CBSE Class 9

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

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 4A.

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:
(i)Angle : When two rays OA and OB meet at a point o, then ∠AOB is called an angle.

(ii) Interior of angle : The interior an angle is a 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.
(iii) Obtuse angle : An angle greater than 90° but less than 180° is called an obtuse angle.
(iv) Reflex angle : An angle more than 180° but less than 360° is called a reflex angle.
(v) Complementary angles : Two angles are said to be complementary angles if their sum is 90°.
(vi) Supplementary angles : Two angles are said to be supplementary angles if their sum is 180°.

Question 2.
Solution:
∠A = 36°27’46”
∠B = 28° 43’39”
Adding, ∠A + ∠B = 64° 70′ 85″
We know that 60″ = 1′ and 60′ = 1°
∠A+ ∠B = 65° 11′ 25″ Ans.

Question 3.
Solution:
36° – 24° 28′ 30″
= 35° 59’60” – 25° 28’30”

= 10° 31′ 30″ Ans.

Question 4.
Solution:
We know that two angles are complementary of their sum is 90°. Each of these two angles is complement to the other, therefore.
(i) Complement of 58° = 90° – 58° = 32°
(ii) Complement of 16° = 90° – 16° = 74°
(iii) Complement of \(\frac { 2 }{ 3 } \) of a right angle i.e.
of \(\frac { 2 }{ 3 } \) x 90° or 60° = 90° – 60° = 30°
= \(\frac { 2 }{ 3 } \) of right angle,
(iv) Complement of 46° 30′
= 90° – 46° 30′
= 43° 30′
(v) Complement of 52° 43′ 20°= 90° – 52° 43′ 20″
= 37° 16′ 40″
(vi) Complement of 68° 35′ 45″
= 90° – 68° 35′ 45″
= 21° 24′ 15″ Ans.

Question 5.
Solution:
We know that two angles are said to be supplement to each other of their sum is 180° therefore
(i) Supplement of 68° = 180° – 68° =112°
(ii) Supplement of 138° = 180° – 138° = 42°
(iii) Supplement of \(\frac { 3 }{ 5 } \) of a right angle or \(\frac { 3 }{ 5 } \) x 90° or 54°
= 180° – 54° = 126°
(iv) Supplement of 75° 36′ = 180° – 75° 36′ = 104° 24′
(v) Supplement of 124° 20′ 40″
= 180° – 124° 20′ 40″
= 55° 39′ 20″
(vi) Supplement of 108° 48′ 32″ = 180° – 108″ 48′ 32″ = 71° 11′ 28″ Ans.

Question 6.
Solution:
(i) Let the measure of required angle = x ,
their its complement = 90° – x
According to the condition,
x = 90° – x => 2x = 90°
=>x = \(\frac { { 90 }^{ o } }{ 2 } \) = 45°
Required angle = 45°
(ii) Let the measure of required angle = x then its supplement = 180° – x
According to the condition,
x = 180° – x => 2x = 180° = 90°
=>x = \(\frac {{ 180 }^{ o }}{ 2 }\) = 90°
Hence required angle = 90° Ans.

Question 7.
Solution:
Let required angle = x
then its complement = 90° – x
According to the condition,
x – (90° – x) = 36°
=> x – 90° + x = 36°
=> 2x = 36° + 90° = 126°
= \(\frac { { 126 }^{ o } }{ 2 } \) = 63°
Required angle = 63° Ans.

Question 8.
Solution:
Let the required angle = x
then its supplement = 180° – x
According to the condition,
(180° – x) – x = 25°
=> 180° – x – x = 25°
=> – 2x = 25° – 180°
=> – 2x = – 155°
=> x = \(\frac { { – 155 }^{ o } }{ – 2 } \)
= 77.5°
Hence required angle = 77.5° Ans.

Question 9.
Solution:
Let required angle = x
Then its complement = 90° – x
According to the condition,
x = 4 (90° – x) => x = 360° – 4x
=> x + 4x = 360° => 5x = 360°
x = \(\frac { { 360 }^{ o } }{ 5 } \) = 72°
Required angle = 72° Ans.

Question 10.
Solution:
Let required angle = x
Then its supplement = 180° – x
According to the condition,
x = 5 (180° – x)
=> x = 900° – 5x
=> x + 5x = 900°
=> 6x = 900°
=> x = \(\frac { { 900 }^{ o } }{ 6 } \) = 150°
Hence, required angle = 150° Ans

Question 11.
Solution:
Let required angle = x
then its supplement = 180° – x
and complement = 90° – x
According to the condition,
180° – x = 4 (90°- x)
=> 180° – x = 360° – 4x
=> – x + 4x — 360° – 180°
=> 3x= 180°
=> x = \(\frac { { 180 }^{ o } }{ 3 } \) = 60°
Required angle = 60° Ans.

Question 12.
Solution:
Let required angle = x
Then, its complement = 90° – x
and its supplement = 180° – x
According to the condition,
90° – x = \(\frac { 1 }{ 3 } \) (180° – x)
=> 90° – x = 60° – \(\frac { 1 }{ 3 } \) x
=> 90° – 60° = x – \(\frac { 1 }{ 3 } \) x
=> \(\frac { 2 }{ 3 } \) x = 30° =>x = \(\frac { { 30 }^{ o }X3\quad \quad }{ 2 } \) => x = 45° Ans.

Question 13.
Solution:
Let one angle = x
Then, its supplement = 180° – x
According to the condition,
x : (180° – x) = 3:2
=> \(\frac { x }{ { 180 }^{ o }-x } =\frac { 3 }{ 2 } \)
=>2x = 3(180°- x)
=> 2x = 540° – 3x
=> 2x + 3x = 540°
=> 5x = 540° => x = \(\frac { { 540 }^{ o } }{ 5 } \) = 108°
Angle = 108° and its supplement = 180° – 108° = 72°
Hence, angles are 108° and 72° Ans.

Question 14.
Solution:
Let angle = x
Then, its complementary angle = 90° – x
According to the condition,
x : (90° – x) = 4 : 5
=> \(\frac { x }{ { 90 }^{ o }-x } =\frac { 4 }{ 5 } \)
=> 5x = 4 (90° – x)
=> 5x = 360° – 4x
=> 5x + 4x = 360°
=> 9x = 360°
=> x = \(\frac { { 360 }^{ o } }{ 9 } ={ 40 }^{ o } \)
and its complement = 90° – 40° = 50°
Hence, angles are 40° and 50° Ans.

Question 15.
Solution:
Let the required angle = x
.’. its complement = 90° – x
and supplement = 180° – x
According to the condition,
7(90° – x) = 3(180° – x) – 10°
=> 630° – 7x = 3 (180° – x) – 10°
=> 630° – 7x = 540° – 3x – 10°
=> – 7x + 3x = 540° – 10° – 630°
– 4x = – 100°
x = \(\frac { { -100 }^{ o } }{ -4 } ={ 25 }^{ o }\)
Hence required angle = 25° Ans.

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

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A.

Question 1.
Solution:
A theorem is a statement that requires a proof while an axiom is the basic fact which is taken for granted without proof.

Question 2.
Solution:
(i) Line segment: The straight line between two points A and B is a called a line segment \(\overline { AB } \)

(ii) Ray : A line segment \(\overline { AB } \) when extended indefinitely is one direction is called a ray \(\overrightarrow { AB } \) It has no definitely length.

(iii) Intersecting lines : Two lines having one common point are called intersecting lines and the common point is called the point of intersection.

(iv) Parallel Lines : If two lines lying in the same plane do not intersect each other when produced on either side, then these two lines are called parallel lines. The distance between two parallel hues always remains the same.

(v) Half line : If we take a point P on a line \(\overleftrightarrow { AB } \), its divides the line into two parts. Each part is called half line or two ray i.e. \(\overrightarrow { PA } \) and \(\overrightarrow { PB } \) .

(vi) Concurrent lines : Three or more lines intersecting at the same point are called concurrent lines.

(vii) Collinear points : Three or more points lying on the same line are called collinear points.

(viii) Plane : A plane is a surface such that every point of the line joining any two points on it, lies on it.

Question 3.
Solution:
(i) Six points are : A, B, C, D, E and F
(ii) Five line segments are : \(\overline { EG }\), \(\overline { FH }\), \(\overline { EF }\), \(\overline { GH }\) and \(\overline { MN }\)
(iii) Four rays are : \(\overrightarrow { EP } \) , \(\overrightarrow { GR } \),\(\overrightarrow { GB } \) and \(\overrightarrow { HD } \)
(iv) Four lines are : \(\overrightarrow { AB } \),\(\overrightarrow { CD } \),\(\overrightarrow { PQ } \) and \(\overrightarrow { RS } \)
(v) Four collinear points are M, E, G, B. Ans

Question 4.
Solution:
(i) \(\overleftrightarrow { EF } \) and \(\overleftrightarrow { GH } \) is a pair of intersecting line whose point of intersection is R
and second pair of intersecting lines is \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { CD } \) and point of intersection is P.
(ii) Three concurrent lines are \(\overleftrightarrow { AB } \), \(\overleftrightarrow { EF } \) and \(\overleftrightarrow { GH } \) and the point of intersection is R.
(iii) Three rays are \(\overleftrightarrow { RB } \),\(\overleftrightarrow { RH } \) and \(\overleftrightarrow { RG } \)
(iv) Two line segments are \(\overleftrightarrow { RQ } \) and \(\overleftrightarrow { RP } \)

Question 5.
Solution:
(i) Through a given point, infinitely many lines can be drawn.
(ii) Only one line can be drawn to pass through two given points.
(iii) Two lines can intersect each other at the most one point
(iv) A, B and C are three collinear points. Then the line segments will be \(\overline { AB } \), \(\overline { BC } \) and \(\overline { AC } \).

Question 6.
Solution:
(iv), (vi), (vii), (viii) and (ix) are true and others are not true.

Hope given RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2K

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2K.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2A
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2B
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2C
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2D
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2E
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2F
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2G
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2H
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2I
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2J
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2K

Question 1.
Solution:
125a³ + b³ + 64c³ – 60abc

Question 2.
Solution:
a³ + 8b³ + 64c³ – 24abc

Question 3.
Solution:
1 + b³ + 8c³ – 6bc

Question 4.
Solution:
216 + 27b³ + 8c³ – 108bc

Question 5.
Solution:
27a³ – b³ + 8c³ + 18abc

Question 6.
Solution:
8a³ + 125b³ – 64c³ + 120abc

Question 7.
Solution:
8 – 27b³ – 343c³ – 126bc

Question 8.
Solution:
125 – 8x³ – 27y³ – 90xy

Question 9.
Solution:
2√2a³ + 16√2b³ + c³ – 12abc

Question 10.
Solution:
x³ + y³ – 12xy + 64

Question 11.
Solution:
(a – b)³ + (b – c)³ + (c – a)³

Question 12.
Solution:
(3a – 2b)³ + (26 – 5c)³ + (5c – 3a)³

Question 13.
Solution:
a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

Question 14.
Solution:
(5a – 7b)³ + (9c – 5a)³ + (7b – 9c)³

Question 15.
Solution:
(x + y – z) (x² + y² + z² – xy + yz + zx)

Question 16.
Solution:
(x – 2y + 3) (x² + 4y² + 2xy -3x + 6y + 9)

Question 17.
Solution:
(x – 2y – z) (x² + 4y² + z² + 2xy + zx- 2yz)

Question 18.
Solution:
x + y + 4 = 0,

Question 19.
Solution:
x = 2y + 6

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2K are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.