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Correct option is   A) ii and iii

(b) 50^o,130^o

We know that, ∠PAB = ∠ABC = 50^o … [because interior alternate angles]

Given, AB || DC so consider it as parallelogram,

In parallelogram adjacent angles of a parallelogram are supplementary.

So, ∠ABC + ∠BCD = 180^o

50^o + ∠BCD = 180^o

∠BCD = 180^o – 50^o

∠BCD = 130^o

∠BCD = ∠CDT = 130^o  … [because interior alternate angles]

Therefore, a = 50^o and b = 130^o

Draw a line ‘l’ parallel to AB and CD through F.

a + 104° = 180° ⇒ a = 180° – 104° = 76°

 b+ 116° = 180° ⇒ -b = 180°- 116° = 64° 

[∵ interior angles on the same side] 

∴ a + b = x = 76° + 64° = 140°

Consider the required angle as x°

We know that the complement can be written as 90° – x° and the supplement can be written as 180° – x°

180° – x° = 4(90° – x°)

We can also write it as

180° – x° = 360° – 4x°

So we get

4x° – x° = 360° – 180°

3x° = 180°

By division we get

x° = 180/3

x° = 60°

Therefore, the angle whose supplement is four times its complement is 60°.

(i) Consider the required angle as x°

We know that the complement can be written as 90° – x°

To find that the measure of an angle is equal to its complement

We get

x° = 90° – x°

We can also write it as

x + x = 90

So we get

2x = 90

By division we get

x = 90/2

x° = 45°

Therefore, the measure of an angle which is equal to its complement is 45°

(ii) Consider the required angle as x°

We know that the supplement can be written as 180° – x°

To find that the measure of an angle is equal to its complement

We get

x° = 180° – x°

We can also write it as

x + x = 180

So we get

2x = 180

By division we get

x = 180/2

x° = 90°

Therefore, the measure of an angle which is equal to its complement is 90°

Correct option is D) adjacent

(a) 60^o, 30^o

When the sum of the measures of two angles is 90^o, the angles are called complementary angles.

So, 60^o + 30^o = 90^o

As per the condition in the question, 60^o – 30^o = 30^o

Let AB and CD perpendicular to the line MN

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

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

Now,

∠ABD = ∠CON = 90°

Therefore,

AB ‖ CD (Since, corresponding angles are equal)

Given l and m are intersecting lines at P. 

∠y = 20° (vertically opposite angles) 

∠x = ∠z (vertically opposite angles) 

∠y + ∠x = 180° (linear pair) 

⇒ 20° + ∠x = 180° 

⇒ 20° + ∠x – 20° = 180°- 20 

⇒ ∠x = 160° 

∴ ∠x = ∠z = 160° ∠x = 160°, ∠y = 20° and ∠z = 160°.

Correct option is A) Supplementary

Correct answer is (c) 75°, 105°.

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°(Linear pair) (i)

Similarly,

∠TOS + ∠SOQ =180°(ii)

∠TOS + ∠SOR + ∠OQR = 180°(iii)

Adding (i) and (iii), we get

∠TOP + ∠POQ + ∠TOS + ∠SOR + ∠QOR = 360°

∠TOP + ∠TOS = ∠POS

Therefore,

∠POQ + ∠QOR + ∠SOR +∠POS = 360°.

Given ∠QOR= 50° 

∠POQ, ∠QOR are linear pair. 

⇒ ∠POQ + ∠QOR = 180° 

⇒ ∠POQ + 50° = 180° 

⇒ ∠POQ - 180° – 50° = 130° 

∠POQ =130°

It is given that ∠1: ∠2 = 2: 3

From the figure we know that ∠1 and ∠2 form a linear pair of angles

So it can be written as

∠1 + ∠2 = 180^o

By substituting the values

2x + 3x = 180^o

On further calculation

5x = 180^o

By division

x = 180^o/5

x = 36^o

By substituting the value of x we get

∠1 = 2x = 2 (36^o) = 72^o

∠2 = 3x = 3 (36^o) = 108^o

From the figure we know that ∠1 and ∠3 are vertically opposite angles

So we get

∠1 = ∠3 = 72^o

From the figure we know that ∠2 and ∠4 are vertically opposite angles

So we get

∠2 = ∠4 = 108^o

It is given that, l || m and t is a transversal

So the corresponding angles according to the figure is written as

∠1 = ∠5 = 72^o

∠2 = ∠6 = 108^o

∠3 = ∠7 = 72^o

∠4 = ∠8 = 108^o

∠QOR + ∠POR = 180°(Linear pair)

2x + 10° + 3x = 180°

5x = 170°

x = 34°

It is given that ∠1 = 120^o

From the figure we know that ∠1 and ∠2 form a linear pair of angles

So it can be written as

∠1 + ∠2 = 180^o

By substituting the values

120^o + ∠2 = 180^o

On further calculation

∠2 = 180^o – 120^o

By subtraction

∠2 = 60^o

From the figure we know that ∠1 and ∠3 are vertically opposite angles

So we get

∠1 = ∠3 = 120^o

From the figure we know that ∠2 and ∠4 are vertically opposite angles

So we get

∠2 = ∠4 = 60^o

It is given that, l || m and t is a transversal

So the corresponding angles according to the figure is written as

∠1 = ∠5 = 120^o

∠2 = ∠6 = 60^o

∠3 = ∠7 = 120^o

∠4 = ∠8 = 60^o

Given that AB // CD. From the figure (2x + 3x) + 80° = 180° 

(∵ interior angles on the same side of the transversal) 

∴ 5x = 180° – 80°

x = 100/5 = 20°

Now 3x = y (∵ alternate interior angles) 

y = 3 x 20° = 60°. 

And y + z = 180° 

(∵ linear pair of angles) 

∴ z = 180°-60° = lg0°

(a) 32

Because,

It is given that the angles of the side be (3x – 8)^o, 50^o and (x + 10)^o.

= (3x – 8) + 50^o + (x + 10) = 180 … [∵ Linear pair]

= 3x – 8 + 50 + x + 10 = 180

= 4x + 52 = 180

= 4x = 180 – 52

= 4x = 128

= x = 128/4

= x = 32

Given,

AOB = Straight line

∠AOC + ∠BOD = 85°

∠AOC + ∠COD + ∠BOD = 180°(Linear pair)

85° + ∠COD = 180°

∠COD = 95°

Let the angle be x°. We know that, sum of two complementary angles is 90°.

x + 62° = 90°

x = 90° - 62° = 28°

Supplement of any angle is (180° - angle)

Supplement of x = 180° - 28° = 152°

Data: In this fiure, ∠BOC = x° 

∠AOC = y° ∠BOD = w° 

∠AOD = z° and x + y = w + z. 

To Prove: AOB is a straight line. 

Proof: x + y + w + z = 360° 

(∵ one completre angle) 

But, x + y = w + z. 

∴ x + y = w + z= \(\frac{360}{2}\) = 180° 

∴ x + y = 180° 

∴ w + z = 180° proved. 

But, ∠AOC and ∠BOC are adjacent angles, 

∠AOC + ∠BOC = 180° 

x = y = 180° 

∴ ZAOB = 180°. 

∴ AOB is a straight line.

Correct option is C) Complementary

Acute angles are those angles which are less than 90°. If we add two angles which are less than 90°, we get the result less than 180°, e.g. If we add 60° and 70°, we get 60°+ 70° = 130° <180°

Hence, two acute angles cannot form a pair of supplementary angles.

Given : OR ⊥ PQ ⇒ ∠ROQ = 90° 

To prove: ∠ROS = 1/2(∠QOS – ∠POS) 

From the figure 

∠ROS = ∠QOS – ∠QOR ……………(1) 

∠ROS = ∠ROP – ∠POS ……………..(2) 

Adding (1) and (2) 

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

[ ∵ ∠QOR = ∠ROP = 90° given] 

⇒ 2∠ROS = ∠QOS – ∠POS

⇒ ∠ROS = 1/2 [∠QOS – ∠POS]

Hence proved.

From given figure . y = 115° 

{∵ Corresponding angles} 

⇒ x + y = 180° {linear pair} 

⇒ x = 180°- y = 180° – 115° = 65° 

⇒ z = x°= 65° { ∵ Corresponding angles} 

c° – 70°

a° + c° = 180° {linear pair} 

⇒ a° + 70° = 180° 

⇒ a° = 180° – 70° = 110° 

b° = c° = 70° 

{Vertically opposite angles} 

So, a = 110°, b = 70°, c = 70°, x = 65°, 

y = 115°, z = 65°

90° + 20° = 1 10° ≠ 90° (or) ≠ 180° 

So, they are not complementary or not supplementary.

∠BOD + ∠BOC = 180°(Linear pair)

63° + ∠BOC = 180°

∠BOC = 117°

Given that AB // CD and ∠2 : ∠1 = 5 : 4 ∠1 + ∠2 

= 180° (-. linear pair of angles) 

Sum of the terms of the ratio ∠2 :∠1 = 5 + 4 = 9

∴ ∠1 = 4/9 x 180° = 80°

∠2= 5/9 x 180° = 100°

Also ∠1, ∠3, ∠5, ∠7 are all equal to 80°. 

Similarly ∠2, ∠4, ∠6, ∠8 are all equal to 100°.

Given,

POQ is a straight line

∠POQ + ∠QOR = 180°(Linear pair)

3x + 2x + 10° = 180°

5x = 170°

x = 34°

Given that x + y = w + z = k say 

From the figure 

x + y + z + w = 360° 

(∵ Angle around a point) 

Also x + y = z + w 

∴ x + y = z + w = 360°/2

∴ x + y = z + w = 180°

OR

k + k = 360° 

2k = 360°

k = 360°/2

(i.e.) (x,y) and (z, w) are pairs of adjacent angles whose sum is 180°. 

In other words (x, y) and (z, w) are linear pair of angles ⇒ AOB is a line.

Two angles are called adjacent angles, if they have a common vertex and a common arm, but no common interior points.

Hence, following are the pairs of adjacent angles taking 1,2, 3, 4 angles only,

i.e. ∠1, ∠2 ; ∠2, ∠3; ∠3, ∠4; ∠4, ∠1.

When two straight lines intersect them,

Adjacent angles are supplementary and opposite angles are equal.

Let, ∠1 = 3x

∠2 = 2x

∠1 + ∠2 = 180°(Linear pair)

3x + 2x = 180°

5x = 180°

x = 36°

Therefore,

∠1 = 3x = 108°

∠2 = 2x = 72°

Vertically opposite angles are equal, so:

∠1 = ∠3 = 108°

∠2 = ∠4 = 72°

∠5 = ∠7 = 108°

∠6 = ∠8 = 72°

Corresponding angles:

∠1 = ∠5 = 108°

∠2 = ∠6 = 72°

Given ∠1 and ∠2 are in the ratio 3: 2

Let us take the angles as 3x, 2x

∠1 and ∠2 are linear pair [from the figure]

3x + 2x = 180°

5x = 180°

x = 180°/5

x = 36°

Therefore, ∠1 = 3x = 3(36) = 108°

∠2 = 2x = 2(36) = 72°

∠1 and ∠5 are corresponding angles

Therefore ∠1 = ∠5

Hence, ∠5 = 108°

∠2 and ∠6 are corresponding angles

So ∠2 = ∠6

Therefore, ∠6 = 72°

∠4 and ∠6 are alternate pair of angles

∠4 = ∠6 = 72°

Therefore, ∠4 = 72°

∠3 and ∠5 are alternate pair of angles

∠3 = ∠5 = 108°

Therefore, ∠5 = 108°

∠2 and ∠8 are alternate exterior of angles

∠2 = ∠8 = 72°

Therefore, ∠8 = 72°

∠1 and ∠7 are alternate exterior of angles

∠1 = ∠7 = 108°

Therefore, ∠7 = 108°

Hence, ∠1 = 108°, ∠2 = 72°, ∠3 = 108°, ∠4 = 72°, ∠5 = 108°, ∠6 = 72°, ∠7 = 108°, ∠8 = 72°

Given that ∠PQR = ∠PRQ 

From the figure 

∠PQR + ∠PQS = 180° ………….. (1) 

∠PRQ + ∠PRT = 180° …………..(2) 

From (1) and (2) 

∠PQR + ∠PQS = ∠PRQ + ∠PRT 

But ∠PQR = ∠PRQ So ∠PQS = ∠PRT 

Hence proved.

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

∠AOD, ∠DOB

∠DOB, ∠BOC

∠COA, ∠AOD

∠BOC, ∠COA

Sum of all the angles around the point = 360°

3x + 3x + 150°+ x = 360°

7x = 360°– 150°

7x = 210°

x = 30°

There are 10 adjacent pairs formed in the given figure, they are

∠EOD and ∠DOC

∠COD and ∠BOC

∠COB and ∠BOA

∠AOB and ∠BOD

∠BOC and ∠COE

∠COD and ∠COA

∠DOE and ∠DOB

∠EOD and ∠DOA

∠EOC and ∠AOC

∠AOB and ∠BOE

Pairs of adjacent angles are:

∠DOA and ∠DOC

∠BOC and ∠COD

∠AOD and ∠BOD

∠AOC and ∠BOC

Linear pairs: [The two adjacent angles are said to form a linear pair of angles if their non–common arms are two opposite rays and sum of the angle is 180°]

∠AOD and ∠BOD

∠AOC and ∠BOC

From figure, pairs of adjacent angles are : 

(∠AOC, ∠COB) ; (∠AOD, ∠BOD) ; (∠AOD, ∠COD) ; (∠BOC, ∠COD) 

And Linear pair of angles are (∠AOD, ∠BOD) and (∠AOC, ∠BOC). 

[As ∠AOD + ∠BOD = 180° and ∠AOC+ ∠BOC = 180°.]

Given that ∠SPR = 135° and ∠PQT =110°

From the figure 

∠SPR + ∠RPQ = 180° 

∠PQT + ∠PQR = 180° 

[∵ linear pair of angles] 

⇒ ∠RPQ = 180° – ∠SPR = 180° – 135° = 45° 

⇒ ∠PQR = 180° – ∠PQT = 180°-110° = 70° 

Now in APQR 

∠RPQ + ∠PQR + ∠PRQ = 180° 

[∵ angle sum property] 

∴ 45° + ’70° + ∠PRQ = 180° 

∴ ∠PRQ = 180°-115° = 65°

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.

Adjacent angles are:

(i) ∠AOC, ∠COB

(ii) ∠AOD, ∠BOD

(iii) ∠AOD, ∠COD

(iv) ∠BOC, ∠COD

Linear pairs are:

∠AOD, ∠BOD and

∠AOC, ∠BOC

(b) 135°

From the figure, PQ is a straight line

We know that, sum of angle on the straight is equal to 180°.

Then,

y + ∠ PQR = 180° 

y + 130 = 180°

y = 50°

Then,

∠ QOS = ∠ TSO [Co-interior angle]

x = 85°

x + y = 135°

(b) 144^o

Sum of all angle about a straight line given in the figure are equal to 180^o.

PQ is a straight line.

Then, ∠POR + ∠ROT + ∠TOQ = 180^o

Given, x : y = 3 : 2

Let us assume x = 3a, y = 2a

90^o + 3a + 2a = 180^o

90^o + 5a = 180^o

5a = 180^o – 90^o

5a = 90^o

a = 90/5

a = 18^o

So, x = 3a = 3 × 18 = 54^o

Y = 2a = 2 × 18 = 36^o

From the figure SOT is a straight line,

Then, Z + Y = 180^o

Z + 36^o = 180^o

Z = 180^o – 36^o

Z = 144^o

The angles that have common vertex and a common arm are known as adjacent angles

Therefore the adjacent angles in given figure are:

∠DOC and ∠BOC

∠COB and ∠BOA

Since,

Vertically opposite angles are equal

So,

∠BOD = z = 90°

∠DOF = y = 50°

Now,

x + y + z = 180°(Linear pair)

x + 90° + 50° = 180°

x = 40°

Given that,

x = 45°

y =?

z =?

u =?

z = x = 45°(Vertically opposite angle)

z + u = 180°(Linear pair)

45° = 180° – u

u = 135°

x + y = 180°(Linear pair)

45° = 180° – y

y = 135°

Therefore,

x = 45°, y = 135°, u = 135° and z = 45°