An angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint called the vertex of the angle. The different kinds of angle are: Right Angle, Acute Angle, Obtuse Angle, Straight Angle, Reflex Angle and Complete Angle.

A line can be straight or curved. The word line usually means a straight line. The different types of lines are: Parallel Lines, Perpendicular lines, Vertical Lines. Horizontal Lines, Skew Lines.

A plane figure bounded by three line segments is called a traingle. The types of triangles on the basis of sides are: Equilateral Triangle, Isosceles Triangles, Scalene Triangle. The types of triangles on the basis of angles are: Right-angled Triangle, Acute-angled Triangle, Obtuse-Angled Triangle.

## Download PDF of RS Aggarwal Class 9 Chapter 4 – Angles, Lines and Triangles

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**Q1: Define the following terms**

**(i)angle**

**(ii) Interior of an angle.**

**(iii) Obtuse angle**

**(iv) Reflex angle**

**(v) Complementary angles**

**(vi) Supplementary angles**

Answer :

(i) Two rays OA and OB, with a common end-point 0, form an angle AOB that is represented as ∠AOB .

(ii) The interior of an angle is the set of all points in its plane, which lie on the same side of OA as B and also on the same side of OB as A.

– ► B

(iii) An angle greater than 90° but less than 180° is called an obtuse angle.

(iv) An angle greater than 180° but less than 360° is called a reflex angle.

Reflex angle

(v) Two angles are said to be complementary if the sum of their measures is 90°.

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

**Q2: If ∠A=36∘27′46” and ∠B=28∘43′39”, find ∠A+∠B.**

Ans:

angle A + angle B = [(36^{o}27’46”) + (28^{o}43’39”)]

We know that 1^{o} = 60 mins or 60’

1’ = 60 sec or 60”

Deg | Min | Sec |

36 | 27 | 46 |

28 | 43 | 39 |

65 | 11 | 25 |

Hence, the measure of the required angle = 65^{o} 11’ 25”

**Q3: Find the difference between two angles measuring 36∘and24∘28′30”.**

Ans:

We know that 36^{o} = 35^{o} 59’ 60”

Deg | Min | Sec |

35 | 59 | 60 |

24 | 28 | 30 |

11 | 31 | 30 |

Hence, the measure of the required angle = 11^{o} 31’ 60”

**Q4: Find the complement of each of the following angles.**

**(i) 58 ^{o}**

**(ii) 16 ^{o}**

**(iii) 23 of a right angle**

**(iv) 46 ^{o} 30’**

**(v) 52 ^{o} 43’ 20”**

**(vi) 68 ^{o} 35’ 45”**

Ans:

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

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

(iii) 23 of a right angle = (90×23)o=60o

Complement of 23 of a right angle = (90 – 60)^{o} = 30^{o}

(iv) We know that 90^{o} = 89^{o} 60’

Complement of 46^{o} 30’ = (89^{o} 60’ – 46^{o} 30’)^{o} = 43^{o} 30’

(v) We know that 90^{o} = 89^{o} 60’

Complement of 52^{o} 43’ 20” = [(89^{o} 59’ 60”) – (52^{o} 43’ 20”)]

Deg | Min | Sec |

89 | 59 | 60 |

52 | 43 | 20 |

37 | 16 | 40 |

Hence, the measure of angle = 37^{o} 16’ 40”

(vi) We know that 90^{o} = 89^{o} 59’ 60”

Complement of 68^{o} 35’ 45” – [(89^{o} 59’ 60”) – (68^{o} 35’ 45”)]

Deg | Min | Sec |

89 | 59 | 60 |

68 | 35 | 45 |

21 | 24 | 15 |

Hence, the measure of the required angle = 21^{o} 24’ 15”

**Q5: Find the supplement of each of the following angles.**

**(i) 63°**

**(ii) 138°**

**(iii) 35 of a right angle**

**(iv) 75° 36′**

**(v) 124° 20′ 40″**

Ans:

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

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

(iii) 35 of a right angle = (35×90) = 54°

Supplement of 35 of a right angle = (180 – 54)^{o} = 126°

(iv) We know that 180° = 179° 60′

Supplement of 75° 36′ = (179° 60′ – 75° 36′) = 104° 24′

(v) We know that 180° = 179° 59′ 60″

Supplement of 124° 20′ 40″ = (179′ 59′ 60″ —124° 20′ 40″)

Deg | Min | Sec |

179 | 59 | 60 |

124 | 20 | 40 |

55 | 39 | 20 |

Hence, the measure of the required angle = 55^{o }39’ 20”

**Q6: Find the measure of an angle which is**

**(i) Equal to its complement.**

**(ii) Equal to its supplement.**

Ans:

(i) Let the measure of the required angle be x°.

Then, in the case of complementary angles:

x + x = 90°

= 2x = 90°

= x = 45°

Hence, the measure of the angle that is equal to its complement is 45°.

(ii) Let the measure of the required angle be x°.

Then, in the case of supplementary angles:

x—x=180°

= 2x = 180°

= x = 90°

Hence, the measure of the angle that is equal to its supplement is 90°.

**Q7:** **Find the measure of an angle which is 36∘ more than its complement.**

Ans:

Let the measure of the required angle be x°.

Then, the measure of its complement = (90 – x)^{o }.

Therefore,

X – (90° – x) = 36°

2x = 126°

X = 63°

Hence, the measure of the required angle is 63°.

**Q8: Find the measure of an angle which is 25∘ less than its supplement.**

Ans:

Let the measure of the required angle be x.

Then, measure of its supplement = (180° – x).

Therefore, (180° – x) – x = 25°

2x = 155°

x = 77.5°

Hence, the measure of the required angle is 77.5°.

**Q9:** **Find the angle which is four times its complement.**

Ans:

Let the measure of the required angle be x.

Then, the measure of its complement = (90° – x).

Therefore,

x = (90° – x) 4

x = 360° – 4x

5x = 360°

x = 72°

Hence, the measure of the required angle is 72°.

**Q10:** **Find the angle which is five times its supplement.**

Ans:

Let the measure of the required angle be x.

Then, measure of its supplement = 180° – x.

Therefore, x = (180° – x) 5

X = 900° – 5x

6x = 900°

x = 150°

Hence, the measure of the required angle is 150°.

**Q11: Find the angle whose supplement is four times its complement.**

Sol:

Let the measure of the required angle be x°.

Then, measure of its complement = (90 – x)^{o}

And, measure of its supplement= (180 – x)^{o}

Therefore,

(180 – x) = 4(90 – x)

180 – x = 360 – 4x

3x = 180

x = 60

Hence, the measure of the required angle is 60°.

**Q12: Find the angle whose complement is one-third of its supplement.**

Answer :

Let the measure of the required angle be x°.

Then, the measure of its complement = (90 – x).

And the measure of its supplement = 180 – x°.

Therefore, 90 – x =13(180 – x)

3(90 – x) = (180 – x)

270 – 3x = 180 – x

2x = 90

x = 45

Hence, the measure of the required angle is 45°.

**Q13: Two supplementary angles are in the ratio 3:2. Find the angles.**

Answer :

Let the two angles be 3x and 2x, respectively.

Then,

3x + 2x = 180

5x = 180

X = 36°

Therefore, the two angles are

3x = 3 x 36° = 108° and 2x = 2 x 36° = 72°.

**Q14: Two complementary angles are in the ratio 4:5. Find the angles.**

Answer :

Let the two angles be 4x and 5x, respectively.

Then, 4x + 5x = 90

9x = 90

X = 10°

Hence, the two angles are

4x = 4 x 10° = 40° and 5x = 5 x 10° = 50°.

**Q15:** **Find the measure of an angle, if 7 times its complement is 10∘ less than three times its supplement.**

Answer :

Let the measure of the required angle be x°.

Then, the measure of its complement = (90 – x)°.

And the measure of its supplement = (180 – x)°.

Therefore, 7(90 – x) = 3(180-x) – 10

= 630 – 7x = 540 – 3x-10

= 4x = 100

= x = 25

Hence, the measure of the required angle is 25°.

**Question 16: In the adjoining figure, AOB is a straight line. Find the value of x.**

** **

Ans: We know that the sum of angles in a linear pair is 180^{o}.

Therefore,

∠AOC+∠BOC = 180^{o}

⇒62o+xo=180o

⇒xo=(180o–62o)

⇒x=118o

Hence, the value of x is 118^{o}

**Question 17: In the adjoining figure, AOB is a straight line. Find the value of x. Hence find ∠AOC and ∠BOD.**

** **

Ans: As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is 180^{o}

Therefore,

∠AOC+∠COD+∠BOD=180o

⇒(3x−6)o+55o+(x+20)o=180o

⇒4x=111o

⇒x=27.5o

Hence,

∠AOC=3x−6

=3 x 27.5-6

=76.5^{o}

And ∠BOD=x+20

= 27.5+20

= 47.5^{o}

**Question 18: In the adjoining figure, AOB is a straight line. Find the value of x. Hence find ∠AOC, ∠COD and ∠BOD.**

** **

Ans: AOB Is a straight line. Therfore,

∠AOC+∠COD+∠BOD

⇒(3x+7)o+(2x−19)o+xo=180o

⇒6x=192o

⇒x=32o

Therefore,

∠AOC = 3×32^{o}+7=103^{o}

∠COD =2×32^{o} – 19 = 45^{o} and

∠BOD = 32^{o}

**Question 19: In the adjoining figure, x:y:z = 5:4:6 . If XOY is a straight line, find the value of x,y,z .**

** **

Ans: Let x = 5a; y = 4a and z = 6a

XOY is a straight line. Therefore

∠XOP+∠POQ+∠YOQ=180o

5a + 4a + 6a = 180^{o}

15a = 180^{o}

a = 12^{o}

Therefore,

x = 5 x 12^{o} = 60^{o}

y = 4 x 12^{o} = 48^{o}

and z = 6 x 12^{o} =72^{o}

**Question 20: In the adjoining figure, what value of x will make AOB, a straight line?**** **

Ans:

AOB will be a straight line, if

3x + 20 + 4x – 36 = 180^{o}

7x = 196^{o}

X = 28^{o}

Hence, x = 28 will make AOB a straight line.

**Question 21: Two lines AB and CD intersect at O. If ∠AOC=50∘, find ∠AOC,∠BODand∠BOC.**** **

Ans: We know that if two line intersects then the vertically opposite angles are equal. Therefore, ∠AOC=∠BOD=50o

Let ∠AOD–∠BOC=xo

Also, we know that the sum of all angles around a point is 360^{o}

Therefore,

∠AOC+∠AOD+∠BOD+∠BOC=360o

50 + x + 50 + x = 360^{o}

2x = 260^{o}

X = 130^{o}

Hence, ∠AOD=∠BOC=130o

Therefore,

∠AOD=130o,∠BOD=50oand∠BOC=130o

**Question 22: In the adjoining figure, three coplanar lines AB,CD and EF intersect at a point O, forming angles as shown. Find the value of x,y,z and t.**** **

Ans: We know that if two lines intersect, then the vertically opposite angles are equal.

therefore, ∠BOD=∠AOC=90o

Hence, t=90^{o}

Also,

∠DOF=∠COE=50o

Since, AOB is a straight line, we have

∠AOC+∠COE+∠BOE=180o

90 + 50 + y = 180^{o}

140 + y =180^{o}

Y = 40^{o}

Also,

∠BOE=∠AOF=40o

Hence, x = 40^{o}

Therefore, x = 40^{o}, y = 40^{o}, z = 50^{o} and t = 90^{o}

**Question 23: In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Hence find ∠AOD,∠COEand∠AOE.**** **

** **

Ans: We know that if two lines intersect, then the vertically opposite angles are equal.

therefore, ∠DOF=∠COE=5xo

∠AOD=∠BOC=2xo and

∠AOE=∠BOF=3xo

Since, AOB is a straight line, we have

∠AOE+∠COE+∠BOC=180o

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

10x = 180^{o}

X = 18^{o}

Therefore,

∠AOD=2×18o=36o

∠COE=5×18o=90o

∠AOE=3×18o=54o

**Question 24: Two adjacent angles on a straight line are in the ration 5:4. Find the measure of each one of these angles.**

Ans: Let the two adjacent angles be 5x and 4x, respectively

Then,

5x + 4x = 180^{o}

9x = 180^{o}

x = 20^{o}

Hence, the two angles are 5 x20^{o} = 100^{o} and 4 x 20^{o} = 80^{o}

**Question 25: If two straight lines intersect each other in such a way that one of the angles formed measures 90∘, show that each of the remaining angles measures 90∘.**

Ans:

We know that if two lines intersect, then the vertically opposite angles are equal.

∠AOC=90o. Then ∠AOC=∠BOD=90o

And let ∠BOC=∠AOD=x

Also, we know that the sum of all angles around a point is 360^{o}.

∠AOC+∠BOD+∠AOD+∠BOC=360o

90^{o} +90^{o} + x + x = 360^{o}

2x = 180^{o}

x =90^{o}

Hence, ∠BOC=∠AOD=90o

Therefore, angle AOC = angle BOD = angle BOC = angle AOD = 90^{o}

Hence, the measure of each of the remaining angles are 90^{o}

**Question 26: Two lines AB and CD intersect at a point O such that angleBOC+angleAOD=280∘, as shown in the figure. Find all the four angles.**** **

** **

Ans: We know that if two lines intersect, then the vertically-opposite angles are equal.

Let ∠BOC=∠AOD

Then,

x + x = 280

2x = 280

x = 140^{o}

therefore, ∠BOC=∠AOD=140o

Also, let ∠AOC=∠BOD=yo

We know that the sum of all angles around a point is 360^{o}.

therefore, ∠AOC+∠BOC+∠BOD+∠AOD=360o

y + 140 + y + 140 = 360^{o}

2y = 80^{o}

Y = 40^{o}

Hence, ∠AOC=∠BOD=40o

Therefore, ∠BOC=∠AOD=140o and ∠AOC=∠BOD=40o

**Question 27: In the given figure, ray OC is the bisector of ∠AOB and OD is the ray opposite to OC. Show that ∠AOD=∠BOD.**

** **

Ans:

Since DOC is a straight line, we have

∠BOD+∠COB=180o …(1)

∠AOC+∠AOD=180o …(2)

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

∠BOD+∠COB=∠AOC+∠AOD

Also, ∠COB=∠AOC [Since OC is a bisector of ∠AOB ]

Therefore, ∠AOD=∠BOD

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

The RS Aggarwal Maths Solutions are considered an extremely vital resource when it comes to exam preparation. The solutions are easy to understand and each step is described to match the understanding of the student. Students will find the complete syllabus in our solutions with a detailed explanation of the answers to the questions provided in the textbook. It will make your exam preparation easier. These solutions assist students to solve the difficult questions with effortless ease and perform well in the exam.