## RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B.

**Other Exercises**

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D

**Question 1.**

**Solution:**

AOB is a straight line

∠AOC + ∠BOC = 180° (Linear pair)

=> 62° + x= 180°

=> x = 180° – 62°

=> x = 118°

Hence, x = 118° Ans.

**Question 2.**

**Solution:**

AOB is straight line

∠AOC + ∠COD + ∠DOB – 180°

=> (3x – 5)° + 55° + (x + 20)° = 180°

=> 3x – 5° + 55° + x + 20° = 180°

=> 4x – 5° + 75° = 180°

=> 4x + 70° = 180°

=> 4x = 180° – 70°

=> 4x = 110°

=> \(x=\frac { { 110 }^{ o } }{ 4 } =27.{ 5 }^{ o }\)

Hence x = 27.5°

and ∠AOC = 3x – 5° = 3 x 27.5° – 5°

= 82.5° – 5° = 77.5°

∠BOD = x + 20° = 27.5° + 20°

= 47.5° Ans.

**Question 3.**

**Solution:**

AOB is a straight line

∠AOC + ∠COD + ∠DOB = 180°

{angles on the same side of line AB}

=> (3x + 7)° + (2x – 19)° + x = 180°

=> 3x + 7° + 2x – 19° + x = 180°

=> 6x – 12° – 180°

=> 6x = 180° + 12° = 192°

=> \(x=\frac { { 192 }^{ o } }{ 6 } =32^{ o }\)

Here x = 32°

∠AOC = 3x + 7° = 3 x 32° + 7°

= 96° + 7°= 103°

∠COD = 2x – 19° = 2 x 32° – 19°

= 64 – 19° = 45°

and ∠BOD = x = 32° Ans.

**Question 4.**

**Solution:**

In the given figure,

x + y + z = 180°

But x : y : z = 5:4:6

Let ∠XOP = x° – 5a

∠POQ =y° = 4a

and ∠QOY = z = 6a

then 5a + 4a + 6a = x + y + z = 180°

=> 15a = 180°

=> a = \(\frac { { 180 }^{ o } }{ 15 } ={ 12 }^{ o }\)

=> x = 5a = 5 x 12° = 60°

y = 4a = 4 x 12° = 48°

and z = 6x = 6 x 12° = 72° Ans.

**Question 5.**

**Solution:**

AOB will be a straight line

If ∠AOC + ∠COB = 180°

If (3x + 20)° + (4x – 36)° = 180°

If 3x + 20 + 4x – 36 = 180°

If 7x – 16 = 180°

If 7x = 180° + 16 = 196°

If \(x=\frac { { 196 }^{ o } }{ 7 } ={ 28 }^{ o }\)

Hence, if x = 28°, then AOB will be a straight line.

**Question 6.**

**Solution:**

AB and CD intersect each other at O

AOC = ∠BOD and ∠BOC = ∠AOD (vertically opposite angles)

But ∠AOC = 50°

∠BOD = ∠AOC = 50°

But ∠AOC + ∠BOC = 180° (Linear pair)

=> 50° + ∠BOC = 180°

=> ∠BOC = 180° – 50° = 130°

∠AOD = ∠BOC = 130°

Hence,∠AOD = 30°,∠BOD = 50° and ∠BOC = 130° Ans.

**Question 7.**

**Solution:**

In the figure,

AB, CD and EF are coplanar lines intersecting at O.

∠AOF = ∠BOE

∠DOF = ∠COE and ∠BOD = ∠AOC (Vertically opposite angles)

x = y,

z = 50°

t = 90°

But AOF + ∠DOF + ∠BOD = 180° (Angles on the same side of a st. line)

=> x + 50° + 90° = 180°

=> x° + 140° + 180°

=> x = 180° – 140° = 40°

Hence, x = 40°, y = x = 40°, z = 50° and t = 90° Ans.

**Question 8.**

**Solution:**

Three coplanar lines AB, CD and EF intersect at a point O

∠AOD = ∠BOC

∠DOF = ∠COE

and ∠AOE = ∠BOF

(Vertically opposite angles)

But ∠AOD = 2x

∠BOC = 2x

and ∠BOF = 3x

∠AOE = 3x

and ∠COE = 5x

∠DOF = 5x

But ∠AOD + ∠DOF + ∠BOF + ∠BOC + ∠COE + ∠AOE = 360° (Angles at a point)

=> 2x + 5x + 3x + 2x + 5x + 3x = 360°

=> 20x = 360° => x = \(\frac { { 360 }^{ o } }{ 20 } \) = 18°

Hence x = 18°

∠AOD = 2x = 2 x 18° = 36°

∠COE = 5x = 5 x 18° = 90°

and ∠AOE = 3x = 3 x 18° = 54° Ans.

**Question 9.**

**Solution:**

AOB is a line and CO stands on it forming ∠AOC and ∠BOC

But ∠AOC : ∠BOC = 5:4

Let ∠AOC = 5x and ∠BOC = 4x

But ∠AOC + ∠BOC = 180° (Linear pair)

=> 5x + 4x = 180° => 9x = 180°

=> x = \(\frac { { 180 }^{ o } }{ 9 } \) = 20°

∠AOC = 5x = 5 x 20° = 100°

and ∠BOC = 4x = 4 x 20° = 80° Ans.

**Question 10.**

**Solution:**

Two lines AB and CD intersect each other at O and

∠AOC = 90°

∠AOC = ∠BOD

(Vertically opposite angles)

∠BOD = 90°

But ∠AOC + ∠BOC = 180° (Linear pair)

=> 90° + ∠BOC – 180°

=> ∠BOC = 180° – 90° = 90°

But ∠AOD = ∠BOC

(Vertically opposite angles)

∠AOD = 90°

Hence each of the remaining angle is 90°.

**Question 11.**

**Solution:**

Two lines AB and CD intersect each other at O and

∠BOC + ∠AOD = 280°

∠AOD = ∠BOC

(vertically opposite angles)

∠BOC + ∠BOC = 280°

(∠AOD = ∠BOC)

=> 2 ∠BOC = 280°

=> ∠BOC = \(\frac { { 280 }^{ o } }{ 2 } \) = 140°

But ∠BOC + ∠AOC = 180° (Linear pair)

=> 140° + ∠AOC = 180°

=> ∠AOC = 180° – 140° = 40°

But ∠BOD = ∠AOC

(vertically opposite angles)

∠BOD = 40°

Hence ∠AOC = 40°, ∠BOC = 140°,

∠BOD = 40°

and ∠AOD = 140° Ans.

**Question 12.**

**Solution:**

OC is the bisector of ∠AOB. and OD is the ray opposite to OC.

Now ∠AOC = ∠BOC (OC is bisector of ∠AOB)

But ∠BOC + ∠BOD = 180° (Linear pair)

Similarly, ∠AOD + ∠AOC = 180°

=> ∠BOC + ∠BOD = ∠AOD + ∠AOC

But ∠AOC = ∠BOC (Given)

∠BOD = ∠AOD

=> ∠AOD = ∠BOD

Hence proved.

**Question 13.**

**Solution:**

AB is the mirror.

PQ is the incident ray, QR is its reflected ray.

=> ∠BQR = ∠PQA

But ∠BQR + ∠PQR + ∠PQA = 180° (Angles on one side of a straight line)

=> ∠PQA + ∠PQA + 112° = 180°

=> 2∠PQA + 112° = 180°

=> 2∠PQA = 180° – 112° = 68°

∠PQA = \(\frac { { 68 }^{ o } }{ 2 } \) = 34° Ans.

**Question 14.**

**Solution:**

**Given.** Two lines AB and CD intersect each other at O.

OE is the bisector of ∠BOD and EO is produced to F.

**To Prove :** OF bisects ∠AOC.

**Proof :** AB and CD intersect each other at O

∠AOC = ∠BOD

(Vertically opposite angles)

OE is the bisector of ∠BOD

∠1 = ∠2

But ∠1 = ∠3

and ∠2 = ∠4 (Vertically opposite angles)

and ∠1 = ∠2 (proved)

∠3 = ∠4

Hence, OF is the bisector of ∠AOC.

Hence proved.

**Question 15.**

**Solution:**

**Given** ∠AOC and ∠BOC are supplementary angles

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

**To Prove :** ∠EOF = 90°

**Proof :** ∠1 = ∠2

∠3 = ∠4

{OE and OF are the bisectors of ∠BOC and ∠AOC respectively}

But ∠AOC + ∠BOC = 180°

(Linear pair)

=> ∠1 + ∠2 + ∠3 + ∠4 = 180°

=> ∠1 + ∠1 + ∠3 + ∠3 = 180°

=> 2∠1 + 2∠3 = 180°

=> 2(∠1 + ∠3) = 180°

=> ∠1 + ∠3 = \(\frac { { 180 }^{ o } }{ 2 } \) 90°

=> ∠EOF = 90°

Hence proved.

Hope given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B are helpful to complete your math homework.

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