## RS Aggarwal Solutions Class 9 Chapter 4 Angles, Lines and Triangles

**Exercise 4A**

**Question 1:**

(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° but less than 180°, is called an obtuse angle.

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

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

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

**Question 2:**

∠A = 36° 27′ 46″ and ∠B = 28° 43′ 39″

∴ Their sum = (36° 27′ 46″) + (28° 43′ 39″)

Therefore, the sum ∠A + ∠B = 65° 11′ 25″

**Question 3:**

Let ∠A = 36° and ∠B = 24° 28′ 30″

Their difference = 36° – 24° 28′ 30″

Thus the difference between two angles is ∠A – ∠B = 11° 31′ 30″

**Read More:**

- Angle Sum Property of a Triangle
- Median and Altitude of a Triangle
- The Angle of An Isosceles Triangle
- Areas of Two Similar Triangles
- Area of A Triangle
- To Prove Triangles Are Congruent
- Criteria For Similarity of Triangles
- Construction of an Equilateral Triangle

**Question 4:**

(i) Complement of 58^{o} = 90^{o }– 58^{o} = 32^{o}

(ii) Complement of 16^{o} = 90 – 16^{o} = 74^{o}

(iii) of a right angle = × 90^{o} = 60^{o}

Complement of 60^{o} = 90^{o} – 60^{o} = 30^{o}

(iv) 1^{o} = 60′

⇒ 90^{o} = 89^{o} 60′

Complement of 46^{o} 30′ = 90^{o} – 46^{o} 30′ = 43^{o} 30′

(v) 90^{o} = 89^{o} 59′ 60″

Complement of 52^{o} 43′ 20″ = 90^{o} – 52^{o} 43′ 20″

= 37^{o} 16′ 40″

(vi) 90^{o} = 89^{o} 59′ 60″

∴ Complement of (68^{o} 35′ 45″)

= 90^{o} – (68^{o} 35′ 45″)

= 89^{o} 59′ 60″ – (68^{o} 35′ 45″)

= 21^{o} 24′ 15″

**Question 5:**

(i) Supplement of 63^{o} = 180^{o} – 63^{o} = 117^{o}

(ii) Supplement of 138^{o} = 180^{o} – 138^{o} = 42^{o}

(iii) of a right angle = × 90^{o} = 54^{o}

∴ Supplement of 54^{o} = 180^{o} – 54^{o} = 126^{o}

(iv) 1^{o} = 60′

⇒ 180^{o} = 179^{o} 60′

Supplement of 75^{o} 36′ = 180^{o} – 75^{o} 36′ = 104^{o} 24′

(v) 1^{o} = 60′, 1′ = 60″

⇒ 180^{o} = 179^{o} 59′ 60″

Supplement of 124^{o} 20′ 40″ = 180^{o} – 124^{o} 20′ 40″

= 55^{o} 39′ 20″

(vi) 1^{o} = 60′, 1′ = 60″

⇒ 180^{o} = 179^{o} 59′ 60″

∴ Supplement of 108^{o} 48′ 32″ = 180^{o} – 108^{o} 48′ 32″

= 71^{o} 11′ 28″.

**Question 6:**

(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}.

**Question 7:**

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}.

**Question 8:**

Let the required angle be x^{o}

Then its supplement is 180^{o} – x^{o}

∴ The measure of an angle which is 25^{o} less than its supplement is

**Question 9:**

Let the required angle be x^{o}

Then, its complement = 90^{o} – x^{o}

∴ The required angle is 72^{o}.

**Question 10:**

Let the required angle be x^{o}

Then, its supplement is 180^{o} – x^{o}

∴ The required angle is 150^{o}.

**Question 11:**

Let the required angle be x^{o}

Then, its complement is 90^{o} – x^{o} and its supplement is 180^{o} – x^{o}

That is we have,

∴ The required angle is 60^{o}.

**Question 12:**

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}.

**Question 13:**

Let the two required angles be x^{o} and 180^{o} – x^{o}.

Then,

⇒ 2x = 3(180 – x)

⇒ 2x = 540 – 3x

⇒ 3x + 2x = 540

⇒ 5x = 540

⇒ x = 108

Thus, the required angles are 108^{o} and 180^{o} – x^{o} = 180^{ o} – 108^{o} = 72^{o}.

**Question 14:**

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

⇒ x = = 40

Thus, the required angles are 40^{o} and 90^{o} – x^{o} = 90^{ o} – 40^{o} = 50^{o}.

**Question 15:**

Let the required angle be x^{o}.

Then, its complementary and supplementary angles are (90^{o} – x) and (180^{o} – x) respectively.

Then, 7(90^{o} – x) = 3 (180^{o} – x) – 10^{o}

⇒ 630^{o} – 7x = 540^{o} – 3x – 10^{o}

⇒ 7x – 3x = 630^{o} – 530^{o}

⇒ 4x = 100^{o}

⇒ x = 25^{o}

Thus, the required angle is 25^{o}.

**Exercise 4B**

**Question 1:**

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}

**Question 2:**

Since, ∠BOD and ∠DOA form a linear pair.

∠BOD + ∠DOA = 180^{o}

∴ ∠BOD + ∠DOC + ∠COA = 180^{o}

⇒ (x + 20)^{o} + 55^{o} + (3x – 5)^{o} = 180^{o}

⇒ x + 20 + 55 + 3x – 5 = 180

⇒ 4x + 70 = 180

⇒ 4x = 180 – 70 = 110

⇒ x = = 27.5

∴ ∠AOC = (3 × 27.5 – 5)^{o} = 82.5-5 = 77.5^{o}

And, ∠BOD = (x + 20)^{o }= 27.5^{o} + 20^{o} = 47.5^{o}.

**Question 3:**

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

⇒ 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}

**Question 4:**

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 x = × 180 = 60

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 y = × 180 = 48

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.

**Question 5:**

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

⇒ x = = 28

∴ The value of x = 28.

**Question 6:**

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}

**Question 7:**

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

**Question 8:**

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}

**Question 9:**

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}

⇒ x = = 20

∴ The required angles are 5x = 5x = 5 20^{o} = 100^{o}

and 4x = 4 × 20^{o} = 80^{o}

**Question 10:**

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}.

**Question 11:**

Since, ∠AOD and ∠BOC are vertically opposite angles.

∴ ∠AOD = ∠BOC

Now, ∠AOD + ∠BOC = 280^{o} [Given]

⇒ ∠AOD + ∠AOD = 280^{o}

⇒ 2∠AOD = 280^{o}

⇒ ∠AOD = = 140^{o}

⇒ ∠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}.

**Question 12:**

Since ∠COB and ∠BOD form a linear pair

So, ∠COB + ∠BOD = 180^{o}

⇒ ∠BOD = 180^{o} – ∠COB …. (1)

Also, as ∠COA and ∠AOD form a linear pair.

So, ∠COA + ∠AOD = 180^{o}

⇒ ∠AOD = 180^{o} – ∠COA

⇒ ∠AOD = 180^{o} – ∠COB …. (2)

[Since, OC is the bisector of ∠AOB, ∠BOC = ∠AOC]

From (1) and (2), we get,

∠AOD = ∠BOD (Proved)

**Question 13:**

Let QS be a perpendicular to AB.

Now, ∠PQS = ∠SQR

Because angle of incident = angle of reflection

⇒ ∠PQS = ∠SQR = = 56^{o}

Since QS is perpendicular to AB, ∠PQA and ∠PQS are complementary angles.

Thus, ∠PQA + ∠PQS = 90^{o}

⇒ ∠PQA + 56^{o} = 90^{o}

⇒ ∠PQA = 90^{o} – 56^{o }= 34^{o}

**Question 14:**

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 and 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.

**Question 15:**

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 = = 90^{o}

∠ECF = 90^{o} (Proved)

**Exercise 4C**

**Question 1:**

Since AB and CD are given to be parallel lines and t is a transversal.

So, ∠5 = ∠1 = 70^{o} [Corresponding angles are equal]

∠3 = ∠1 = 70^{o} [Vertically opp. Angles]

∠3 + ∠6 = 180^{o} [Co-interior angles on same side]

∴ ∠6 = 180^{o} – ∠3

= 180^{o} – 70^{o }= 110^{o}

∠6 = ∠8 [Vertically opp. Angles]

⇒ ∠8 = 110^{o}

⇒ ∠4 + ∠5 = 180^{o} [Co-interior angles on same side]

∠4 = 180^{o} – 70^{o} = 110^{o}

∠2 = ∠4 = 110^{o} [ Vertically opposite angles]

∠5 = ∠7 [Vertically opposite angles]

So, ∠7 = 70^{o}

∴ ∠2 = 110^{o}, ∠3 = 70^{o} , ∠4 = 110^{o}, ∠5 = 70^{o}, ∠6 = 110^{o}, ∠7 = 70^{o} and ∠8 = 110^{o}.

**Question 2:**

Since ∠2 : ∠1 = 5 : 4.

Let ∠2 and ∠1 be 5x and 4x respectively.

Now, ∠2 + ∠1 = 180^{o} , because ∠2 and ∠1 form a linear pair.

So, 5x + 4x = 180^{o}

⇒ 9x = 180^{o}

⇒ x = 20^{o}

∴ ∠1 = 4x = 4 × 20^{o} = 80^{o}

And ∠2 = 5x = 5 × 20^{o} = 100^{o}

∠3 = ∠1 = 80^{o }[Vertically opposite angles]

And ∠4 = ∠2 = 100^{o} [Vertically opposite angles]

∠1 = ∠5 and ∠2 = ∠6 [Corresponding angles]

So, ∠5 = 80^{o} and ∠6 = 100^{o}

∠8 = ∠6 = 100^{o }[Vertically opposite angles]

And ∠7 = ∠5 = 80^{o} [Vertically opposite angles]

Thus, ∠1 = 80^{o}, ∠2 = 100^{o}, ∠3 = ∠80^{o}, ∠4 = 100^{o}, ∠5 = 80^{o}, ∠6 = 100^{o}, ∠7 = 80^{o} and ∠8 = 100^{o}.

**Question 3:**

Given: AB || CD and AD || BC

To Prove: ∠ADC = ∠ABC

Proof: Since AB || CD and AD is a transversal. So sum of consecutive interior angles is 180^{o}.

⇒ ∠BAD + ∠ADC = 180^{o} ….(i)

Also, AD || BC and AB is transversal.

So, ∠BAD + ∠ABC = 180^{o} ….(ii)

From (i) and (ii) we get:

∠BAD + ∠ADC = ∠BAD + ∠ABC

⇒ ∠ADC = ∠ABC (Proved)

**Question 4:**

(i) 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.

(ii) 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.

(iii) 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.

**Question 5:**

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

**Question 6:**

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

**Question 7:**

Given: AB || CD

To Prove: ∠BAE – ∠DCE = ∠AEC

Construction : Through E draw EF || AB

Proof : Since EF || AB, AE is a transversal.

So, ∠BAE + ∠AEF = 180^{O} ….(i)

[sum of consecutive interior angles is 180^{o}]

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

So, EF || CD and EC is a transversal.

So, ∠FEC + ∠DCE = 180^{o} ….(ii)

[sum of consecutive interior angles is 180^{o}]

From (i) and (ii) we get,

∠BAE + ∠AEF = ∠FEC + ∠DCE

⇒ ∠BAE – ∠DCE = ∠FEC – ∠AEF = ∠AEC [Proved]

**Question 8:**

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.

**Question 9:**

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

**Question 10:**

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.

**Question 11:**

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}

⇒ 105^{o} + x^{o} + 30^{o} = 180^{o}

⇒ x = 180 – 30 – 105 = 45

_{Hence, x = 45.}

**Question 12:**

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

**Question 13:**

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,

∠AOB + ∠BAO + ∠BOA = 180^{o}

⇒ x^{o} + 75^{o} + y^{o} = 180^{o}

⇒ 35 + 75 + y = 180

⇒ y = 180 – 110 = 70

∴ x = 35, y = 70 and z = 75.

**Question 14:**

Since AB || CD and PQ is a transversal.

So, y = 75 [Alternate angle]

Since PQ is a transversal and AB || CD, so x + APQ = 180^{o}

[Sum of consecutive interior angles]

⇒ x^{o} = 180^{o} – APQ

⇒ x = 180 – 75 = 105

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

So, ∠APR = ∠PRD [Alternate angle]

⇒ ∠APQ + ∠QPR = ∠PRD [Since ∠APR = ∠APQ + ∠QPR]

⇒ 75^{o} + z^{o} = 125^{o}

⇒ z = 125 – 75 = 50

∴ x = 105, y = 75 and z = 50.

**Question 15:**

∠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, ∠PRD = ∠APR [Alternate angles]

⇒ ∠PRQ + ∠QRD = ∠APR [since ∠PRD = ∠PRQ + ∠QRD]

⇒ 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

**Question 16:**

(i) 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

(ii) Lines will be parallel if (3x + 5)^{o} + 4x^{o} = 180^{o}

[if sum of pairs of consecutive interior angles is 180^{o}, the lines are parallel]

So, (3x + 5) + 4x = 180

⇒ 3x + 5 + 4x = 180

⇒ 7x = 180 – 5 = 175

⇒ x = = 25

**Question 17:**

Given: Two lines m and n are perpendicular to a given line l.

To Prove: m || n

Proof : Since m ⊥ l

So, ∠1 = 90^{o}

Again, since n ⊥ l

∠2 = 90^{o}

∴ ∠1 = ∠2 = 90^{o}

But ∠1 and ∠2 are the corresponding angles made by the transversal l with lines m and n and they are proved to be equal.

Thus, m || n.

**Exercise 4D**

**Question 1:**

Since, sum of the angles of a triangle is 180^{o}

∠A + ∠B + ∠C = 180^{o}

⇒ ∠A + 76^{o} + 48^{o} = 180^{o}

⇒ ∠A = 180^{o} – 124^{o} = 56^{o}

∴ ∠A = 56^{o}

**Question 2:**

Let the measures of the angles of a triangle are (2x)^{o}, (3x)^{o} and (4x)^{o}.

Then, 2x + 3x + 4x = 180 [sum of the angles of a triangle is 180^{o} ]^{ }

⇒ 9x = 180

⇒ x = = 20

∴ The measures of the required angles are:

2x = (2 × 20)^{o} = 40^{o}

3x = (3 × 20)^{o} = 60^{o}

4x = (4 × 20)^{o} = 80^{o}

**Question 3:**

Let 3∠A = 4∠B = 6∠C = x (say)

Then, 3∠A = x

⇒ ∠A =

4∠B = x

⇒ ∠B =

and 6∠C = x

⇒ ∠C =

As ∠A + ∠B + ∠C = 180^{o}

**Question 4:**

∠A + ∠B = 108^{o} [Given]

But as ∠A, ∠B and ∠C are the angles of a triangle,

∠A + ∠B + ∠C = 180^{o}

⇒ 108^{o} + ∠C = 180^{o}

⇒ C = 180^{o} – 108^{o} = 72^{o}

Also, ∠B + ∠C = 130^{o} [Given]

⇒ ∠B + 72^{o} = 130^{o}

⇒ ∠B = 130^{o} – 72^{o} = 58^{o}

Now as, ∠A + ∠B = 108^{o}

⇒ ∠A + 58^{o} = 108^{o}

⇒ ∠A = 108^{o} – 58^{o} = 50^{o}

∴ ∠A = 50^{o}, ∠B = 58^{o} and ∠C = 72^{o}.

**Question 5:**

Since. ∠A , ∠B and ∠C are the angles of a triangle .

So, ∠A + ∠B + ∠C = 180^{o}

Now, ∠A + ∠B = 125^{o} [Given]

∴ 125^{o} + ∠C = 180^{o}

⇒ ∠C = 180^{o} – 125^{o} = 55^{o}

Also, ∠A + ∠C = 113^{o} [Given]

⇒ ∠A + 55^{o} = 113^{o}

⇒ ∠A = 113^{o} – 55^{o} = 58^{o}

Now as ∠A + ∠B = 125^{o}

⇒ 58^{o} + ∠B = 125^{o}

⇒ ∠B = 125^{o} – 58^{o} = 67^{o}

∴ ∠A = 58^{o}, ∠B = 67^{o} and ∠C = 55^{o}.

**Question 6:**

Since, ∠P, ∠Q and ∠R are the angles of a triangle.

So, ∠P + ∠Q + ∠R = 180^{o} ….(i)

Now, ∠P – ∠Q = 42^{o} [Given]

⇒ ∠P = 42^{o} + ∠Q ….(ii)

and ∠Q – ∠R = 21^{o} [Given]

⇒ ∠R = ∠Q – 21^{o} ….(iii)

Substituting the value of ∠P and ∠R from (ii) and (iii) in (i), we get,

⇒ 42^{o} + ∠Q + ∠Q + ∠Q – 21^{o} = 180^{o}

⇒ 3∠Q + 21^{o} = 180^{o}

⇒ 3∠Q = 180^{o} – 21^{o} = 159^{o}

∠Q = = 53^{o}

∴ ∠P = 42^{o} + ∠Q

= 42^{o} + 53^{o} = 95^{o}

∠R = ∠Q – 21^{o}

= 53^{o} – 21^{o} = 32^{o}

∴ ∠P = 95^{o}, ∠Q = 53^{o} and ∠R = 32^{o}.

**Question 7:**

Given that the sum of the angles A and B of a ABC is 116^{o}, i.e., ∠A + ∠B = 116^{o}.

Since, ∠A + ∠B + ∠C = 180^{o}

So, 116^{o} + ∠C = 180^{o}

⇒ ∠C = 180^{o} – 116^{o} = 64^{o}

Also, it is given that:

∠A – ∠B = 24^{o}

⇒ ∠A = 24^{o} + ∠B

Putting, ∠A = 24^{o} + ∠B in ∠A + ∠B = 116^{o}, we get,

⇒ 24^{o} + ∠B + ∠B = 116^{o}

⇒ 2∠B + 24^{o} = 116^{o}

⇒ 2∠B = 116^{o} – 24^{o} = 92^{o}

∠B = = 46^{o}

Therefore, ∠A = 24^{o} + 46^{o} = 70^{o}

∴ ∠A = 70^{o}, ∠B = 46^{o} and ∠C = 64^{o}.

**Question 8:**

Let the two equal angles, A and B, of the triangle be x^{o} each.

We know,

∠A + ∠B + ∠C = 180^{o}

⇒ x^{o} + x^{o} + ∠C = 180^{o}

⇒ 2x^{o} + ∠C = 180^{o} ….(i)

Also, it is given that,

∠C = x^{o} + 18^{o }….(ii)

Substituting ∠C from (ii) in (i), we get,

⇒ 2x^{o} + x^{o} + 18^{o} = 180^{o}

⇒ 3x^{o} = 180^{o} – 18^{o} = 162^{o}

x = = 54^{o}

Thus, the required angles of the triangle are 54^{o}, 54^{o} and x^{o} + 18^{o} = 54^{o} + 18^{o} = 72^{o}.

**Question 9:**

Let ∠C be the smallest angle of ABC.

Then, ∠A = 2∠C and B = 3∠C

Also, ∠A + ∠B + ∠C = 180^{o}

⇒ 2∠C + 3∠C + ∠C = 180^{o}

⇒ 6∠C = 180^{o}

⇒ ∠C = 30^{o}

So, ∠A = 2∠C = 2 (30^{o}) = 60^{o}

∠B = 3∠C = 3 (30^{o}) = 90^{o}

∴ The required angles of the triangle are 60^{o}, 90^{o}, 30^{o}.

**Question 10:**

Let ABC be a right angled triangle and ∠C = 90^{o}

Since, ∠A + ∠B + ∠C = 180^{o}

⇒ ∠A + ∠B = 180^{o} – ∠C = 180^{o} – 90^{o} = 90^{o}

Suppose ∠A = 53^{o}

Then, 53^{o} + ∠B = 90^{o}

⇒ ∠B = 90^{o} – 53^{o} = 37^{o}

∴ The required angles are 53^{o}, 37^{o} and 90^{o}.

**Question 11:**

Let ABC be a triangle.

Given, ∠A + ∠B = ∠C

We know, ∠A + ∠B + ∠C = 180^{o}

⇒ ∠C + ∠C = 180^{o}

⇒ 2∠C = 180^{o}

⇒ ∠C = = 90^{o}

So, we find that ABC is a right triangle, right angled at C.

**Question 12:**

Given : ∆ABC in which ∠A = 90^{o}, AL ⊥ BC

To Prove: ∠BAL = ∠ACB

Proof :

In right triangle ∆ABC,

⇒ ∠ABC + ∠BAC + ∠ACB = 180^{o}

⇒ ∠ABC + 90^{o} + ∠ACB = 180^{o}

⇒ ∠ABC + ∠ACB = 180^{o} – 90^{o}

∴ ∠ABC + ∠ACB = 90^{o}

⇒ ∠_{ }ACB = 90^{o} – ∠ABC ….(1)

Similarly since ∆ABL is a right triangle, we find that,

∠BAL = 90^{o} – ∠ABC …(2)

Thus from (1) and (2), we have

∴ ∠BAL = ∠ACB (Proved)

**Question 13:**

Let ABC be a triangle.

So, ∠A < ∠B + ∠C

Adding A to both sides of the inequality,

⇒ 2∠A < ∠A + ∠B + ∠C

⇒ 2∠A < 180^{o} [Since ∠A + ∠B + ∠C = 180^{o}]

⇒ ∠A < = 90^{o}

Similarly, ∠B < ∠A + ∠C

⇒ ∠B < 90^{o}

and ∠C < ∠A + ∠B

⇒ ∠C < 90^{o}

∆ABC is an acute angled triangle.

**Question 14:**

Let ABC be a triangle and ∠B > ∠A + ∠C

Since, ∠A + ∠B + ∠C = 180^{o}

⇒ ∠A + ∠C = 180^{o} – ∠B

Therefore, we get

∠B > 180^{o }– ∠B

Adding ∠B on both sides of the inequality, we get,

⇒ ∠B + ∠B > 180^{o} – ∠B + ∠B

⇒ 2∠B > 180^{o}

⇒ ∠B > = 90^{o}

i.e., ∠B > 90^{o} which means ∠B is an obtuse angle.

∆ABC is an obtuse angled triangle.

**Question 15:**

Since ∠ACB and ∠ACD form a linear pair.

So, ∠ACB + ∠ACD = 180^{o}

⇒ ∠ACB + 128^{o} = 180^{o}

⇒ ∠ACB = 180^{o} – 128 = 52^{o}

Also, ∠ABC + ∠ACB + ∠BAC = 180^{o}

⇒ 43^{o} + 52^{o} + ∠BAC = 180^{o}

⇒ 95^{o} + ∠BAC = 180^{o}

⇒ ∠BAC = 180^{o} – 95^{o} = 85^{o}

∴ ∠ACB = 52^{o} and ∠BAC = 85^{o}.

**Question 16:**

As ∠DBA and ∠ABC form a linear pair.

So, ∠DBA + ∠ABC = 180^{o}

⇒ 106^{o} + ∠ABC = 180^{o}

⇒ ∠ABC = 180^{o} – 106^{o} = 74^{o}

Also, ∠ACB and ∠ACE form a linear pair.

So, ∠ACB + ∠ACE = 180^{o}

⇒ ∠ACB + 118^{o} = 180^{o}

⇒ ∠ACB = 180^{o} – 118^{o} = 62^{o}

In ∠ABC, we have,

∠ABC + ∠ACB + ∠BAC = 180^{o}

74^{o} + 62^{o} + ∠BAC = 180^{o}

⇒ 136^{o} + ∠BAC = 180^{o}

⇒ ∠BAC = 180^{o} – 136^{o} = 44^{o}

∴ In triangle ABC, ∠A = 44^{o}, ∠B = 74^{o} and ∠C = 62^{o}

**Question 17:**

(i) ∠EAB + ∠BAC = 180^{o} [Linear pair angles]

110^{o} + ∠BAC = 180^{o}

⇒ ∠BAC = 180^{o} – 110^{o} = 70^{o}

Again, ∠BCA + ∠ACD = 180^{o }[Linear pair angles]

⇒ ∠BCA + 120^{o} = 180^{o}

⇒ ∠BCA = 180^{o} – 120^{o} = 60^{o}

Now, in ∆ABC,

∠ABC + ∠BAC + ∠ACB = 180^{o}

x^{o} + 70^{o} + 60^{o} = 180^{o}

⇒ x + 130^{o} = 180^{o}

⇒ x = 180^{o} – 130^{o} = 50^{o}

∴ x = 50

(ii)

In ∆ABC,

∠A + ∠B + ∠C = 180^{o}

⇒ 30^{o} + 40^{o} + ∠C = 180^{o}

⇒ 70^{o} + ∠C = 180^{o}

⇒ ∠C = 180^{o} – 70^{o} = 110^{o}

Now ∠BCA + ∠ACD = 180^{o} [Linear pair]

⇒ 110^{o} + ∠ACD = 180^{o}

⇒ ∠ACD = 180^{o} – 110^{o} = 70^{o}

In ∆ECD,

⇒ ∠ECD + ∠CDE + ∠CED = 180^{o}

⇒ 70^{o} + 50^{o} + ∠CED = 180^{o}

⇒ 120^{o} + ∠CED = 180^{o}

∠CED = 180^{o} – 120^{o} = 60^{o}

Since ∠AED and ∠CED from a linear pair

So, ∠AED + ∠CED = 180^{o}

⇒ x^{o} + 60^{o} = 180^{o}

⇒ x^{o} = 180^{o} – 60^{o} = 120^{o}

∴ x = 120

(iii)

∠EAF = ∠BAC [Vertically opposite angles]

⇒ ∠BAC = 60^{o}

In ∆ABC, exterior ∠ACD is equal to the sum of two opposite interior angles.

So, ∠ACD = ∠BAC + ∠ABC

⇒ 115^{o} = 60^{o} + x^{o}

⇒ x^{o} = 115^{o} – 60^{o} = 55^{o}

∴ x = 55

(iv)

Since AB || CD and AD is a transversal.

So, ∠BAD = ∠ADC

⇒ ∠ADC = 60^{o}

In ∠ECD, we have,

∠E + ∠C + ∠D = 180^{o}

⇒ x^{o} + 45^{o} + 60^{o} = 180^{o}

⇒ x^{o} + 105^{o} = 180^{o}

⇒ x^{o} = 180^{o} – 105^{o} = 75^{o}

∴ x = 75

(v)

In ∆AEF,

Exterior ∠BED = ∠EAF + ∠EFA

⇒ 100^{o} = 40^{o} + ∠EFA

⇒ ∠EFA = 100^{o} – 40^{o} = 60^{o}

Also, ∠CFD = ∠EFA [Vertically Opposite angles]

⇒ ∠CFD = 60^{o}

Now in ∆FCD,

Exterior ∠BCF = ∠CFD + ∠CDF

⇒ 90^{o} = 60^{o} + x^{o}

⇒ x^{o} = 90^{o} – 60^{o} = 30^{o}

∴ x = 30

(vi)

In ∆ABE, we have,

∠A + ∠B + ∠E = 180^{o}

⇒ 75^{o} + 65^{o} + ∠E = 180^{o}

⇒ 140^{o} + ∠E = 180^{o}

⇒ ∠E = 180^{o} – 140^{o} = 40^{o}

Now, ∠CED = ∠AEB [Vertically opposite angles]

⇒ ∠CED = 40^{o}

Now, in ∆CED, we have,

∠C + ∠E + ∠D = 180^{o}

⇒ 110^{o} + 40^{o} + x^{o} = 180^{o}

⇒ 150^{o} + x^{o} = 180^{o}

⇒ x^{o} = 180^{o} – 150^{o }= 30^{o}

∴ x = 30

**Question 18:**

Produce CD to cut AB at E.

Now, in ∆BDE, we have,

Exterior ∠CDB = ∠CEB + ∠DBE

⇒ x^{o} = ∠CEB + 45^{o } …..(i)

In ∆AEC, we have,

Exterior ∠CEB = ∠CAB + ∠ACE

= 55^{o} + 30^{o} = 85^{o}

Putting ∠CEB = 85^{o} in (i), we get,

x^{o} = 85^{o} + 45^{o} = 130^{o}

∴ x = 130

**Question 19:**

The angle ∠BAC is divided by AD in the ratio 1 : 3.

Let ∠BAD and ∠DAC be y and 3y, respectively.

As BAE is a straight line,

∠BAC + ∠CAE = 180^{o } [linear pair]

⇒ ∠BAD + ∠DAC + _{ ∠}CAE = 180^{o}

⇒ y + 3y + 108^{o} = 180^{o}

⇒ 4y = 180^{o} – 108^{o} = 72^{o}

⇒ y = = 18^{o}

Now, in ∆ABC,

∠ABC + ∠BCA + ∠BAC = 180^{o}

y + x + 4y = 180^{o}

[Since, ∠ABC = ∠BAD (given AD = DB) and ∠BAC = y + 3y = 4y]

⇒ 5y + x = 180

⇒ 5 × 18 + x = 180

⇒ 90 + x = 180

∴ x = 180 – 90 = 90

**Question 20:**

Given : A ∆ABC in which BC, CA and AB are produced to D, E and F respectively.

To prove : Exterior ∠DCA + Exterior ∠BAE + Exterior ∠FBD = 360^{o}

Proof : Exterior ∠DCA = ∠A + ∠B ….(i)

Exterior ∠FAE = ∠B + ∠C ….(ii)

Exterior ∠FBD = ∠A + ∠C ….(iii)

Adding (i), (ii) and (iii), we get,

Ext. ∠DCA + Ext. ∠FAE + Ext. ∠FBD

= ∠A + ∠B + ∠B + ∠C + ∠A + ∠C

= 2∠A + 2∠B + 2∠C

= 2 (∠A + ∠B + ∠C)

= 2 × 180^{o}

[Since, in triangle the sum of all three angle is 180^{o}]

= 360^{o}

Hence, proved.

**Question 21:**

In ∆ACE, we have,

∠A + ∠C + ∠E = 180^{o }….(i)

In ∆BDF, we have,

∠B + ∠D + ∠F = 180^{o} ….(ii)

Adding both sides of (i) and (ii), we get,

∠A + ∠C +∠E + ∠B + ∠D + ∠F = 180^{o} + 180^{o}

⇒ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360^{o}.

**Question 22:**

Given : In ∆ABC, bisectors of ∠B and ∠C meet at O and ∠A = 70^{o}

In ∆BOC, we have,

∠BOC + ∠OBC + ∠OCB = 180^{o}

= 180^{o} – 55^{o} = 125^{o}

∴ ∠BOC = 125^{o}.

**Question 23:**

We have a ∆ABC whose sides AB and AC have been procued to D and E. A = 40^{o} and bisectors of ∠CBD and ∠BCE meet at O.

In ∆ABC, we have,

Exterior ∠CBD = C + 40^{o}

And exterior ∠BCE = B + 40^{o}

Now, in ∆BCO, we have,

= 50^{o} + 20^{o}

= 70^{o}

Thus, ∠BOC = 70^{o}

**Question 24:**

In the given ∆ABC, we have,

∠A : ∠B : ∠C = 3 : 2 : 1

Let ∠A = 3x, ∠B = 2x, ∠C = x. Then,

∠A + ∠B + ∠C = 180^{o}

⇒ 3x + 2x + x = 180^{o}

⇒ 6x = 180^{o}

⇒ x = 30^{o}

∠A = 3x = 3 30^{o} = 90^{o}

∠B = 2x = 2 30^{o} = 60^{o}

and, ∠C = x = 30^{o}

Now, in ∆ABC, we have,

Ext ∠ACE = ∠A + ∠B = 90^{o} + 60^{o} = 150^{o}

∠ACD + ∠ECD = 150^{o}

⇒ ∠ECD = 150^{o} – ∠ACD

⇒ ∠ECD = 150^{o} – 90^{o} [since , AD ⊥ CD, ∠ACD = 90^{o}]

⇒ ∠ECD= 60^{o}

**Question 25:**

In ∆ABC, AN is the bisector of ∠A and AM ⊥ BC.

Now in ∆ABC we have;

∠A = 180^{o} – ∠B – ∠C

⇒ ∠A = 180^{o} – 65^{o} – 30^{o}

= 180^{o} – 95^{o}

= 85^{o}

Now, in ∆ANC we have;

Thus, ∠MAN =

**Question 26:**

(i) False (ii) True (iii) False (iv) False (v) True (vi) True.

Yuvraj Jaiswal says

We have unit test topic also in rs agarwal after the chapters.

Jivan Kumar says

Thanks sir please New addition of solutions in syllabus

Akarshit Pandey says

Thanks,it is very helpful for me.

Akarshit Pandey says

Ya

vaibhav says

what is triangles

Gupt says

Please Explain

pratap kr mandal says

thanks

riya agrawal says

Thanks ????☺

Sarthak Anand says

Thanks, it is very helpful for me .

Thanks thanks