Lines And Angles

(i) Given angle is 20^o

Since the sum of an angle and its compliment is 90^o

Therefore, its compliment will be:

90^o – 20^o = 70^o

(ii) Given angle is 35^o

Since the sum of an angle and its compliment is 90^o

Therefore, its compliment will be:

90^o – 35^o = 55^o

(iii) Given angle is 90^o

Since the sum of an angle and its compliment is 90^o

Therefore, its compliment will be:

90^o – 90^o = 0^o

(iv) Given angle is 77^o

Since the sum of an angle and its compliment is 90^o

Therefore, its compliment will be:

90^o – 77^o = 13^o

(v) Given angle is 30^o

Since the sum of an angle and its compliment is 90^o

Therefore, its compliment will be:

90^o – 30^o = 60^o

(i) Given angle is 54^o

Since the sum of an angle and its supplement is 180^o

Therefore, its compliment will be:

180^o – 54^o = 126^o

(ii) Given angle is 132^o

Since the sum of an angle and its supplement is 180^o

Therefore, its compliment will be:

180^o – 132^o = 48^o

(iii) Given angle is 138^o

Since the sum of an angle and its supplement is 180^o

Therefore, its compliment will be:

180^o – 138^o = 42^o

Angle measured will be ‘x’ say

Therefore, its compliment will be (90^o – x)

It is given that angle = Compliment – 28^o

x = (90^o – x) – 28^o

2x = 62^o

x = 31^o

Let the angle be "x"

The, its complement will be (90^o – x)

Note: Complementary angles: When the sum of 2 angles is 90°.

It is given that angle = 30^o + Complement

x = 30^o + (90^o – x)

x = 30^o + 45^o - x/2

x + x/2 = 30^o + 45^o

= 75^o

3x = 150^o

x= 50⁰

Thus, the angle is 50⁰

Supplementary angles are in the ratio 4: 5

Let the angles be 4x and 5x.

It is given that they are supplementary angles

Therefore,

4x + 5x = 180^o

x = 20^o

Hence, 4x = 80^o

5x = 100^o

Therefore, angles are 80^o and 100^o.

Given that,

Two supplementary angles are differ by 48^o

Let, the angle measured be x^o

Therefore, its supplementary angle will be (180^o – x)

It is given that,

(180^o – x) – x = 48^o

2x = 180^o – 48^o

x = 66^o

Hence, 180^o – x = 114^o

Therefore, angles are 66^o and 114^o.

It is given that,

Angle = 8 times its compliment

Let x be the measured angle

Angle = 8 (Compliment)

Angle = 8 (90^o – x^o)

x = 720^o – 8x

9x = 720^o

x = 80^o

Given that,

(2x – 10)^o and (x – 5)^o are compliment angles.

Let x be measured angle

Since, angles are complimentary

Therefore,

(2x – 10)^o + (x – 5)^o = 90⁰

3x – 15^o = 90^o

x = 35^o

Let the angle measured be x

Compliment angle = (90^o – x)

Supplement angle = (180^o – x)

Given that,

Supplementary of thrice of the angle = (180^o – 3x)

According to question,

(90^o – x) = (180^o – 3x)

2x = 90^o

x = 45^o

Let the angle measured be x

Given that,

The angles measured will be differ by 10^o

x^o – (90^o – x) = 10^o

2x = 100^o

x = 50^o

Given that,

Supplement angle = 3 times its compliment angle

Let the angle measured be x

According to the question

(180^o – x) = 3 (90^o – x)

2x = 90^o

x = 45^o

Given that,

Supplement = of the angle itself

Let the angle be x

Therefore,

Supplement = (180^o – x)

According to the question

(180^o – x) =

5x = 540^o

x = 108^o

Hence, supplement = 72^o

Therefore, the angle will be 108^o and its supplement will be 72^o.

Given that,

An angle is 14 more than its compliment

Let the angle be x

Compliment = (90^o – x)

According to the question,

x – (90^o – x) = 14

2x = 90^o + 14^o

x = 52^o

Given that,

Angle measured is twice its supplement

Let the angle measured be x

Therefore,

Supplement = (180^o – x)

According to the question

x^o = 2 (180^o – x)

3x = 360^o

x = 120^o

(i) Given that,

x = 25^o

∠AOC +∠BOC = 180^o (Linear pair)

(2y + 5) + 3x = 180^o

(2y + 5) + 3 (25) = 180^o

2y = 100^o

y = 50^o

(ii) Given that,

If y = 35^o

∠AOC +∠BOC = 180^o

(2y + 5) + 3x = 180^o

2 (35) + 5 + 3x = 180^o

3x = 105^o

x = 35^o

Adjacent angles are:

(i) ∠AOC, ∠COB

(ii) ∠AOD, ∠BOD

(iii) ∠AOD, ∠COD

(iv) ∠BOC, ∠COD

Linear pairs are:

∠AOD,∠BOD and

∠AOC,∠BOC

∠AOD +∠BOD = 180^o (Linear pair)

∠AOD +∠COD +∠BOC = 180^o (Linear pair)

Given that,

∠AOD = (x + 10)

∠COD = x

∠BOC = (x + 20)

(x + 10) + x + (x + 20) = 180⁰

3x = 150^o

x = 50^o

Therefore,

∠AOD = 60^o

∠COD = 50^o

∠BOC = 70^o

Given that,

The rays OA, OB, OC, OD, and OE have the common endpoint O.

A ray of opposite to OA is drawn.

Since,

∠AOB,∠BOF are linear pair

∠AOB +∠BOF = 180^o

∠AOB +∠BOC +∠COF = 180^o (i)

Also,

∠AOE +∠EOF = 180^o

∠AOE +∠DOF +∠DOE = 180^o (ii)

Adding (i) and (ii), we get

∠AOB +∠BOC +∠COF +∠AOE +∠DOF +∠DOE = 360^o

∠AOB +∠BOC +∠COD +∠DOE +∠EOA = 360^o

Hence, proved

Given,

If (a – 2b) = 30^o

∠AOC = a

∠BOC = b

Therefore,

a + b = 180^o (i)

Given,

(a – 2b) = 30^o (ii)

Now,

Subtracting (i) and (ii), we get

a + b – a + 2b = 180^o – 30^o

b = 50^o

Hence,

(a – 2b) = 30^o

a – 2 (50) = 30^o

a = 130^o

Four pairs of adjacent angles formed when two lines intersect any point. They are:

∠AOD,∠DOB

∠DOB,∠BOC

∠COA,∠AOD

∠BOC,∠COA

Pairs of adjacent angles are:

∠EOC,∠DOC

∠EOD,∠DOB

∠DOC,∠COB

∠EOD,∠DOA

∠DOC,∠COA

∠BOC,∠BOA

∠BOA,∠BOD

∠BOA,∠BOE

∠EOC,∠COA

∠EOC,∠COB

Hence, ten pairs of adjacent angles.

Sum of all the angles around the point = 360^o

3x + 3x + 150^o + x = 360^o

7x = 360^o – 150^o

7x = 210^o

x = 30^o

∠AOB +∠BOC = 180^o (Linear pair)

70^o + 2x = 180^o

2x = 110^o

x = 55^o

∠POQ +∠QOS = 180^o (Linear pair)

∠POQ +∠QOR +∠SOR = 180^o

60^o + 4x + 40^o = 180^o

4x = 80^o

x = 20^o

∠ACD +∠BCD = 180^o(Linear pair)

5x + 4x = 180^o

9x = 180^o

x = 20^o

∠QOR +∠POR = 180^o(Linear pair)

2x + 10^o + 3x = 180^o

5x = 170^o

x = 34^o

a +b = 180^o (Linear pair)

a =180^o - b (i)

Now, given that

a = b + * 90^o

a = b + 30^o (ii)

a – b = 30^o

Equating (i) and (ii), we get

180^o – b = b + 30^o

150^o = 2b

b = 75^o

Hence, a = 180^o – b

= 105^o

∠AOC +∠BOC = 180^o

6y + 30^o + 4y = 180^o

10y = 150^o

y = 15^o

(i) Say,

∠FOE = x

∠DOE = y

∠BOC = z

∠AOF + 30^o = 180^o (∠AOF +∠FOG = 180^o)

∠AOF = 150^o

∠AOB +∠BOC +∠COD +∠DOE +∠EOF = 150^o

30^o + z + 30^o + y + x = 150^o

x + y + z = 90^o(i)

Now,

∠FOC = 90^o

∠FOE +∠EOD +∠DOC = 90^o

x + y + 30^o = 90^o

x + y = 60^o (ii)

Using (ii) in (i), we get

x + y + z = 90^o

60^o + z = 90^o

z = 30^o (∠BOC = 30^o)

∠BOE = 90^o

∠BOC +∠COD +∠DOE = 90^o

30^o + 30^o +∠DOE = 90^o

∠DOE = 30^o

Now, we have

x + y = 60^o

y = 30^o

∠FOE = 30^o

(ii) Right angles are:

∠DOG,∠COF,∠BOF,∠AOD

(iii) Three pairs of adjacent complimentary angles are:

∠AOB,∠BOD

∠AOC,∠COD

∠BOC,∠COE

(iv) Three pairs of adjacent supplementary angles are:

∠AOB,∠BOG

∠AOC,∠COG

∠AOD,∠DOG

(v) Three pairs of adjacent angles are:

∠BOC,∠COD

∠COD,∠DOE

∠DOE,∠EOF

Given that,

OP, OQ, OR and OS are four rays

You need to produce any of the rays OP, OQ, OR and OS backwards to a point T so that TOQ is a line.

Ray OP stands on line TOQ

∠TOP +∠POQ = 180^o (Linear pair) (i)

Similarly,

∠TOS +∠SOQ =180^o(ii)

∠TOS +∠SOR +∠OQR = 180^o (iii)

Adding (i) and (iii), we get

∠TOP +∠POQ +∠TOS +∠SOR +∠QOR = 360^o

∠TOP +∠TOS =∠POS

Therefore, ∠POQ +∠QOR +∠SOR +∠POS = 360^o.

Given that,

Ray OS stand on a line POQ

Ray OR and OT are angle bisector of ∠POS and ∠SOQ respectively.

∠POS = x

∠POS +∠QOS = 180^o(Linear pair)

∠QOS = 180^o – x

∠ROS =∠POS (Given)

=x

∠ROS =

Similarly,

∠TOS = (90^o -)

Therefore,

∠ROT =∠ROS +∠ROT

=+ 90^o -

= 90^o

Therefore, ∠ROT = 90^o

Given,

∠POR and∠ROP is linear pair

∠POR +∠ROP = 180^o

Given that,

∠POR =∠ROQ = 5: 7

Therefore,

∠POR = × 180^o = 75^o

Similarly,

∠ROQ = 125^o

Now,

∠POS =∠ROQ = 125^o(Vertically opposite angles)

Therefore,

∠SOQ =∠POR = 75^o(Vertically opposite angles)

OR perpendicular to PQ

Therefore,

∠POR = 90^o

∠POS +∠SOR = 90^o [Therefore, ∠POR =∠POS +∠SOR]

∠ROS = 90^o -∠POS (i)

∠QOR = 90^o (Therefore, OR perpendicular to PQ)

∠QOS -∠ROS = 90^o

∠ROS =∠QOS – 90^o(ii)

By adding (i) and (ii), we get

2∠ROS =∠QOS -∠POS

∠ROS =(∠QOS -∠POS)

Given that,

x = 45^o

y =?

z =?

u =?

z = x = 45^o (Vertically opposite angle)

z + u = 180^o (Linear pair)

45^o = 180^o – u

u = 135^o

x + y = 180^o (Linear pair)

45^o = 180^o – y

y = 135^o

Therefore, x = 45^o, y = 135^o, u = 135^o and z = 45^o.

Since,

Vertically opposite angles are equal

So,

∠BOD = z = 90^o

∠DOF = y = 50^o

Now,

x + y + z = 180^o (Linear pair)

x + 90^o + 50^o = 180⁰

x = 40^o

From the given figure,

∠y = 25^o(Vertically opposite angle)

(x + y) = 180^o (Linear pair)

x + 25^o = 180^o

x = 155^o

Also,

z = x = 155^o (Vertically opposite angle)

y = 25^o

∠AOE =∠BOF = 5x (Vertically opposite angle)

By Linear pair,

∠COA +∠AOE +∠EOD = 180^o

3x + 5x + 2x = 180^o

x = 18^o

Given that,

Lines AOB and COD intersect at point O such that,

∠AOC =∠BOD

Also,

OF is the bisector of ∠AOC and OE is the bisector of ∠BOD

To prove: EOF is a straight line.

∠AOD =∠BOC = 2x (Vertically opposite angle) (i)

∠AOD +∠BOD = 180° (linear pair)

From (1)

Given that,

AB and CD intersect at O

OF bisects ∠COB

To prove: ∠AOF =∠DOF

Proof:
OF bisects ∠COB [given]

(Vertically opposite angle)

∠BOE =∠AOF = x (i)

∠COE =∠DOF = x (ii)

From (i) and (ii), we get

∠AOF =∠DOF = x

Hence, proved

Given that,

AB and CD are two lines intersecting at O

To prove: ∠BOC = 90^o

∠AOC = 90^o

∠AOD = 90^o

∠BOD = 90^o

Proof: Given that,

∠BOC = 90^o

∠BOC =∠AOD = 90^o(Vertically opposite angle)

∠AOC +∠BOC = 180^o(Linear pair)

∠AOC + 90^o = 180^o

∠AOC = 90^o

∠AOC =∠BOD = 90^o(Vertically opposite angles)

Therefore,

∠AOC =∠BOC =∠BOD =∠AOD = 90^o

Hence, proved

(i) Given that,

x = 60^o

y =?

∠AOC +∠DOC = 180^o(Linear pair)

2x + y = 180^o

120^o + y = 180⁰

y = 60^o

(ii) Given that,

y = 40^o

x =?

∠AOC +∠BOC = 180^o (Linear pair)

2x + y = 180^o

2x = 140^o

x = 70^o

∠AOE +∠EOB = 180^o(Linear pair)

∠AOE +∠DOE +∠BOD = 180^o

∠DOE = 105^o

∠DOE =∠COF = 105^o(Vertically opposite angle)

∠AOE +∠AOF = 180^o(Linear pair)

40^o +∠AOC + 105^o= 180^o

∠AOC = 35^o

Also,

∠BOF =∠AOE = 40^o(Vertically opposite angle)

OF bisects ∠BOD

∠BOF = 35^o

∠BOC =?

∠AOD =?

∠BOD =∠BOF = 70^o(Therefore, OF bisects ∠BOD)

∠BOD =∠AOC = 70^o(Vertically opposite angle)

∠BOC +∠AOC = 180^o

∠BOC + 70^o = 180^o

∠BOC = 110^o

∠AOD =∠BOC = 110^o(Vertically opposite angle)

Given that,

∠AOC +∠BOE = 70^o

And,

∠BOD = 40^o

∠BOF =?

∠BOD =∠AOC = 40^o (Vertically opposite angle)

Given,

∠AOC +∠BOE = 70^o

40^o +∠BOE = 70^o

∠BOE = 70^o – 40^o

= 30^o

∠AOC and∠BOC are linear pair angle

∠AOC +∠COE +∠BOE = 180^o

∠COE = 180^o – 30^o – 40^o

= 110^o

Therefore,

∠COE = 360^o – 110^o

= 250^o

(i) True: Since, the angles form a sum of 180^o.

(ii) False: Since, the two angles unless are on the line are necessarily equal to 90^o.

(iii) False: Since, acute are less than 90^o and hence two acute angles cannot give a sum of 180^o

(iv) True: Since, sum of angles of linear pair is 180^o hence, if both the angles are equal they would measure 90^o.

(i) Obtuse angle

(ii) 180^o

(iii) Uncommon

Let, ∠1 = 3x

∠2 = 2x

∠1 +∠2 = 180^o(Linear pair)

3x + 2x = 180^o

5x = 180^o

x = 36^o

Therefore,

∠1 = 3x = 108^o

∠2 = 2x = 72^o

Vertically opposite angles are equal, so:

∠1 =∠3 = 108^o

∠2 =∠4 = 72^o

∠5 =∠7 = 108^o

∠6 =∠8 = 72^o

Corresponding angles:

∠1 =∠5 = 108^o

∠2 =∠6 = 72^o

From the given figure,

∠3 +∠myz = 180^o(Linear pair)

∠3 = 60^o

Now,

Line l ‖ m

∠1 =∠3 (Corresponding angles)

∠1 = 60^o

Now, m ‖ n

∠2 = 120^o(Alternate interior angle)

Therefore,

∠1 =∠3 = 60^o

∠2 = 120^o

Produce LK to meet GF at N

Now,

∠HGN = ∠CHG= 60^o(Alternate angle)

∠HGN =∠KNF = 60^o(Corresponding angles)

Therefore,

∠KNG = 120^o

∠GNK =∠AKL = 120^o(Corresponding angle)

∠AKH =∠KHD = 25^o(Alternate angles)

Therefore,

∠HKL =∠AKH +∠AKL

= 25^o + 120^o

= 145^o

Produce EF to intersect AC at K

Now,

∠DCE +∠CEF = 35^o + 145^o

= 180^o

Therefore, EF‖CD (Since, sum of co-interior angles is 180^o) (i)

Now,

∠BAC =∠ACD = 57^o

BA‖CD (Therefore, alternate angles are equal) (ii)

From (i) and (ii), we get

AB‖EF (Lines parallel to the same line are parallel to each other)

Hence, proved

Since,

EF ‖ CD

Therefore,

∠EFC +∠LEC = 180^o(Co. interior angles)

∠ECD = 180^o – 130^o

= 50^o

Also, BA ‖ CD

∠BAC =∠ACD = 70^o(Alternate angles)

But, ∠ACE +∠ECD = 70^o

∠ACE = 70^o -50^o

=20^o

AB is produced to meet PR at K

Since, PQ ‖ AB

∠QPR =∠BKR = 102^o (Corresponding angles)

Since, PR ‖ BC

Therefore,

∠RKB +∠CBK = 180^o (Co. interior angles)

∠CKB = 180^o – 102^o

= 78^o

∠EOC =∠DOK = 100^o(Vertically opposite angle)

∠DOK =∠ACO = 100^o(Vertically opposite angle)

Here two lines, EK and CA cut by a third line L and the corresponding angles to it are equal.

Therefore,

EK ‖ AC

Corresponding angles are equal so,

∠1 =∠3 = 85^o

By using co-interior angle property,

∠2 +∠3 = 180^o

∠2 + 85^o = 180^o

∠2 = 95^o

Given m and l perpendicular to t

∠1 =∠2 = 90^o

Since,

l and m are two lines and t is transversal and the corresponding angles are equal.

Therefore,

l ‖ m

Hence, proved

Consider the angles,

∠AOB and∠ACB

Given that,

OA perpendicular AO and OB perpendicular BO

To prove: ∠AOB =∠ACB or,

∠AOB +∠ACB = 180^o

Proof: In a quadrilateral

∠A +∠O +∠B +∠C = 360^o(Sum of angles of a quadrilateral)

180^o + ∠O +∠C = 360^o

∠O +∠C = 180^o

Hence, ∠AOB +∠AOC = 180^o (i)

Also,

∠B +∠ACB = 180^o

∠ACB = 180^o – 90^o

∠ACB = 90^o (ii)

From (i) and (ii), we get

∠ACB =∠AOB = 90^o

Hence, the angles are equal as well as supplementary.

Given that,

AB ‖ CD

Let, EF be the parallel line to AB and CD which passes through P.

It can be seen from the figure that alternate angles are equal

∠ABP =∠BPF

∠CDP =∠DPF

∠ABP +∠CDP =∠BPF +∠DPF

∠ABP +∠CDP =∠DPB

Hence, proved

AB is parallel to CD, P is any point

To prove: ∠ABP +∠BPD +∠CDP = 360^o

Construction: Draw EF ‖ AB passing through F

Proof: Since,

AB ‖ EF and AB ‖ CD

Therefore,

EF ‖ CD (Lines parallel to the same line are parallel to each other)

∠ABP +∠EPB = 180^o(Sum of co interior angles is 180^o, AB ‖ EF and BP is transversal)

∠EPD +∠COP = 180^o (Sum of co. interior angles is 180^o, EF ‖ CD and DP is transversal) (i)

∠EPD +∠CDP = 180^o(Sum of co. interior angles is 180^o, EF ‖ CD and DP is the transversal) (ii)

Adding (i) and (ii), we get

∠ABP +∠EPB +∠EPD +∠CDP = 360^o

∠ABP +∠EPD +∠COP = 360^o

Let, ∠A = 2x and

∠B = 3x

Now,

∠AHB = 180^o (Co. interior angles are supplementary)

2x + 3x = 180^o

5x = 180^o

x = 36^o

∠A = 2x = 72^o

∠B = 3x = 108^o

Now,

∠A =∠C = 72^o(Opposite sides angles of a parallelogram are equal)

∠B =∠D = 108^o(Opposite sides angles of a parallelogram are equal)

Let AB and CD be perpendicular to MN

∠ABD = 90^o(AB perpendicular to MN) (i)

∠CON = 90^o (CD perpendicular to MN) (ii)

Now,

∠ABD =∠CON = 90^o

Since, AB ‖ CP

Therefore, corresponding angles are equal.

Given,

∠1 = 60^o

∠2 = of right angle

To prove: l ‖ m

Proof: ∠1 = 60^o

∠2 =* 90^o = 60^o

Since, ∠1 =∠2 = 60^o

Therefore, l ‖ m as pair of corresponding angles are equal.

Since,

l ‖ m and P is transversal

Therefore,

Given that,

l ‖ m ‖ n

∠1 = 60^o

∠1 =∠3 = 60^o(Corresponding angles)

Now,

∠3 + ∠4 = 180^o(Linear pair)

60^o +∠4 = 180^o

∠4 = 120^o

Also,

m ‖ n and P is the transversal

Therefore,

∠4 =∠2 = 120^o(Alternate interior angles)

Let AB and CD perpendicular to the line MN

∠ABD = 90^o(Since, AB perpendicular MN) (i)

∠CON = 90^o(Since, CD perpendicular MN) (ii)

Now,

∠ABD =∠CON = 90^o

Therefore,

AB ‖ CD (Since, corresponding angles are equal)

AB ‖ CD and AD is transversal

AD ‖ BC

Therefore,

∠A +∠D = 180^o(Co. interior angles are supplementary)

60^o +∠D = 180^o

∠D = 120^o

Now,

AD ‖ BC and AB is transversal

∠A +∠B = 180^o(Co. interior angles are supplementary)

60^o +∠B = 180^o

∠B = 120^o

Hence,

∠A =∠C = 60^o

∠B =∠D = 120^o

∠AOC +∠COB +∠BOP = 270^o

To find: ∠AOC,∠COB,∠BOD and ∠BOA

Here, ∠AOC +∠COB +∠BOD +∠AOD = 360^o(Complete angle)

270^o + ∠AOD = 360^o

∠AOD = 360^o – 270^o

= 90^o

Now,

∠AOD +∠BOD = 180^o(Linear pair)

90^o +∠BOD = 180^o

Therefore,

∠BOD = 90^o

∠AOD =∠BOC = 90^o(Vertically opposite angle)

∠BOD =∠AOC = 90^o(Vertically opposite angle)

Given that,

∠2 = 120^o

∠5 = 60^o

To prove: ∠2 +∠1 = 180^o(Linear pair)

120^o + ∠1 = 180^o

∠1 = 180^o – 120^o

= 60^o

Since,

∠1 =∠5 = 60^o

Therefore,

m ‖ n (As pair of corresponding angles are equal)

Given,

∠4 = 110^o,

∠7 = 65^o

To find: m ‖ n

Here,

∠7 =∠5 = 65^o(Vertically opposite angle)

Now,

∠4 +∠5 = 110^o + 65^o

= 175^o

Therefore, m is parallel to n as the pair of co interior angles is not supplementary.

∠A +∠B = 115^o + 65^o

= 180^o

Therefore,

AB ‖ BC, as sum of co interior angles are supplementary.

∠B +∠C = 65^o + 115^o

= 180^o

Therefore,

AB ‖ CD, as sum of co interior angles are supplementary.

Given that,

l ‖ m and n perpendicular to m

Since, l ‖ m and n intersects them at G and H respectively

Therefore,

∠1 =∠2 (Corresponding angles)

But, ∠1 = 90^o as n is perpendicular to l

Therefore, ∠2 = 90^o

Hence, n is perpendicular to m.

Given that,

AB ‖ DE and EC ‖ EF

To prove: ∠ABC =∠DEF

Construction: Produce BC to X such that it intersects DE at M

Proof: Since, AB ‖ DE and BX is the transversal

Therefore,

∠ABC =∠DMX (Corresponding angles) (i)

Also,

BX ‖ EF and DE is transversal

Therefore,

∠DMX =∠DEF (Corresponding angles) (ii)

From (i) and (ii), we get

∠ABC =∠DEF

Hence, proved

Given that,

AB ‖ DE and BC ‖ EF

To prove: ∠ABC +∠DEF = 180^o

Construction: Produce BC to intersect DE at M

Proof: Since, AB ‖ EM and BL is the transversal

∠ABC =∠EML (Corresponding angles) (i)

Also,

EF ‖ ML and EM is the transversal

By the property co interior angles are supplementary

∠DEF +∠EML = 180^o(ii)

From (i) and (ii), we have

∠DEF +∠ABC = 180^o

Hence, proved

(i) False: The corresponding angles can only be equal if the two lines that are intersected by the transversal are parallel in nature.

(ii) True: Since, if two parallel lines are intersected by a transversal, then alternate interior angles are equal.

(iii) False: Two lines perpendicular to the same line are parallel to each other.

(iv) True: Since, two lines parallel to the same line are parallel to each other.

(v) False: If two parallel lines are intersected by a transversal, then the interior angles on the same side of the transversal sums up to 180^o.

(i) Equal

(ii) Supplementary

(iii) Parallel

(iv) Parallel

(v) Parallel

(vi) Parallel

Two Angles are Complementary when they add up to 90 degrees (a right angle). They don't have to be next to each other, just so long as the total is 90 degrees. Examples: 60° and 30° are complementary angles.

Let the required angle be x

Supplement = 180^o – x

According to question,

x = 3 (180^o – x)

x = 540^o – 3x

x = 135^o

Two Angles are Supplementary when they add up to 180 degrees. They don't have to be next to each other, just so long as the total is 180 degrees. Examples: 60° and 120° are supplementary angles.

Let x and (90^o – x) be two complimentary angles

According to question,

2x = 3 (90^o – x)

2x = 270^o – 3x

x = 54^o

The angles are:

54^o and 90^o – 54^o = 36^o

Thus, smallest angle is 36^o

Two angles are Adjacent when they have a common side and a common vertex (corner point) and don't overlap. Angle ABC is adjacent to angle CBD. Because: they have a common side (line CB) they have a common vertex (point B).

Given,

∠AOC + ∠COB + ∠BOD = 274^o (i)

∠AOD + ∠AOC + ∠COB + ∠BOD = 360^o (Angles at a point)

∠AOD + 274^o = 360^o

∠AOD = 86^o

The complement of an acute angle is an acute angle. Since complementary angles add to 90 degrees, the only angles that add to 90 are acute angles.

∠BOD + ∠BOC = 180^o (Linear pair)

63^o + ∠BOC = 180^o

∠BOC = 117^o

When two straight lines intersect them,

Adjacent angles are supplementary and opposite angles are equal.

Supplement is defined as the other angle that adds up to 180 degrees. So therefore if you have an acute angle you know by definition that the angle is less than 90 degrees. In order for both of them to be supplementary (add up to 180 degrees) the other angle must be greater than 90 degrees. Angles that are greater than 90 degrees are obtuse angles. So an obtuse angle is the supplement of an acute angle.

Given,

POQ is a straight line

∠POQ + ∠QOR = 180^o (Linear pair)

3x + 2x + 10^o = 180^o

5x = 170^o

x = 34^o

It is also a right angle

Supplement = 180^o - angle = 180^o – 90^o

= 90^o = Right angle

Given,

AOB = Straight line

∠AOC + ∠BOD = 85^o

∠AOC + ∠COD + ∠BOD = 180^o (Linear pair)

85^o + ∠COD = 180^o

∠COD = 95^o

Compliments are the angle that adds up to give 90°.

Hence, compliment of x° = 90° - x°

3x + y + 2x = 180^o (Linear pair)

5x + y = 180^o (i)

From figure,

y = x (Vertically opposite angles)

Using it in (i), we get

5x + x = 180^o

6x = 180^o

x = 30^o

Thus,

Y = x = 30^o

Supplementary angles' measures have a sum of 180^o

Hence, supplementary angles of 2y° = 180 - 2y°

Total measure of angles of the wheel = 360°

6 spokes = 360°

1 spoke = 60°

So, measure of the angle between 2 adjacent spokes = 2 x 60° = 120°

Given,

=

y = 5x

And,

= 4

z = 4x

From figure,

x + y + z = 180^o (Linear pair)

x + 5x + 4x = 80^o

10x = 180^o

x = 18^o

Supplementary angles are pairs of angles whose measures add up to 180 degrees.

Hence, let one angle be x. Since they are equal, therefore the other angle is also equal to x.

So, x + x = 180°

2x = 180°

x = 180/2

Therefore, x = 90°

So, the angle which is its supplement is 90°.

Let,

AB, CD and EF intersect at O

∠COB = ∠AOD (Vertically opposite angle)

∠AOD = 3x + 10 (i)

∠AOE + ∠AOD + ∠DOF = 180^o (Linear pair)

x + 3x + 10^o + 90^o = 180^o

4x + 100^o = 180^o

4x = 80^o

x = 20^o

Let AB, CD and EF intersect at O

∠AOD = ∠COB (Vertically opposite angle)

b = e (i)

∠EOC = ∠DOF (Vertically opposite angle)

f = c (ii)

Adding (i) and (ii), we get

b + f = c + e (iii)

Now,

∠ADE + ∠EOC + ∠COB = 180^o

a + f + e = 180^o

a + c + e = 180^o [From (ii)]

Let the complement be x then the number = 5x

Now, according to question

x + 5x = 90°

6x = 90°

x =

x = 30°

Hence, measure of the angle is 30°.

Let,

l and m be two parallel lines and transversal p cuts them

∠1: ∠2 = 2: 3 (Interior angles on same side)

Let,

∠1 = 2k

∠2 = 3k

∠1 + ∠2 = 180^o (Interior angle)

2k + 3k = 180^o

k = 36^o

So, ∠1 = 2k = 72^o

∠2 = 3k = 108^o

Hence, larger angle is 108^o

When two lines intersect each other then four adjacent pairs of angles are formed. Hence, there are four basic pairs of adjacent angles formed.

Given,

AB ‖ CD ‖ EF and GH ‖ KL

Produce HG to M and KL to N

∠MHD and ∠CHG = 60^o (Vertically opposite angle)

Since,

MG ‖ NL and transversal cuts them

So,

∠MHD + ∠1 = 180^o (Interior angles)

60^o + ∠1 = 180^o

∠1 = 120^o

∠3 = ∠HKD = 25^o (Alternate angles) (i)

∠1 = ∠MKL = 120^o (Corresponding angles) (ii)

Now,

∠HKL = ∠3 + ∠MKL

= 25^o + 120^o

= 145^o

Given that,

AB ‖ CD and transversal cuts them

Let,

∠1 = 120^o + x and

∠2 = x

∠1 = ∠3 (Alternate angles)

∠3 = 120^o + x (i)

∠2 + ∠3 = 180^o (Linear pair)

x + 120^o + x = 180^o

2x = 60^o

x = 30^o

Given that,

AB ‖ CD and PQ cuts them

EL is bisector of ∠FEB

∠LEB = ∠FEL = 35^o

∠FEB = ∠LEB + ∠FEL

= 35^o + 35^o

= 70^o

∠FEB = ∠EFC = 70^o (Alternate angles)

∠EFC + ∠CFQ = 180^o (Linear pair)

70^o + ∠CFQ = 180^o

∠CFQ = 110^o

Given that,

AB and CD intersect at O

∠AOC +∠COB +∠BOD = 270° (i)

∠COB + ∠BOD = 180^o (Linear pair) (ii)

Using (ii) in (i), we get

∠AOC + 180^o = 270^o

∠AOC = 90^o

Given that,

PQ ‖ RS

∠AEF = 95^o

∠BHS = 110^o

∠ABC = x^o

∠AEF = ∠AGH = 95^o (Corresponding angles)

∠AGH + ∠HGB = 180^o (Linear pair)

95^o + ∠HGB = 180^o

∠HGB = 85^o

∠BHS + ∠BHG = 180^o (Linear pair)

110^o + ∠BHG = 180^o

∠BHG = 70^o

In BHG,

∠BHG + ∠HGB + ∠GBH = 180^o

70^o + 95^o + ∠GBH = 180^o

∠GBH = 25^o

Thus,

∠ABC = ∠GBH = 25^o

Given that,

l₁‖ l₂

Let transversal P and Q cuts them

∠1 = 37^o

∠4 = 58^o

∠5 = x^o

∠1 = ∠2 = 37^o (Corresponding angles) (i)

∠2 = ∠3 (Vertically opposite angle)

∠3 = 37^o

∠3 + ∠4 + ∠5 = 180^o (Linear pair)

37^o + 58^o + x = 180^o

x = 85^o

Given that,

l₁ ‖ l₂

Let m and n be two transversal cutting them

∠w + ∠x = 180^o (Consecutive interior angle)

x = 180^o – w (i)

z = y (Alternate angles) (ii)

From (i) and (ii), we get

x + y = 180^o – w + z

Given that,

l₁ ‖ l �₂ and l₃ is transversal

∠1 = 3x (Vertically opposite angle)

y = ∠1 (Corresponding angle)

y = 3x (i)

y + x = 180^o (Linear pair)

3x + x = 180^o [From (i)]

4x = 180^o

x = 45^o

Therefore,

y = 3x = 3 * 45^o

= 135^o

Given that,

l₁ ‖ l₂ and l₃ ‖ l₄

Let,

∠1 = x

∠2 = y

∠3 = y

∠1 = ∠4 (Alternate angle)

∠4 = x

∠5 = ∠2 (Vertically opposite angle)

∠6 = ∠3 (Vertically opposite angle)

∠5 = ∠6 = y

Now,

∠4 + ∠5 + ∠6 = 180^o (Consecutive interior angle)

y = 90^o -

Given that,

l ‖ m

Let,

∠1 = 3y

∠2 = 2y + 25^o

∠3 = x + 15^o

∠1 = ∠2 (Alternate angle)

3y = 2y + 25^o

y = 25^o

∠2 = ∠3 (Vertically opposite angle)

x + 15^o = 2 (25^o) + 25^o

x = 60^o

Since, AB ‖ CD

And, BD cuts them

y + 2y + y + 5y = 180^o (Consecutive interior angle)

9y = 180^o

y = 20^o

Given that,

CP ‖ BQ

Produce CP to E

So, PE ‖ BQ and AB cuts them

∠QBE = ∠CBA = 105^o (Corresponding angles)

In

∠CEA + ∠ECA + ∠EAC = 180^o

105^o + ∠ECA + 25^o = 180^o

∠ECA = 50^o

∠PCA + ∠ECA = 180^o (Linear pair)

x + 50^o = 180^o

x = 130^o

Given that,

AB ‖ HF and CD cuts them

∠HFC = ∠FDA (Corresponding angle)

∠FDA = 28^o

∠FDA + ∠FDE + ∠EDB = 180^o (Linear pair)

28^o + ∠FDE + 72^o = 180^o

∠FDE = 80^o

l ‖ m

Let transversal be n and ∠1 = 65^o

∠2 = 20^o

∠3 = x

Since,

l ‖ m and n cuts them so,

∠1 + ∠4 = 180^o (Co. interior angle)

65^o + ∠4 = 180^o

∠4 = 115^o (i)

∠4 = ∠5 = 115^o (Vertically opposite angle)

∠2 + ∠5 + ∠3 = 180^o

20^o + 115⁰ + x = 180^o

x = 45^o

Given that,

AB ‖ CD

Produce P to Q so that PQ ‖ AB ‖ CD

∠BAP + ∠APQ = 180^o (Interior angle)

132^o + ∠APQ = 180^o

∠APQ = 48^o (i)

∠APC = ∠APQ + ∠QPC

148^o = 48^o + ∠QPC [From (i)]

∠QPC = 100^o

∠QPC + ∠PCD = 180^o (Interior angles)

100^o + ∠PCD = 180^o

∠PCD = 80^o

∠PCD + x = 180^o (Linear pair)

80^o + x = 180^o

x = 100^o

Given that,

l ‖ m

Let, l ‖ m and transversal cuts them and

∠1 = 70^o

∠3 = 20^o

∠4 = 30^o

∠1 + ∠2 = 180^o (Interior angle)

∠2 = 110^o (i)

∠2 = ∠5 (Vertically opposite angle)

∠5 = 110^o (ii)

∠5 + ∠3 + ∠4 = 180^o (Sum of angles of a triangle is 180^o)

110^o + x + 30^o = 180^o

x = 40^o

Given that,

l ‖ m and n cuts them

Let,

∠1 = 65^o

∠2 = x

∠3 = 40^o

∠1 = ∠4 = 65^o (Alternate angle) (i)

∠3 + ∠4 + ∠5 = 180^o (Angle sum property)

40^o + 65^o + ∠5 = 180^o

∠5 = 75^o

Now,

∠2 + ∠5 = 180^o (Linear pair)

x + 75^o = 180^o

x = 105^o

Given that,

l ‖ m and n cuts them

Let,

∠1 = x

∠2 = 90^o

∠3 = 125^o

∠3 + ∠5 = 180^o (Linear pair)

125^o + ∠5 = 180^o

∠5 = 55^o (i)

∠4 = 90^o (ii)

Now,

∠1 + ∠4 + ∠5 = 180^o (Angle sum property)

x + 90^o + 55^o = 180^o

x = 35^o