Lines And Angles

The angle between North and West is 90^° and between South and East is also 90^°.

As 90^°+ 90^°= 180^°

So they are supplementary angles.

The angle between South and West is 90^° and between South and East is also 90^°.

So the sum of this two angles is equal to 180^°.

Since this two angles are adjacent to each other so they form a linear pair.

Angle of incidence(i) = Angle of reflection(r)

∠ABC = Angle of incidence(i) + Angle of reflection(r)

∠ABC = 46° (Given)

⇒ 2i = 46^°

⇒ i = 23^°

∠ABP + i = 90^°

⇒ ∠ABP = 67^°

Complement means the sum of angles is 90^°

Complement of 79^° = 90^°-79^° = 11^°

Vertically opposite angles are always equal to each other. So they can be supplementary if and only if their value is half of 180^° i.e. 90^°

Linear pair means the sum of the two angle is 180^°

Angle which makes a linear pair = 180^°-61^° = 119^°

The sum of the two angles x and 90° – x is 90^°

Hence they are complementary

The sum of the two angles is 180^°.Hence they are supplementary.

Sum of all four angles about the point = 360^°

⇒ x + 100^° + 46^° + 64^° = 360^°

⇒ x + 210^° = 360^°

⇒ x = 150^°

∠APQ = 50°(Given)

⇒ ∠PQR = 50°(Alternate interior angles)

∠PRD = 130°(Given)

∠PRD = (180^°- 130°) = 50^° (Forms a Linear pair)

Sum of interior angles of a triangle = 180^°

∠ QPR = 180^°- ∠PQR-∠PRD

∠ QPR = 180^°-100^°

∠ QPR = 80^°

l and m intersect at a point

∠a = ∠b (Vertically Opposite)

∠d = ∠c (Vertically Opposite)

∠a + ∠d = 180° (Forms a linear pair)

Hence Option A, B & C are correct

∠a ≠ ∠d

Sum of two angles P and Q when they are supplementary is 180^°

∠P + ∠Q = 180^°

⇒ ∠Q = 180^°-60^°

⇒ ∠Q = 120^°

∠ POQ and ∠ QOR forms a linear pair

(3a + 5)^° + (2a-25)^° = 180^°

⇒ 5a - 20^° = 180^°

⇒ 5a = 200^°

⇒ a = 40^°

∠ POS, ∠ SOR & ∠ ROQ forms a linear pair

x + 5y = 180^°

⇒ 5y = 150^°

⇒ y = 30^°

∠QOR = 3y

⇒ ∠QOR = 90^°

Let one angle be x and the other angle 4x

Since the two angles are supplementary, hence

x + 4x = 180^°

⇒ 5x = 180^°

⇒ x = 36^°

Measure of angle = 4x = 144^°

The three angles forms a linear pair

6y + y + 2y = 180^°

⇒ 9y = 180^°

⇒ y = 20^°

a = ∠PAB = 50^° (Alternate interior angles)

a = b (Adjacent angles of a parallelogram are supplementary)

a + b = 180^°

⇒ b = 130^°

Let one angle be x and the other angle be (x + 30^°)

Sum of complementary angles is 90^°

x + (x + 30^°) = 90^°

⇒ 2x = 60^°

⇒ x = 30^°

x + 30^° = 60^°

a = ∠ SRP = 20° (Alternate interior angle)

b = ∠ QRP = 50° (Alternate interior angle)

l is a transversal of the two line segments m and n

The angles a and b lies on the inner side of each line segment but on opposite sides of transversal. So the angles are known as alternate interior angles.

Ratio = 1 : 2

Let the angles be x and 2x

Sum of supplementary angles is 180°

x + 2x = 180°

⇒ 3x = 180°

⇒ x = 60°

The bigger angle = 2 × 60° = 120°

Since ∠ROS, ∠POR and ∠QOS forms a linear pair

∠ROS is a right angle

So ∠POR and ∠QOS are complementary angles

Let ∠POR be x, ∠QOS be 5x

x + 5x = 90°

⇒ 6x = 90°

⇒ x = 15°

∠QOS = 15°× 5 = 75°

When two lines intersect then the vertically opposite angles are always equal.

When a transversal intersects, two other lines, then the sum of two interior angles on the same side of the transversal is 180° if and only if the two lines are parallel. The pair of angles are known as co-interior angles.

So both a is true and b is false.

Summation of a and b is 180⁰, hence they form a linear pair.

Angle a and b have a common vertex O and a common side OC. So they form a pair of adjacent angles.

So both p and q statements are true.

Sum of ∠AOC and ∠BOC is 180°

Hence they forms supplementary angles

Let the point of intersection of all the three lines be O

∠ AOF = ∠ COD = 90°

∠ BOC, ∠ COD and ∠ DOE forms a linear pair

Since ∠ COD = 90°

So ∠ BOC and ∠ DOE are complementary angles

40° + 5a = 90°

⇒ 5a = 50°

⇒ a = 10°

We draw an imaginary straight line UT ||QP || SR

∠ UTS = ∠ TSR = 30° (Alternate Interior angle)

∠ UTQ = ∠TQP = 60° (Alternate Interior angle)

a = ∠ UTQ + ∠UTS

⇒ a = 60° + 30° = 90°

Adjacent angles share a common vertex and a common side but no common interior points which is found only in option D.

Adjacent angles share a common vertex and a common side but no common interior points. But it is not mandatory for the uncommon arms to have opposite rays.

x and y are complementary pairs while y and z are supplementary pairs.

∠POR + ∠QOR = 180⁰( Forms a linear pair)

⇒ ∠QOR = 90⁰

Let x = 3t, y = 2t

∠QOR = x + y

⇒ ∠QOR = 5t

⇒ 90⁰ = 5t

⇒ t = 18⁰

y = 2t = 36⁰

y + z = 180⁰

⇒ z = 180⁰-36⁰

⇒ z = 144⁰

∠ POR and ∠ ROQ forms a linear pair

a + ∠ POR = 180⁰

⇒ a = 180⁰-100⁰ = 80⁰

Vertically Opposite angles are formed when two lines intersect at a point. This angles are always equal.

Angle 5b and 2a forms a linear pair

5b + 2a = 180⁰

⇒ 5b = 180⁰-80⁰

⇒ 5b = 100⁰

⇒ b = 20⁰

Let one angle be x other (180⁰- x)

According to the problem:

2x- (180⁰-x) = 60⁰

⇒ 3x = 240⁰

⇒ x = 80⁰

Supplement of x = 180⁰ - 80⁰ = 100⁰

∠1 = (2a + b)° …Equation (i)

∠6 = (3a–b)°

Since ∠5 and ∠6 forms a linear pair so

∠5 = (180-3a + b)°

∠5 = ∠1 = (180-3a + b)° (Corresponding angles) …Equation (ii)

Equating Equation (i) and Equation (ii) we get

2a + b = 180-3a + b

⇒ 5a = 180

⇒ a = 36⁰

Since ∠1 and ∠2 forms a linear pair so

∠2 = 180⁰- 2a-b

⇒ ∠2 = (108-b)⁰

Let a be 3x and b be 2x

Since a and b forms a linear pair, so

a + b = 180⁰

⇒ 5x = 180⁰

⇒ x = 36⁰

⇒ a = 3x = 108⁰

a = f = 108⁰ (Corresponding angle)

According to this figure

∠ 1 = ∠ 3 (Corresponding angles)

∠2 = ∠4 (Corresponding angles)

∠6 = ∠5 (Vertically Opposite angle)

∠5 = ∠7 (Corresponding angles)

So we can say

∠6 = ∠7

∠2 and ∠8 are supplementary pair

Since ∠2 = ∠4

So ∠4 and ∠8 are also supplementary pair

Hence ∠4≠∠8

∠1 = ∠3 (Corresponding angles)

∠3 + ∠5 = 180⁰ (Supplementary pairs)

So ∠1 + ∠5 = 180°

∠5 = ∠7 (Corresponding angles)

∠2 + ∠7 = 180⁰ (Supplementary pairs)

So ∠2 + ∠5 = 180°

∠1 = ∠3 (Corresponding angles)

∠1 + ∠8 = 180⁰ (Supplementary pairs)

So ∠3 + ∠8 = 180°

∠2 = ∠4 (Corresponding angles)

∠3 = ∠4 (Vertically Opposite angle)

So ∠3 = ∠2

Hence ∠3 and ∠2 are not supplementary pairs

∠5 = ∠7 (Corresponding angle)

∠7 = ∠8 (Vertically Opposite angle)

Combining the above result we can say that

∠5 = ∠8

y + ∠ PQR = 180⁰ (Forms a linear pair)

⇒ y + 130⁰ = 180⁰

⇒ y = 50⁰

∠ QOS = ∠ TSO (Co-interior angle)

⇒ x = 85⁰

⇒x + y = 135⁰

∠ RQP = ∠ TSR = 85⁰ (Corresponding angles)

a + ∠ TSR = 180⁰

⇒ a = 95⁰

Complementary

Complementary, since the sum of measures of two angles is 90⁰

Supplementary

Supplementary, since the sum of measures of two angles is 180⁰

Distinct

A transversal intersects two or more than two lines at distinct points.

180⁰

Sum of interior angles on the same side of a transversal is 180⁰

Vertex

Alternate interior angle have one common vertex.

Same

Corresponding angles are on the same side of the transversal.

Opposite

Alternate interior angles are on the opposite side of the transversal.

Parallel

Two lines in a plane which do not meet at a point anywhere are called parallel lines

Linear

Two angles forming a linear pair are supplementary.

Obtuse

The supplement of an acute angle is always obtuse angle since acute angle is always less than 90⁰

Right

The supplement of a right angle is always a right angle since a right angle = 90⁰

Supplement = 180⁰-90⁰ = 90⁰

Acute

The supplement of an obtuse angle is always acute angle since obtuse angle is always more than 90⁰

90⁰

In a pair of complementary angles, each angle cannot be more than 90⁰

45⁰

Complementary angle = (90⁰-45⁰) = 45⁰

60⁰

Let the angle be x, supplement be 2x

x + 2x = 180⁰

⇒ 3x = 180⁰

⇒ x = 60⁰

False

Sum of two right angles = 180⁰. So they are always supplementary to each other. Only two acute angles can be complementary to each other since their sum has to be 90⁰

False

An Obtuse angle is always more than 90⁰, so it can never form a complementary pair.

False

An Obtuse angle is always more than 90⁰, so sum of angles of two obtuse angles will always be greater than 180⁰ and can never form a supplementary pair.

True

Sum of two right angles is always 180⁰ and so they are always supplementary.

True

Obtuse angles are more than 90⁰ and acute angles are less than 90⁰. So they can form a supplementary pair.

True

Acute angles are less than 90⁰, so they can’t form a supplementary pair.

False

All linear pairs forms two supplementary angles but the reverse is not always true.

True

All linear pairs forms two supplementary angles but the reverse is not always true.

True

The angles are on a straight line so they share a common vertex and arm. Hence they are always adjacent.

False

Vertically Opposite angles are always equal. Linear pairs can only be formed by adjacent angles.

False

Interior angles on the same side of a transversal with two distinct parallel lines are supplementary angles.

True

Vertically Opposite angles are always equal.

False

The linear pair has one acute and one obtuse angle which are adjacent to each other to add up to 180⁰.

True

Complementary angles always add up to 90⁰. When one angle is more than 45⁰ the other angle is always less than 45⁰ to make the sum up to 90⁰.

False

Two angles forming a linear pair are always adjacent but the reverse is not always true.

(a) The adjacent pairs are:

● ∠ DOC and ∠ COB

● ∠ COB and ∠ AOB

● ∠ DOC and ∠ COA

● ∠ DOB and ∠ AOB

(b) The adjacent pairs are:

● ∠TQP and ∠PQR

● ∠PRQ and ∠QRU

● ∠SPR and ∠PRQ

(c) The adjacent pairs are:

● ∠TSV and ∠VSU

● ∠TVS and ∠SVU

(d) The adjacent pairs are:

● ∠ AOC and ∠AOD

● ∠ AOC and ∠BOC

● ∠ BOD and ∠BOC

● ∠ BOD and ∠AOD

(i) Pair of vertically opposite angles:

● ∠ 1 and ∠ 3

● ∠2 and ∠4

● ∠6 and ∠8

● ∠5 and ∠7

Linear Pairs:

● ∠ 1 and ∠ 2

● ∠ 2 and ∠ 3

● ∠ 3 and ∠ 4

● ∠ 4 and ∠ 1

● ∠5 and ∠6

● ∠6 and ∠7

● ∠7 and ∠8

● ∠8 and ∠5

(ii) Pair of vertically opposite angles:

No vertically Opposite angles are present

Linear Pairs:

● ∠ABD and ∠DBC

● ∠ABE and ∠EBC

(iii) Pair of vertically opposite angles:

No vertically Opposite angles are present

Linear Pairs:

No linear pairs

(iv) Pair of vertically opposite angles:

● ∠ ROP and ∠QOS

● ∠ROQ and ∠ POS

Linear pairs:

● ∠ ROP and ∠QOR

● ∠ ROQ and ∠QOS

● ∠ SOP and ∠QOS

● ∠ ROP and∠ SOP

● ∠ ROT and ∠ TOS

● ∠ POT and ∠ TOQ

(i) The pairs of Supplementary angles are :

● ∠ AOC and ∠AOD

● ∠ AOC and ∠BOC

● ∠ BOD and ∠BOC

● ∠ BOD and ∠AOD

(ii) The pairs of Supplementary angles are :

● ∠ POR and ∠ QOR

● ∠ POS and ∠ QOS

(iii) The pairs of Supplementary angles are :

● ∠ 1 and ∠ 2

● ∠ 3 and ∠ 4

● ∠ 5 and ∠ 6

∠PTR = ∠TRU = 42° (Alternate Interior angle)

∠ TRU + ∠QUR = 180⁰ (Co interior angle)

∠ QUR = (180⁰ -42°) = 138⁰

(a) The player 4 has the best kicking angle since its position is the best.

(b) The player 4 has the best kicking angle since its position is the midway between all the players.

c) When the angles are complementary the sum is always 90⁰

The two different positions of the player may be

i.) 30⁰ and 60⁰

ii.) 0⁰ and 90⁰

Vertically Opposite angles are always equal

Sum of two vertically opposite angles = 166°

Each angle

∠QPS = ∠PQA = 35° (Alternate Interior angle)

∠QRT = ∠RQA = 55° (Alternate Interior angle)

∠PQA + ∠RQA = ∠PQR

⇒ ∠PQR = 90⁰

Pair of complementary angles:

i) ∠TQS and ∠SQR

Two Pairs of Supplementary angles:

i) ∠PQT and ∠TQR

ii) ∠PQS and ∠SQR

Four pairs of adjacent angles:

i) ∠TQS and ∠SQR

ii) ∠PQT and ∠TQR

iii) ∠PQS and ∠SQR

iv) ∠PQT and ∠TQS

Pairs of Adjacent angles:

i) ∠ x and ∠y

ii) ∠y and ∠z

iii) ∠x + y and ∠z

iv) ∠z + y and ∠x

Pair of complementary angle:

∠y and ∠x

A. There are all total 13 angles.

B. The four types of angles formed are

● Vertically Opposite Angles.

● Linear Pair

● Adjacent Angles

● Supplementary angles

C. The Pair of vertically opposite angles are:

● ∠ 1 and ∠ 3

● ∠ 2 and ∠ 4

i) ∠a and ∠b are adjacent since they share a common vertex and a common arm.

ii) ∠a and ∠b are not adjacent since they does not share a common arm.

iii) ∠a and ∠b are not adjacent since they does not share a common vertex

iv) ∠a and ∠b are not adjacent since they share common interior points.

The pairs of supplementary angles are :

i) 7 and 2

ii) 1 and 8

iii) 3 and 6

iv) 6 and 5

v) 5 and 4

vi) 4 and 3

A. If one angle is acute the other angle to form a linear pair will always be obtuse.

B. If one angle is obtuse the other angle to form a linear pair will always be acute.

C. If one angle is right angle then the other angle to form a linear pair will always be a right angle.

Two acute angles can never form a pair of supplementary angles.

Acute angles are always less than 90⁰. So their sum will always be less than 2 × 90⁰ = 180⁰.Hence they can never can form a pair of supplementary angles.

The pairs of adjacent angles are:

i) 1 and 2

ii) 2 and 3

iii) 3 and 4

iv) 4 and 1

Complement of an angle = 62⁰

Supplement of that angle = (90⁰ + 62⁰) = 152⁰

152⁰

Let ∠ x = 30⁰

⇒ ∠ y = ∠ x = 30⁰ (Corresponding angle)

∠ c = 180⁰-30⁰ = 150⁰( Forms a linear pair with ∠ y)

∠ 1 = ∠ y = 30⁰ (Vertically Opposite)

∠ a = ∠ 1 = 30⁰ (Corresponding angle)

∠ 2 = 180⁰-30⁰ = 150⁰( Forms a linear pair with ∠ a)

∠ b = ∠ 2 = 150⁰ (Alternate interior angle)

∠ a = 30⁰, ∠ b = 150⁰, ∠ c = 150⁰

∠ x = ∠ OQR = 35⁰ (Alternate Interior angle)

∠ x and ∠ y forms a linear pair

∠ y = 180⁰-35⁰ = 145⁰

∠ x = 35⁰ and ∠ y = 145⁰

∠ 1 = 30⁰ (Vertically Opposite angle)

∠4 = 75⁰ (Alternate Interior angle)

∠ 2 + ∠ QPT = 180⁰ (Co interior angles)

∠ 2 = 180⁰-75⁰ = 105⁰

∠ 3 = 75⁰ (Alternate Interior angle)

(i) b and c = ∠ 1 = 30⁰

(ii) d and e = ∠ 2 = 105⁰

(iii) d and f = ∠ 3 = 75⁰

(iv) c and f = ∠4 = 75⁰

∠A = 120⁰ (Vertically Opposite )

∠A + ∠ x = 180⁰ (Co-interior angle)

⇒ ∠x = 60⁰

∠B + ∠ x = 180⁰ (Co-interior angle)

∠B = 120⁰

⇒ ∠y = 120⁰ (Vertically Opposite angle)

∠ z = ∠x = 60⁰ (Alternate interior angle)

∠x = 60⁰, ∠y = 120⁰, ∠z = 60⁰

∠FED = ∠AFE = 42° (Alternate interior angle)

∠AFC, ∠AFE and ∠EFD are supplementary

∠EFD = 180⁰-(68⁰ + 42⁰) = 70⁰

∠BOC = 49° (Given)

∠AOC = 90⁰-49⁰ = 41⁰ (Complementary angles)

∠AOC and ∠AOD forms a linear pair

∠AOD = 180⁰-41⁰ = 139⁰

∠AOE, ∠DOB and ∠ EOD forms a linear pair

∠AOE = 30°

∠DOB = 40°

∠ EOD = 180⁰-∠AOE- ∠DOB

⇒ ∠ EOD = 110⁰

∠COF = ∠ EOD = 110⁰( Vertically Opposite)

∠COF = 110⁰

Let one angle be x other angle x + 2⁰

Since the two angles are complementary their sum is 90⁰

x + x + 2 = 90⁰

⇒ 2x = 88⁰

⇒ x = 44⁰

The two angles are 44⁰ and 46⁰

Let one angle be x and the other angle be x + 20⁰

Sum of two interior angles on the same side of a transversal = 180⁰

x + x + 20⁰ = 180⁰

⇒ 2x = 160⁰

⇒ x = 80⁰

The two angles are 80⁰ and 100⁰

Let one angle be x and the other angle be 3x

Since the two angles forms a linear pair their sum is 180⁰

x + 3x = 180⁰

⇒ 4x = 180⁰

⇒ x = 45⁰

The two angles are 45⁰, 135⁰

Let one angle be x other angle x + 2

Since the two angles are supplementary their sum is 180⁰

x + x + 2 = 180⁰

⇒ 2x = 178⁰

⇒ x = 89⁰

The two angles are 89⁰ and 91⁰

∠ HCB = ∠CHE = 120⁰ (Alternate Interior angles)

∠ HCB + ∠ ABC = 180⁰ ( Co interior angles)

⇒ ∠ ABC = 60⁰

∠ HCB = ∠CDE = 120⁰ (Corresponding angle)

∠ ABC = 60⁰ and ∠CDE = 120⁰

∠ AOD, ∠ DOC and ∠ COB are linear pairs.

So the sum of this three angles is 180⁰

(x-10) + (4x-25) + (x + 5) = 180

⇒ 6x = 180 + 30

⇒ 6x = 210

⇒ x = 35⁰

∠BOC = (x + 5)⁰

= 40⁰

∠ x = 180⁰-132⁰ = 48⁰(Forms a linear pair)

∠a = 180⁰-(65⁰ + 48⁰) = 67⁰ (Interior angles of a triangle are supplementary)

∠a, ∠b and ∠y are supplementary

∠ y = 65⁰ (Corresponding angle)

∠b = 180⁰-(67⁰ + 65⁰) = 48⁰

∠a = 67⁰, ∠b = 48⁰

∠a = 132⁰ (Corresponding angle)

∠b = 132⁰ (Vertically Opposite angle)

2a + b = 396⁰

Since QP || RS

∠a = 65⁰ (Alternate Interior angle)

∠b = 70⁰ (Corresponding angle)

Since PQ || RT

∠b = 55⁰ (Alternate Interior angle)

∠a = 45⁰ (Corresponding angle)

∠a + ∠b = 100⁰

(i) a + b = c

(Given)

⇒ b = 38⁰

d + b = 180⁰(Co- interior angle)

⇒ d = 142⁰

(ii) c = 75⁰(Given)

a = c

⇒ a = 30⁰

a + b = c

⇒ b = c - a

⇒ b = (75⁰- 30⁰) = 45⁰

∠ 2 + ∠ FGD = 180⁰(Co Interior angle)

∠ 2 = 180⁰-135⁰ = 45⁰

∠ 1 = ∠ AEF = 34⁰(Alternate Interior angle)

∠ 1 + ∠ 2 = 79⁰

Reflex ∠EFG = (360⁰-(∠ 1 + ∠ 2)) = 281⁰

x + 66⁰ = 180⁰(Co Interior angle)

⇒ x = 114⁰

y + 48⁰ = 180⁰(Co Interior angle)

⇒ y = 132⁰

x = 114⁰ and y = 132⁰

∠ 6a = ∠3b = ∠4c = 120⁰ (Corresponding angle)

∠ 6a = 120⁰

⇒ ∠ a = 20⁰

∠3b = 120⁰

⇒ ∠b = 40⁰

∠4c = 120⁰

⇒ ∠c = 30⁰

∠ a = 20⁰, ∠b = 40⁰, ∠c = 30⁰

Supplement of 120⁰ = 60⁰

So the transversal l meets the line m and n forming an angle of 60⁰ on the same side of transversal

Therefore this two angles are corresponding pairs

Hence m and n are parallel

i) ∠ CQF≠∠ SRD (Alternate Exterior angle)

Hence EF is not || GH

ii) ∠ CQF = ∠ RQF = 65⁰ (Vertically Opposite)

∠ RQF + ∠ SPQ = 115⁰ + 65⁰ = 180⁰

Hence they form a pair of corresponding interior angle. So AB || CD

∠ VUF + ∠ TUV = 180⁰

∠ TUV = 180⁰-123⁰ = 57⁰

∠ TUV = ∠UVH

So they form alternate interior pairs

Hence EF||GH

∠UVH≠∠ VQP

Hence PK is not parallel to EF and GH

∠ VUF≠∠RSV

Hence AB is not parallel to CD as they don’t form corresponding pairs.

EF||GH

Let P,Q,R,S be the four intersecting point

∠ FQS and ∠ SQP forms a linear pair

∠ SQP = 180⁰-50⁰ = 130⁰

∠ APQ = 130⁰

∠ APQ = ∠ SQP = 130⁰

∠ APQ & ∠ SPQ are alternate interior angles

Hence AB || CD

∠ APQ and ∠ QPR forms a linear pair

∠ QPR = 180⁰-130⁰ = 50⁰

∠ QPR = ∠ GRP = 50⁰

So ∠ QPR and ∠ GRP are alternate interior angles

Hence EF || GH

y + 80⁰ = 180⁰ (Co- interior angles)

⇒ y = 100⁰

x = 110⁰ (Alternate interior angles)

x = 110⁰, y = 100⁰