 # RS Aggarwal Class 9 Solutions Chapter 4 - Lines And Triangles

## RS Aggarwal Class 9 Chapter 4 - Lines And Triangles Solutions Free PDF

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

With the help of these RS Aggarwal Class 9 solutions scoring desirable marks in Class 9 exams becomes easy. We cover all the chapters and provide the step by step solutions to all the questions asked in the RS Aggarwal maths textbook. The solutions provided below are accurately solved by our team of subject experts.

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$\angle AOB$ .

(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=362746$\angle A=36^{\circ}{27}'{46}”$ and B=284339$\angle B=28^{\circ}{43}'{39}”$, find A+B$\angle A+\angle B$.

Ans:

angle A + angle B = [(36o27’46”) + (28o43’39”)]

We know that    1o = 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 = 65o 11’ 25”

Q3: Find the difference between two angles measuring 36and242830$36^{\circ}\; and \; 24^{\circ}{28}'{30}”$.

Ans:

We know that 36o = 35o 59’ 60”

 Deg Min Sec 35 59 60 24 28 30 11 31 30

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

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

(i) 58o

(ii) 16o

(iii) 23$\frac{2}{3}$ of a right angle

(iv) 46o 30’

(v) 52o 43’ 20”

(vi) 68o 35’ 45”

Ans:

(i) Complement of 58o = (90 -58)o = 32o

(ii) Complement if 16o = (90 – 16)o = 74o

(iii) 23$\frac{2}{3}$ of a right angle = (90×23)o=60o$(90 \times \frac{2}{3})^{o} = 60^{o}$

Complement of 23$\frac{2}{3}$ of a right angle = (90 – 60)o = 30o

(iv) We know that 90o = 89o 60’

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

(v) We know that 90o = 89o 60’

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

 Deg Min Sec 89 59 60 52 43 20 37 16 40

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

(vi) We know that 90o = 89o 59’ 60”

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

 Deg Min Sec 89 59 60 68 35 45 21 24 15

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

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

(i) 63°

(ii) 138°

(iii) 35$\frac{3}{5}$ of a right angle

(iv) 75° 36′

(v) 124° 20′ 40″

Ans:

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

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

(iii) 35$\frac{3}{5}$ of a right angle = (35×90)$(\frac{3}{5} \times 90)$ = 54°

Supplement of 35$\frac{3}{5}$ 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 = 55o 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$36^{\circ}$ 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$25^{\circ}$ 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$\frac{1}{3}$(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$10^{\circ}$ 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 180o.

Therefore,

AOC+BOC$\angle AOC + \angle BOC$ = 180o

62o+xo=180o$\Rightarrow 62^{o} + x^{o} = 180^{o}$
xo=(180o62o)$\Rightarrow x^{o} = (180^{o} – 62^{o})$
x=118o$\Rightarrow x = 118^{o}$

Hence, the value of x is 118o

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

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

Therefore,

AOC+COD+BOD=180o$\angle AOC + \angle COD + \angle BOD = 180^{o}$
(3x6)o+55o+(x+20)o=180o$\Rightarrow (3x-6)^{o} +55^{o} +(x+20)^{o} = 180^{o}$
4x=111o$\Rightarrow 4x=111^{o}$
x=27.5o$\Rightarrow x = 27.5^{o}$

Hence,

AOC=3x6$\angle AOC = 3x-6$

=3 x 27.5-6

=76.5o

And BOD=x+20$\angle BOD = x + 20$

= 27.5+20

= 47.5o

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

Ans: AOB Is a straight line. Therfore,

AOC+COD+BOD$\angle AOC + \angle COD + \angle BOD$
(3x+7)o+(2x19)o+xo=180o$\Rightarrow (3x+7)^{o} + (2x -19)^{o} +x^{o} =180^{o}$
6x=192o$\Rightarrow 6x=192^{o}$
x=32o$\Rightarrow x = 32^{o}$

Therefore,

AOC$\angle AOC$ = 3×32o+7=103o

COD$\angle COD$ =2×32o – 19 = 45o and

BOD$\angle BOD$ = 32o

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$\angle XOP + \angle POQ + \angle YOQ = 180^{o}$

5a + 4a + 6a = 180o

15a = 180o

a = 12o

Therefore,

x = 5 x 12o = 60o

y = 4 x 12o = 48o

and z = 6 x 12o =72o

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 = 180o

7x = 196o

X = 28o

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

Question 21: Two lines AB and CD intersect at O. If AOC=50$\angle AOC=50^{\circ}$, find AOC,BODandBOC$\angle AOC,\angle BOD\; and \; \angle BOC$.

Ans: We know that if two line intersects then the vertically opposite angles are equal. Therefore, AOC=BOD=50o$\angle AOC = \angle BOD = 50^{o}$

Let AODBOC=xo$\angle AOD – \angle BOC = x^{o}$

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

Therefore,

AOC+AOD+BOD+BOC=360o$\angle AOC + \angle AOD + \angle BOD + \angle BOC = 360^{o}$

50 + x + 50 + x = 360o

2x = 260o

X = 130o

Hence, AOD=BOC=130o$\angle AOD = \angle BOC = 130^{o}$

Therefore,

AOD=130o,BOD=50oandBOC=130o$\angle AOD = 130^{o} , \angle BOD = 50^{o} and \angle BOC =130^{o}$

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$\angle BOD = \angle AOC = 90^{o}$

Hence, t=90o

Also,

DOF=COE=50o$\angle DOF = \angle COE = 50^{o}$

Since, AOB is a straight line, we have

AOC+COE+BOE=180o$\angle AOC + \angle COE + \angle BOE = 180^{o}$

90 + 50 + y = 180o

140 + y =180o

Y = 40o

Also,

BOE=AOF=40o$\angle BOE = \angle AOF = 40^{o}$

Hence, x = 40o

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

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,COEandAOE$\angle AOD,\angle COE\; and \; \angle AOE$.

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

therefore, DOF=COE=5xo$\angle DOF = \angle COE = 5x^{o}$

AOD=BOC=2xo$\angle AOD = \angle BOC = 2x^{o}$ and

AOE=BOF=3xo$\angle AOE = \angle BOF = 3x^{o}$

Since, AOB is a straight line, we have

AOE+COE+BOC=180o$\angle AOE + \angle COE + \angle BOC = 180^{o}$

3x + 5x + 2x =180o

10x = 180o

X = 18o

Therefore,

AOD=2×18o=36o$\angle AOD = 2 \times 18^{o} = 36^{o}$
COE=5×18o=90o$\angle COE = 5 \times 18^{o} = 90^{o}$
AOE=3×18o=54o$\angle AOE = 3 \times 18^{o} = 54^{o}$

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 = 180o

9x = 180o

x = 20o

Hence, the two angles are 5 x20o = 100o and 4 x 20o = 80o

Question 25: If two straight lines intersect each other in such a way that one of the angles formed measures 90$90^{\circ}$, show that each of the remaining angles measures 90$90^{\circ}$.

Ans:

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

AOC=90o. Then AOC=BOD=90o$\angle AOC = 90^{o}.\ Then\ \angle AOC = \angle BOD = 90^{o}$

And let BOC=AOD=x$\angle BOC = \angle AOD = x$

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

AOC+BOD+AOD+BOC=360o$\angle AOC + \angle BOD + \angle AOD + \angle BOC = 360o$

90o +90o + x + x = 360o

2x = 180o

x =90o

Hence, BOC=AOD=90o$\angle BOC = \angle AOD = 90^{o}$

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

Hence, the measure of each of the remaining angles are 90o

Question 26: Two lines AB and CD intersect at a point O such that angleBOC+angleAOD=280$angle BOC+angle AOD =280^{\circ}$, 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$\angle BOC = \angle AOD$

Then,

x + x = 280

2x = 280

x = 140o

therefore, BOC=AOD=140o$\angle BOC = \angle AOD = 140^{o}$

Also, let AOC=BOD=yo$\angle AOC = \angle BOD = y^{o}$

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

therefore, AOC+BOC+BOD+AOD=360o$\angle AOC + \angle BOC + \angle BOD + \angle AOD = 360^{o}$

y + 140 + y + 140 = 360o

2y = 80o

Y = 40o

Hence, AOC=BOD=40o$\angle AOC = \angle BOD = 40^{o}$

Therefore, BOC=AOD=140o and AOC=BOD=40o$\angle BOC = \angle AOD = 140^{o}\ and\ \angle AOC = \angle BOD = 40^{o}$

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

Ans:

Since DOC is a straight line, we have

BOD+COB=180o$\angle BOD + \angle COB = 180^{o}$ …(1)

AOC+AOD=180o$\angle AOC + \angle AOD = 180^{o}$ …(2)

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

BOD+COB=AOC+AOD$\angle BOD + \angle COB = \angle AOC + \angle AOD$

Also, COB=AOC$\angle COB = \angle AOC$ [Since OC is a bisector of AOB$\angle AOB$ ]

Therefore, AOD=BOD$\angle AOD = \angle BOD$

### 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.