R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 7 - Lines And Angles

(i) Angle: Two rays having a common end point form an angle.

(ii) Interior of an angle: The interior of AOB is the set of all points in its plane, which lie on the same side of OA as B and also on same side of OB as A.

(iii) Obtuse angle: An angle whose measure is more than 90^o but less than 180^o, is called an obtuse angle.

(iv) Reflex angle: An angle whose measure is more than 180^o but less than 360^o is called a reflex angle.

(v) Complementary angles: Two angles are said to be complementary, if the sum of their measures is 90^o.

(vi) Supplementary angles: Two angles are said to be supplementary, if the sum of their measures is 180^o.

Let the required angle be x^o

Then, its complement is 90^o - x^o and its supplement is 180^o - x^o

The required angle is 45^o.

Let the two required angles be x^o and 90^o - x^o.

Then

5x = 4(90 - x)

5x = 360 - 4x

5x + 4x = 360

9x = 360

Thus, the required angles are 40^o and 90^o - x^o = 90 ^o - 40^o = 50^o.

(2x - 5)° and (x - 10)° are complementary angles.

∴ (2x - 5)° + (x - 10)° = 90° 

⇒ 2x - 5° + x - 10° = 90° 

⇒ 3x - 15° = 90° 

⇒ 3x = 105° 

⇒ x = 35° 

Complement of 55° = 90° - 55° = 35°  

Complement of 16^o = 90 - 16^o = 74^o

Complement of 90° = 90° - 90° = 0° 

Complement of 46^o 30' = 90^o - 46^o 30' = 43^o 30'

Supplement of 42° = 180° - 42° = 138° 

Supplement of 90° = 180° - 90° = 90° 

Supplement of 124° = 180° - 124° = 56° 

Supplement of 75^o 36' = 180^o - 75^o 36' = 104^o 24'

(i) Let the required angle be x^o

Then, its complement = 90^o - x^o

The measure of an angle which is equal to its complement is 45^o.

(ii) Let the required angle be x^o

Then, its supplement = 180^o - x^o

The measure of an angle which is equal to its supplement is 90^o.

Let the required angle be x^o

Then its complement is 90^o - x^o

The measure of an angle which is 36^o more than its complement is 63^o.

Let the measure of the required angle = x° 

Then, measure of its supplement = (180 - x)° 

It is given that

x° = (180 - x)° - 30° 

⇒ x° = 180° - x° - 30° 

⇒ 2x° = 150° 

⇒ x° = 75° 

Hence, the measure of the required angle is 75°. 

Let the required angle be x^o

Then, its complement = 90^o - x^o

The required angle is 72^o.

Let the required angle be x^o

Then, its supplement is 180^o - x^o

The required angle is 150^o.

Let the required angle be x^o

Then, its complement is 90^o - x^o and its supplement is 180^o - x^o

The required angle is 60^o.

Since BOC and COA form a linear pair of angles, we have

BOC + COA = 180^o

x^o + 62^o = 180^o

x = 180 - 62

x = 118^o

Let two straight lines AB and CD intersect at O and let AOC = 90^o.

Now, AOC = BOD [Vertically opposite angles]

BOD = 90^o

Also, as AOC and AOD form a linear pair.

90^o + AOD = 180^o

AOD = 180^o - 90^o = 90^o

Since, BOC = AOD [Verticallty opposite angles]

BOC = 90^o

Thus, each of the remaining angles is 90^o.

Since, AOD and BOC are vertically opposite angles.

AOD = BOC

Now, AOD + BOC = 280^o [Given]

AOD + AOD = 280^o

2AOD = 280^o

AOD =

BOC = AOD = 140^o

As, AOC and AOD form a linear pair.

So, AOC + AOD = 180^o

AOC + 140^o = 180^o

AOC = 180^o - 140^o = 40^o

Since, AOC and BOD are vertically opposite angles.

AOC = BOD

BOD = 40^o

BOC = 140^o, AOC = 40^o , AOD = 140^o and BOD = 40^o.

Let ∠AOC = 5x and ∠AOD = 7x

Now, ∠AOC + ∠AOD = 180° (linear pair of angles)

⇒ 5x + 7x = 180° 

⇒ 12x = 180° 

⇒ x = 15° 

⇒ ∠AOC = 5x = 5(15°) = 75° and ∠AOD = 7x = 7(15°) = 105° 

Now, ∠AOC = ∠BOD (vertically opposite angles)

⇒ ∠BOD = 75° 

Also, ∠AOD = ∠BOC (vertically opposite angles)

⇒ ∠BOC = 105° 

∠BOD = 40° 

⇒ AOC = ∠BOD = 40° (vertically opposite angles)

∠AOE = 35° 

⇒ ∠BOF = ∠AOE = 35° (vertically opposite angles)

∠AOB is a straight angle.

⇒ ∠AOB = 180° 

⇒ ∠AOE + ∠EOD + ∠BOD = 180° 

⇒ 35° + ∠EOD + 40° = 180° 

⇒ ∠EOD + 75° = 180° 

⇒ ∠EOD = 105° 

Now, ∠COF = ∠EOD = 105° (vertically opposite angles)

∠AOC + ∠BOC = 180° (linear pair of angles)

⇒ x + 125 = 180° 

⇒ x = 55° 

Now, ∠AOD = ∠BOC  (vertically opposite angles)

⇒ y = 125° 

Also, ∠BOD = ∠AOC (vertically opposite angles)

⇒ z = 55° 

Given : AB and CD are two lines which are intersecting at O. OE is a ray bisecting the BOD. OF is a ray opposite to ray OE.

To Prove: AOF = COF

Proof : Since are two opposite rays, is a straight line passing through O.

AOF = BOE

and COF = DOE

[Vertically opposite angles]

But BOE = DOE (Given)

AOF = COF

Hence, proved.

Given: is the bisector of BCD and is the bisector of ACD.

To Prove: ECF = 90^o

Proof: Since ACD and BCD forms a linear pair.

ACD + BCD = 180^o

ACE + ECD + DCF + FCB = 180^o

ECD + ECD + DCF + DCF = 180^o

because ACE = ECD

and DCF = FCB

2(ECD) + 2 (CDF) = 180^o

2(ECD + DCF) = 180^o

ECD + DCF =

ECF = 90^o (Proved)

∠AOB is a straight angle.

⇒ ∠AOB = 180° 

⇒ ∠AOC + ∠COD + ∠BOD = 180° 

⇒ (3x - 7)° + 55° + (x + 20)° = 180° 

⇒ 4x + 68° = 180° 

⇒ 4x = 112° 

⇒ x = 28° 

Thus, ∠AOC = (3x - 7)° = 3(28°) - 7° = 84° - 7° = 77° 

And, ∠BOD = (x + 20)° = 28° + 20° = 48° 

Since BOD and DOA from a linear pair of angles.

BOD + DOA = 180^o

BOD + DOC + COA = 180^o

x^o + (2x - 19)^o + (3x + 7)^o = 180^o

6x - 12 = 180

6x = 180 + 12 = 192

x = 32

AOC = (3x + 7)^o = (3 32 + 7)^o = 103^o

COD = (2x - 19)^o = (2 32 - 19)^o = 45^o

and BOD = x^o = 32^o

x: y: z = 5: 4: 6

The sum of their ratios = 5 + 4 + 6 = 15

But x + y + z = 180^o

[Since, XOY is a straight line]

So, if the total sum of the measures is 15, then the measure of x is 5.

If the sum of angles is 180^o, then, measure of 

And, if the total sum of the measures is 15, then the measure of y is 4.

If the sum of the angles is 180^o, then, measure of

And z = 180^o - x - y

= 180^o - 60^o - 48^o

= 180^o - 108^o = 72^o

x = 60, y = 48 and z = 72.

AOB will be a straight line, if two adjacent angles form a linear pair.

BOC + AOC = 180^o

(4x - 36)^o + (3x + 20)^o = 180^o

4x - 36 + 3x + 20 = 180

7x - 16 = 180^o

7x = 180 + 16 = 196

The value of x = 28.

Since AOC and AOD form a linear pair.

AOC + AOD = 180^o

50^o + AOD = 180^o

AOD = 180^o - 50^o = 130^o

AOD and BOC are vertically opposite angles.

AOD = BOC

BOC = 130^o

BOD and AOC are vertically opposite angles.

BOD = AOC

BOD = 50^o

Since COE and DOF are vertically opposite angles, we have,

COE = DOF

z = 50^o

Also BOD and COA are vertically opposite angles.

So, BOD = COA

t = 90^o

As COA and AOD form a linear pair,

COA + AOD = 180^o

COA + AOF + FOD = 180^o [t = 90^o]

t + x + 50^o = 180^o

90^o + x^o + 50^o = 180^o

x + 140 = 180

x = 180 - 140 = 40

Since EOB and AOF are vertically opposite angles

So, EOB = AOF

y = x = 40

Thus, x = 40 = y = 40, z = 50 and t = 90

Since COE and EOD form a linear pair of angles.

COE + EOD = 180^o

COE + EOA + AOD = 180^o

5x + EOA + 2x = 180

5x + BOF + 2x = 180

[EOA and BOF are vertically opposite angles so, EOA = BOF]

5x + 3x + 2x = 180

10x = 180

x = 18

Now AOD = 2x^o = 2 18^o = 36^o

COE = 5x^o = 5 18^o = 90^o

and, EOA = BOF = 3x^o = 3 18^o = 54^o

Let the two adjacent angles be 5x and 4x.

Now, since these angles form a linear pair.

So, 5x + 4x = 180^o

9x = 180^o

The required angles are 5x = 5x = 5 20^o = 100^o

and 4x = 4 20^o = 80^o

Given, ∠1 = 120° 

Now, ∠1 + ∠2 = 180° (linear pair)

⇒ 120° + ∠2 = 180° 

⇒ ∠2 = 60° 

∠1 = ∠3  (vertically opposite angles)

⇒ ∠3 = 120° 

Also, ∠2 = ∠4  (vertically opposite angles)

⇒ ∠4 = 60° 

Line l ∥ line m and line t is a transversal.

⇒ ∠5 = ∠1 = 120° (corresponding angles)

 ∠6 = ∠2 = 60° (corresponding angles)

 ∠7 = ∠3 = 120° (corresponding angles)

 ∠8 = ∠4 = 60° (corresponding angles)

Through C draw FG || AE

Now, since CG || BE and CE is a transversal.

So, GCE = CEA = 20^o            [Alternate angles]

DCG = 130^o - GCE

= 130^o - 20^o = 110^o

Also, we have AB || CD and FG is a transversal.

So, BFC = DCG = 110^o          [Corresponding angles]

As, FG || AE, AF is a transversal.

BFG = FAE                           [Corresponding angles]

x^o = FAE = 110^o.

Hence, x = 110

Since AB || PQ and EF is a transversal.

So, CEB = EFQ                 [Corresponding angles]

EFQ = 75^o

EFG + GFQ = 75^o

25^o + y^o = 75^o

y = 75 - 25 = 50

Also, BEF + EFQ = 180^o   [sum of consecutive interior angles is 180^o]      

BEF = 180^o - EFQ

           = 180^o - 75^o

  BEF = 105^o

FEG + GEB = BEF = 105^o

FEG = 105^o - GEB = 105^o - 20^o = 85^o

In EFG we have,

x^o + 25^o + FEG = 180^o

Hence, x = 70.

Since AB || CD and AC is a transversal.

So, BAC + ACD = 180^o   [sum of consecutive interior angles is 180^o]

ACD = 180^o - BAC

= 180^o - 75^o = 105^o

ECF = ACD                     [Vertically opposite angles]

ECF = 105^o

Now in CEF,

ECF + CEF + EFC =180^o

x = 180 - 30 - 105 = 45

_{Hence, x = 45.}

Since AB || CD and PQ a transversal.

So, PEF = EGH [Corresponding angles]

EGH = 85^o

EGH and QGH form a linear pair.

So, EGH + QGH = 180^o

QGH = 180^o - 85^o = 95^o

Similarly, GHQ + 115^o = 180^o

GHQ = 180^o - 115^o = 65^o

In GHQ, we have,

x^o + 65^o + 95^o = 180^o

x = 180 - 65 - 95 = 180 - 160

x = 20

Since AB || CD and BC is a transversal.

So, ABC = BCD

x = 35

Also, AB || CD and AD is a transversal.

So, BAD = ADC

z = 75

In ABO, we have,

x^o + 75^o + y^o = 180^o

35 + 75 + y = 180

y = 180 - 110 = 70

x = 35, y = 70 and z = 75.

Through F, draw KH || AB || CD

Now, KF || CD and FG is a transversal.

KFG = FGD = r^o (i)

[alternate angles]

Again AE || KF, and EF is a transversal.

So,AEF + KFE = 180^o

KFE = 180^o - p^o (ii)

Adding (i) and (ii) we get,

KFG + KFE = 180 - p + r

EFG = 180 - p + r

q = 180 - p + r

i.e.,p + q - r = 180

PRQ = x^o = 60^o            [vertically opposite angles]

Since EF || GH, and RQ is a transversal.

So, x = y                   [Alternate angles]

y = 60

AB || CD and PR is a transversal.

So,         [Alternate angles]

     [since ]

x + QRD = 110^o

QRD = 110^o - 60^o = 50^o

In QRS, we have,

QRD + t^o + y^o = 180^o

50 + t + 60 = 180

t = 180 - 110 = 70

Since, AB || CD and GH is a transversal

So, z^o = t^o = 70^o [Alternate angles]

x = 60 , y = 60, z = 70 and t = 70

AB ∥ CD and a transversal t cuts them at E and F respectively.

⇒ ∠BEF + ∠DFE = 180° (interior angles)

⇒ ∠GEF + ∠GFE = 90° ….(i)

Now, in ΔGEF, by angle sum property

∠GEF + ∠GFE + ∠EGF = 180° 

⇒ 90° + ∠EGF = 180° ….[From (i)]

⇒ ∠EGF = 90° 

Since AB ∥ CD and t is a transversal, we have

∠AEF = ∠EFD (alternate angles)

⇒ ∠PEF = ∠EFQ

But, these are alternate interior angles formed when the transversal EF cuts EP and FQ.

∴ EP ∥ FQ

Given, ∠7 = 80° 

Now, ∠7 + ∠8 = 180° (linear pair)

⇒ 80° + ∠8 = 180° 

⇒ ∠8 = 100° 

∠7 = ∠5 (vertically opposite angles)

⇒ ∠5 = 80° 

Also, ∠6 = ∠8 (vertically opposite angles)

⇒ ∠6 = 100° 

Line l ∥ line m and line t is a transversal.

⇒ ∠1 = ∠5 = 80°  (corresponding angles)

 ∠2 = ∠6 = 100° (corresponding angles)

 ∠3 = ∠7 = 80°  (corresponding angles)

 ∠4 = ∠8 = 100° (corresponding angles) 

Construction: Produce DE to meet BC at Z.

Now, AB ∥ DZ and BC is the transversal.

⇒ ∠ABC = ∠DZC (corresponding angles) ….(i)

Also, EF ∥ BC and DZ is the transversal.

⇒ ∠DZC = ∠DEF (corresponding angles) ….(ii)

From (i) and (ii), we have

∠ABC = ∠DEF 

Construction: Produce ED to meet BC at Z.

Now, AB ∥ EZ and BC is the transversal.

⇒ ∠ABZ + ∠EZB = 180° (interior angles)

⇒ ∠ABC + ∠EZB = 180° ….(i)

Also, EF ∥ BC and EZ is the transversal.

⇒ ∠BZE = ∠ZEF (alternate angles)

⇒ ∠BZE = ∠DEF ….(ii)

From (i) and (ii), we have

∠ABC + ∠DEF = 180° 

Let the normal to mirrors m and n intersect at P.

Now, OB ⊥ m, OC ⊥ n and m ⊥ n.

⇒ OB ⊥ OC

⇒ ∠APB = 90° 

⇒ ∠2 + ∠3 = 90° (sum of acute angles of a right triangle is 90°)

By the laws of reflection, we have

∠1 = ∠2 and ∠4 = ∠3 (angle of incidence = angle of reflection)

⇒ ∠1 + ∠4 = ∠2 + ∠3 = 90° 

⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180° 

⇒ ∠CAB + ∠ABD = 180° 

But, ∠CAB and ∠ABD are consecutive interior angles formed, when the transversal AB cuts CA and BD.

∴ CA ∥ BD 

In the given figure,

∠BAC = ∠ACD = 110° 

But, these are alternate angles when transversal AC cuts AB and CD.

Hence, AB ∥ CD. 

Let the two parallel lines be m and n.

Let p ⊥ m.

⇒ ∠1 = 90° 

Let q ⊥ n.

⇒ ∠2 = 90° 

Now, m ∥ n and p is a transversal.

⇒ ∠1 = ∠3 (corresponding angles)

⇒ ∠3 = 90° 

⇒ ∠3 = ∠2 (each 90°)

But, these are corresponding angles, when transversal n cuts lines p and q.

∴ p ∥ q. 

Hence, two lines which are perpendicular to two parallel lines, are parallel to each other. 

Given, ∠1 : ∠2 = 2 : 3

Now, ∠1 + ∠2 = 180° (linear pair)

⇒ 2x + 3x = 180° 

⇒ 5x = 180° 

⇒ x = 36° 

⇒ ∠1 = 2x = 72° and ∠2 = 3x = 108° 

∠1 = ∠3 (vertically opposite angles)

⇒ ∠3 = 72° 

Also, ∠2 = ∠4  (vertically opposite angles)

⇒ ∠4 = 108° 

Line l ∥ line m and line t is a transversal.

⇒ ∠5 = ∠1 = 72°  (corresponding angles)

 ∠6 = ∠2 = 108° (corresponding angles)

 ∠7 = ∠3 = 72°  (corresponding angles)

 ∠8 = ∠4 = 108° (corresponding angles) 

Lines l and m will be parallel if 3x - 20 = 2x + 10

[Since, if corresponding angles are equal, lines are parallel]

3x - 2x = 10 + 20

x = 30

For lines l and m to be parallel to each other, the corresponding angles (3x + 5)° and (4x)° should be equal.

⇒ (3x + 5)° = 4x° 

⇒ x = 5° 

Since AB || CD and BC is a transversal.

So, BCD = ABC = x^o     [Alternate angles]

As BC || ED and CD is a transversal.

BCD + EDC = 180^o    

  BCD + 75^o =180^o

BCD = 180^o - 75^o = 105^o 

ABC = 105^o                 [since BCD = ABC]

x^o = ABC = 105^o

Hence, x = 105.

Since AB || CD and BC is a transversal.

So, ABC = BCD                [atternate interior angles]

70^o = x^o + ECD(i)

Now, CD || EF and CE is transversal.

So,ECD + CEF = 180^o    [sum of consecutive interior angles is 180^o]

ECD + 130^o = 180^o

ECD = 180^o - 130^o = 50^o

Putting ECD = 50^o in (i) we get,

70^o = x^o + 50^o

x = 70 - 50 = 20

AB ∥ CD and EF is transversal.

⇒ ∠AEF = ∠EFG (alternate angles)

Given, ∠AEF = 75° 

⇒ ∠EFG = y = 75° 

Now, ∠EFC + ∠EFG = 180° (linear pair)

⇒ x + y = 180° 

⇒ x + 75° = 180° 

⇒ x = 105° 

∠EGD = ∠EFG + ∠FEG (Exterior angle property)

⇒ 125° = y + z

⇒ 125° = 75° + z

⇒ z = 50° 

Thus, x = 105°, y = 75° and z = 50° 

Through E draw EG || CD. Now since EG||CD and ED is a transversal.

So,GED = EDC = 65^o[Alternate interior angles]

Since EG || CD and AB || CD,

EG||AB and EB is transversal.

So,BEG = ABE = 35^o[Alternate interior angles]

So,DEB = x^o

BEG + GED = 35^o + 65^o = 100^o.

Hence, x = 100.

Through O draw OF||CD.

Now since OF || CD and OD is transversal.

CDO + FOD = 180^o

[sum of consecutive interior angles is 180^o]

25^o + FOD = 180^o

FOD = 180^o - 25^o = 155^o

As OF || CD and AB || CD [Given]

Thus, OF || AB and OB is a transversal.

So,ABO + FOB = 180^o    [sum of consecutive interior angles is 180^o]

55^o + FOB = 180^o

FOB = 180^o - 55^o = 125^o

Now, x^o = FOB + FOD = 125^o + 155^o = 280^o.

Hence, x = 280.

Through E, draw EF || CD.

Now since EF || CD and EC is transversal.

FEC + ECD = 180^o

[sum of consecutive interior angles is 180^o]

FEC + 124^o = 180^o

FEC = 180^o - 124^o = 56^o

Since EF || CD and AB ||CD

So, EF || AB and AE is a trasveral.

So,BAE + FEA = 180^o

[sum of consecutive interior angles is 180^o]

116^o + FEA = 180^o

FEA = 180^o - 116^o = 64^o

Thus,x^o = FEA + FEC

= 64^o + 56^o = 120^o.

Hence, x = 120.

Correct option: (d)

In a right triangle, one angle is 90° and the sum of acute angles of a right triangle is 90°.

Correct option: (a)

Option (a) is false, since through a given point we can draw an infinite number of straight lines.

Correct option: (b)

Let each interior opposite angle be x.

Then, x + x = 110° (Exterior angle property of a triangle)

⇒ 2x = 110° 

⇒ x = 55° 

Correct option: (c)

Let ∠AOC = x° 

Draw YOZ ∥ CD ∥ AB.

Now, YO ∥ AB and OA is the transversal.

⇒ ∠YOA = ∠OAB = 60° (alternate angles)

Again, OZ ∥ CD and OC is the transversal.

⇒ ∠COZ + ∠OCD = 180° (interior angles)

⇒ ∠COZ + 110° = 180° 

⇒ ∠COZ = 70° 

Now, ∠YOZ = 180° (straight angle)

⇒ ∠YOA + ∠AOC + ∠COZ = 180° 

⇒ 60° + x + 70° = 180° 

⇒ x = 50° 

⇒ ∠AOC = 50° 

Correct option: (d)

Let ∠A = 130° 

In ΔABC, by angle sum property,

∠B + ∠C + ∠A = 180° 

⇒ ∠B + ∠C + 130° = 180° 

⇒ ∠B + ∠C = 50° 

Correct option: (b)

AOB is a straight line.

⇒ ∠AOB = 180° 

⇒ 60° + 5x° + 3x° = 180° 

⇒ 60° + 8x° = 180° 

⇒ 8x° = 120° 

⇒ x = 15° 

Correct option: (c)

By angle sum property,

2x + 3x + 4x = 180° 

⇒ 9x = 180° 

⇒ x = 20° 

Hence, largest angle = 4x = 4(20°) = 80° 

Correct option: (c)

Through B draw YBZ ∥ OA ∥ CD.

Now, OA ∥ YB and AB is the transversal.

⇒ ∠OAB + ∠YBA = 180° (interior angles are supplementary)

⇒ 110° + ∠YBA = 180° 

⇒ ∠YBA = 70° 

Also, CD ∥ BZ and BC is the transversal.

⇒ ∠DCB + ∠CBZ = 180° (interior angles are supplementary)

⇒ 130° + ∠CBZ = 180° 

⇒ ∠CBZ = 50° 

Now, ∠YBZ = 180° (straight angle)

⇒ ∠YBA + ∠ABC + ∠CBZ = 180° 

⇒ 70° + x + 50° = 180° 

⇒ x = 60° 

⇒ ∠ABC = 60° 

Correct option: (a)

Two angles are said to be complementary, if the sum of their measures is 90°.

Clearly, the measures of each of the angles have to be less than 90°.

Hence, each angle is an acute angle.

Correct option: (d)

An angle which measures more than 180^o but less than 360^o is called a reflex angle.