Class 9 - Maths - Lines and Angles

**Exercise 6.1**

**Question 1:**

In the figure lines AB and CD intersect at O. If ∠ AOC + ∠ BOE = 70^{0} and ∠ BOD = 40^{0},find ∠ BOE and reflex ∠ COE.

Answer:

Lines AB and CD intersect at O.

∠AOC + ∠BOE = 70^{0} (Given) ………………1

∠BOD = 40^{0} (Given) ………………2

Since, ∠AOC = ∠BOD (Vertically opposite angles)

Therefore, ∠AOC = 40^{0} [From equation 2]

and 40^{0} + ∠BOE = 70^{0} [From equation 1]

=> ∠BOE = 70^{0} – 40^{0} = 30^{0}

Also, ∠AOC + ∠BOE + ∠COE = 180^{0} [Since AOB is a straight line]

=> 70^{0} + ∠COE = 180^{0} [Form equation 1]

=> ∠COE = 180^{0} – 70^{0} = 110^{0}

Now, reflex ∠COE = 360^{0} – 110^{0} = 250^{0}

Hence, ∠BOE = 30^{0} and reflex ∠COE = 250^{0}

**Question 2:**

In the figure, lines XY and MN intersect at O. If ∠POY = 90^{0} and a : b = 2 : 3, find c.

Answer:

In the figure, lines XY and MN intersect at O and ∠ POY = 90^{0}.

Also, given a : b = 2 : 3

Let a = 2x and b = 3x.

Since, ∠XOM + ∠POM + ∠POY = 180^{0} [Linear pair axiom]

=> 3x + 2x + 90^{0} = 180^{0}

=> 5x = 180^{0} – 90^{0}

=> x = 90^{0}/5^{0} = 18^{0}

So, ∠XOM = b = 3x = 3 * 18^{0} = 54^{0}

and ∠POM = a = 2x = 2 * 18^{0} = 36^{0}

Now, ∠XON = c = ∠MOY = ∠POM + ∠POY [Vertically opposite angles]

=> c = 36^{0} + 90^{0} = 126^{0}

Hence, c = 126^{0}

**Question 3:**

In the figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠ PRT.

Answer:

∠PQS + ∠PQR = 180° ……………1 (Linear pair axiom)

∠PRQ + ∠PRT = 180° ……………2 (Linear pair axiom)

But, ∠PQR = ∠PRQ (Given)

From equation 1 and 2, we get

∠PQS = ∠PRT

**Question 4:**

In the figure, if x + y = w + z, then prove that AOB is a line.

Answer:

Assume AOB is a line.

Therefore, x + y = 180° …………….1 [Linear pair axiom]

and w + z = 180° .......................2 [Linear pair axiom]

Now, from equation 1 and 2, we get

x + y = w + z

**Question 5:**

In the figure, POQ is a line. Ray OR is perpendicular to line PQ.OS is another ray lying between rays OP and OR.

Prove that ∠ROS = (∠QOS – ∠ POS)/2

Answer:

From the figure,

∠ROS = ∠ROP – ∠POS ……………..1

and ∠ROS = ∠QOS – ∠QOR .......2

Adding equation 1 and 2, we get

∠ROS + ∠ROS = ∠QOS – ∠QOR + ∠ROP – ∠POS

=> 2∠ROS = ∠QOS – ∠POS [Since ∠QOR = ∠ROP = 90^{0}]

=> ∠ROS = (∠QOS – ∠POS)/2

**Question 6:**

It is given that ∠ XYZ = 64^{0} and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ ZYP, find ∠ XYQ and reflex ∠QYP.

Answer:

From the figure,

∠XYZ = 64^{0} (Given)

Now, ∠ZYP + ∠XYZ = 180^{0} [Linear pair axiom]

=> ∠ZYP + 64^{0} = 180^{0}

=> ∠ZYP = 180^{0} – 64^{0} – 116^{0}

Also, given that ray YQ bisects ∠ZYP.

But, ∠ZYP = ∠QYP = ∠QYZ = 116^{0}

Therefore, ∠QYP = 58^{0} and ∠QYZ = 58^{0}

Also, ∠XYQ = ∠XYZ + ∠QYZ

=> ∠XYQ = 64^{0} + 58^{0} = 122^{0}

and reflex ∠QYP = 360^{0} – ∠QYP = 360^{0} – 58^{0} = 302^{0} [Since ∠QYP = 58^{0}]

Hence, ∠XYQ = 122^{0} and reflex ∠QYP = 302^{0}

**Exercise 6.2**

**Question 1:**

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

Answer:

In the given figure, a transversal intersects two lines AB and CD such that

x + 50^{0} = 180^{0} [Linear pair axiom]

=> x = 180^{0} – 50^{0} = 130^{0}

and y = 130^{0} [Vertically opposite angles]

Therefore, ∠x = ∠y = 130^{0} [Alternate angles]

Hence, AB || CD [Converse of alternate angles axiom]

**Question 2:**

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

Answer:

In the given figure, AB || CD, CD || EF and y : z = 3 : 7.

Let y = 3a and z = 7a

∠DHI = y [vertically opposite angles]

∠DHI + ∠FIH = 180^{0 } [Interior angles on the same side of the transversal]

=> y + z = 180^{0 }

=> 3a + 7a = 180^{0}

=> 10a = 180^{0}

=> a = 18^{0}

So, y = 3 * 18° = 54^{0}

and z = 18^{0} * 7 = 126^{0}

Also, x + y = 180^{0}

=> x + 54^{0} = 180^{0}

=> x = 180^{0} – 54^{0} = 126^{0}

Hence, x = 126^{0}

**Question 3:**

In the figure, if AB || CD, EF ⊥ CD and ∠GED = 126°. Find ∠ AGE, ∠GEF and ∠FGE.

Answer:

In the given figure, AB || CD, EF ⊥ CD and ∠GED = 126^{0}

^{}

∠AGE = ∠LGE [Alternate angle]

So, ∠AGE = 126^{0}

Now, ∠GEF = ∠GED – ∠DEF

= 126^{0} – 90^{0} = 36^{0} [Since ∠DEF = 90^{0}]

Also, ∠AGE + ∠FGE = 180^{0} [Linear pair axiom]

=> 126^{0} + FGE = 180^{0}

=> ∠FGE = 180^{0} – 126^{0} = 54^{0}

**Question 4:**

In the figure, if PQ || ST, ∠PQR = 110^{0 }and ∠ RST = 130^{0}, find ∠QRS.

Answer:

Extend PQ to Y and draw LM || ST through R.

∠TSX = ∠QXS [Alternate angles]

=> ∠QXS = 130^{0}

∠QXS + ∠RXQ = 180^{0} [Linear pair axiom]

=> ∠RXQ = 180^{0} – 130^{0} = 50^{0} …………..1

∠PQR = ∠QRM [Alternate angles]

=> ∠QRM = 110^{0} ………………2

∠RXQ = ∠XRM [Alternate angles]

=> ∠XRM = 50^{0} [from equation 1]

∠QRS = ∠QRM – ∠XRM

= 110^{0} – 50^{0} = 60^{0}

**Question 5:**

In the figure, if AB || CD, ∠APQ = 50^{0} and ∠PRD = 127°, find x and y.

Answer:

In the given figure, AB || CD, ∠APQ = 50^{0} and ∠PRD = 127^{0}

∠APQ + ∠PQC = 180^{0} [Pair of consecutive interior angles are supplementary]

=> 50^{0} + ∠PQC = 180^{0 }

=> ∠PQC = 180^{0} – 50^{0} = 130^{0}

Now, ∠PQC + ∠PQR = 180^{0} [Linear pair axiom]

=> 130^{0} + x = 180^{0}

=> x = 180^{0} – 130^{0} = 50^{0}

Also, x + y = 127^{0} [Exterior angle of a triangle is equal to the sum of the two interior opposite

angles]

=> 50^{0} + y = 127^{0}

=> y = 127^{0} – 50^{0} = 77^{0}

Hence, x = 50^{0} and y = 77^{0}

**Question 6:**

In the 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 || CD.

Answer:

At point B, draw BE ⊥ PQ and at point C, draw CF ⊥ RS.

∠1 = ∠2 …………..1 [Angle of incidence is equal to angle of reflection]

∠3 = ∠4 …………..2 [Angle of incidence is equal to angle of reflection]

Also, ∠2 = ∠3 ......3 [Alternate angles]

=> ∠1 = ∠4 [From equation 1, 2, and 3]

=> 2∠1 = 2∠4

=> ∠1 + ∠1 = ∠4 + ∠4

=> ∠1 + ∠2 = ∠3 + ∠4 [From (i) and (ii)]

=> ∠BCD = ∠ABC

Hence, AB || CD. [Alternate angles are equal]

Hence, Proved.

**Exercise 6.3**

**Question 1:**

In the figure, sides QP and RQ of ∆ PQR are produced to points S and T respectively. If ∠SPR = 135^{0} and ∠PQT = 110^{0}, find ∠PRQ.

Answer:

In the given figure, ∠SPR = 135^{0} and ∠PQT = 110^{0}.

∠PQT + ∠PQR = 180^{0} [Linear pair axiom]

=> 110^{0} + ∠PQR = 180^{0}

=> ∠PQR = 180^{0} – 110^{0} = 70^{0}

Also, ∠SPR + ∠QPR = 180^{0} [Linear pair axiom]

=> 135^{0} + ∠QPR = 180^{0}

=> ∠QPS = 180^{0} – 135^{0} = 45^{0}

Now, in the triangle PQR

∠PQR + ∠PRQ + ∠QPR = 180^{0} [Angle sum property of a triangle]

=> 70^{0} + ∠PRQ + 45^{0} = 180^{0}

=> ∠PRQ + 115^{0} = 180^{0}

=> ∠PRQ = 180^{0} – 115^{0} = 65^{0}

Hence, ∠ PRQ = 65^{0}

**Question 2:**

In the figure, ∠ X = 62^{0}, ∠ XYZ = 54^{0}. If YO and ZO are the bisectors of ∠ XYZ and ∠ XZY respectively of

∆ XYZ, find ∠OZY and ∠YOZ.

Answer:

In the given figure,

∠X = 62^{0} and ∠XYZ = 54^{0}.

∠XYZ + ∠XZY + ∠YXZ = 180^{0} ………1 [Angle sum property of a triangle]

=> 54^{0} + ∠XZY + 62^{0} = 180^{0}

=> ∠XZY + 116^{0} = 180^{0}

=> ∠XZY = 180^{0} – 116^{0} = 64^{0}

Now, ∠OZY = 1/2 * ∠XZY [Since ZO is bisector of ∠XZY]

= 1/2 * 64^{0} = 32^{0}

Similarly, ∠OYZ = 1/2 * 54^{0} = 27^{0}

Now, in ∆OYZ, we have

∠OYZ + ∠OZY + ∠YOZ = 180^{0} [Angle sum property of a triangle]

=> 27^{0} + 32^{0} + ∠YOZ = 180^{0}

=> ∠YOZ = 180^{0} – 59^{0} = 121^{0}

Hence, ∠OZY = 32^{0} and ∠YOZ = 121^{0}

**Question 3:**

In the figure, if AB || DE, ∠ BAC = 35^{0 }and ∠CDE = 53^{0}, find ∠ DCE.

Answer:

In the given figure,

∠BAC = ∠CED [Alternate angles]

=> ∠CED = 35^{0}

In ∆CDE,

∠CDE + ∠DCE + ∠CED = 180^{0} [Angle sum property of a triangle]

=> 53^{0} + ∠DCE + 35^{0} = 180^{0}

=> ∠DCE + 88^{0} = 180^{0}

=> ∠DCE = 180^{0} – 88^{0} = 92^{0}

Hence, ∠DCE = 92^{0}

**Question 4:**

In the figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40^{0}, ∠ RPT = 95^{0}

and ∠TSQ = 75^{0}, find ∠SQT.

Answer:

In the given figure, lines PQ and RS

intersect at point T, such that ∠PRT = 40^{0},

∠RPT = 95^{0} and ∠TSQ = 75^{0}.

In ∆PRT

∠PRT + ∠RPT + ∠PTR = 180^{0} [Angle sum property of a triangle]

=> 40^{0} + 95^{0} + ∠PTR = 180^{0}

=> 135^{0} + ∠PTR = 180^{0}

=> ∠PTR = 180^{0} – 135^{0} = 45^{0}

Also, ∠PTR = ∠STQ [Vertical opposite angles]

So, ∠STQ = 45^{0}

Now, in ∆STQ,

∠STQ + ∠TSQ + ∠SQT = 180° [Angle sum property of a triangle]

=> 45^{0} + 75^{0} + ∠SQT = 180^{0}

=> 120^{0} + ∠SQT = 180^{0}

=> ∠SQT = 180^{0} – 120^{0} = 60^{0}

Hence, ∠SQT = 60^{0}

**Question 5:**

In the figure, if PT ⊥ PS, PQ || SR, ∠SQR = 28^{0 }and ∠QRT = 65^{0}, then find the values of x and y.

Answer:

In the given figure, lines PQ ⊥ PS, PQ || SR,

∠SQR = 28^{0} and ∠QRT = 65^{0}

∠PQR = ∠QRT [Alternate angles]

=> x + 28^{0} = 65^{0}

=> x = 65^{0} – 28^{0} = 37^{0}

In ∆PQS,

∠SPQ + ∠PQS + ∠QSP = 180^{0} [Angle sum property of a triangle]

=> 90^{0} + 37^{0} + y = 180^{0} [Since PQ ⊥ PS, ∠PQS = x = 37^{0} and ∠QSP = y)

=> 127^{0} + y = 180^{0}

=> y = 180^{0} – 127^{0} = 53^{0}

Hence, x = 37^{0} and y = 53^{0}

**Question 6:**

In the figure, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠ PQR and ∠ PRS meet at point T, then prove that ∠QTR = ∠QPR/2.

Answer:

Exterior ∠PRS = ∠PQR + ∠QPR [Exterior angle property]

Therefore, ∠PRS/2 = ∠PQR/2 + ∠QPR/2

=> ∠TRS = ∠TQR + ∠QPR/2 ……………1

But in ∆QTR,

Exterior ∠TRS = ∠TQR + ∠QTR ……….2 [Exterior angles property]

Therefore, from equation 1 and 2, we get

∠TQR + ∠QTR = ∠TQR + ∠QPR/2

=> ∠QTR = ∠QPR/2

.