \]
Since ZO is the bisector of \( \angle XZY \):
\[ \angle OZY = \frac{1}{2} \angle XZY = \frac{64^{\circ}}{2} = 32^{\circ} \]
Since YO is the bisector of \( \angle XYZ \):
\[ \angle OYZ = \frac{1}{2} \angle XYZ = \frac{54^{\circ}}{2} = 27^{\circ} \]
In \( \Delta OYZ \):
\[ \angle OYZ + \angle OZY + \angle YOZ = 180^{\circ} \]
\[ 27^{\circ} + 32^{\circ} + \angle YOZ = 180^{\circ} \]
\[ 59^{\circ} + \angle YOZ = 180^{\circ} \]
\[ \angle YOZ = 180^{\circ} - 59^{\circ} = 121^{\circ} \]
Answer: \( \angle OZY = 32^{\circ} \) and \( \angle YOZ = 121^{\circ} \).
Q3
In the given figure, if \( AB \parallel DE \), \( \angle BAC = 35^{\circ} \) and \( \angle CDE = 53^{\circ} \), find \( \angle DCE \).
Solution:
Since \( AB \parallel DE \) and AE is a transversal:
\[ \angle BAC = \angle CED \quad \text{(Alternate interior angles)} \]
\[ \therefore \angle CED = 35^{\circ} \]
In \( \Delta CDE \), using the angle sum property:
\[ \angle CDE + \angle CED + \angle DCE = 180^{\circ} \]
\[ 53^{\circ} + 35^{\circ} + \angle DCE = 180^{\circ} \]
\[ 88^{\circ} + \angle DCE = 180^{\circ} \]
\[ \angle DCE = 180^{\circ} - 88^{\circ} = 92^{\circ} \]
Answer: \( \angle DCE = 92^{\circ} \).
Q4
In the given figure, if lines PQ and RS intersect at point T, such that \( \angle PRT = 40^{\circ} \), \( \angle RPT = 95^{\circ} \) and \( \angle TSQ = 75^{\circ} \), find \( \angle SQT \).
Solution:
In \( \Delta PRT \):
\[ \angle PRT + \angle RPT + \angle PTR = 180^{\circ} \]
\[ 40^{\circ} + 95^{\circ} + \angle PTR = 180^{\circ} \]
\[ \angle PTR = 180^{\circ} - 135^{\circ} = 45^{\circ} \]
Also, \( \angle STQ = \angle PTR \) (Vertically opposite angles).
\[ \therefore \angle STQ = 45^{\circ} \]
In \( \Delta STQ \):
\[ \angle STQ + \angle SQT + \angle QST = 180^{\circ} \]
\[ 45^{\circ} + \angle SQT + 75^{\circ} = 180^{\circ} \]
\[ \angle SQT + 120^{\circ} = 180^{\circ} \]
\[ \angle SQT = 60^{\circ} \]
Answer: \( \angle SQT = 60^{\circ} \).
Q5
In the given figure, if \( PQ \perp PS \), \( PQ \parallel SR \), \( \angle SQR = 28^{\circ} \) and \( \angle QRT = 65^{\circ} \), then find the values of \( x \) and \( y \).
Solution:
It is given that \( PQ \parallel SR \) and QR is a transversal.
Therefore, alternate interior angles are equal:
\[ \angle PQR = \angle QRT \]
From the figure, \( \angle PQR = x + 28^{\circ} \).
\[ x + 28^{\circ} = 65^{\circ} \]
\[ x = 65^{\circ} - 28^{\circ} = 37^{\circ} \].
Now consider \( \Delta SPQ \). We know \( \angle SPQ = 90^{\circ} \) (since \( PQ \perp PS \)).
\[ \angle SPQ + x + y = 180^{\circ} \]
\[ 90^{\circ} + 37^{\circ} + y = 180^{\circ} \]
\[ 127^{\circ} + y = 180^{\circ} \]
\[ y = 180^{\circ} - 127^{\circ} = 53^{\circ} \]
Answer: \( x = 37^{\circ} \) and \( y = 53^{\circ} \).
Q6
In the given figure, the side QR of \( \Delta PQR \) is produced to a point S. If the bisectors of \( \angle PQR \) and \( \angle PRS \) meet at point T, then prove that \( \angle QTR = \frac{1}{2}\angle QPR \).
Solution:
For \( \Delta QTR \), \( \angle TRS \) is an exterior angle.
\[ \angle TRS = \angle QTR + \angle TQR \]
\[ \Rightarrow \angle QTR = \angle TRS - \angle TQR \quad \dots(1) \]
For \( \Delta PQR \), \( \angle PRS \) is an exterior angle.
\[ \angle PRS = \angle QPR + \angle PQR \]
Since RT is the angle bisector of \( \angle PRS \) and QT is the angle bisector of \( \angle PQR \):
\[ 2\angle TRS = \angle QPR + 2\angle TQR \]
\[ \angle QPR = 2\angle TRS - 2\angle TQR \]
\[ \angle QPR = 2(\angle TRS - \angle TQR) \]
Substitute the value from equation (1) into this equation:
\[ \angle QPR = 2\angle QTR \]
\[ \Rightarrow \angle QTR = \frac{1}{2}\angle QPR \]
Hence proved.