RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles
RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles

Class 10: Maths Chapter 11 solutions. Complete Class 10 Maths Chapter 11 Notes.

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles

RS Aggarwal 10th Maths Chapter 11, Class 10 Maths Chapter 11 solutions

Question 1.
Solution:
On substituting the value of various T-ratios, we get
sin60° cos30° + cos60° sin30°

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 1

Question 2.
Solution:
On substituting the value of various T-ratios, we get
cos60° cos30° – sin60° sin30°

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 2

Question 3.
Solution:
On substituting the value of various Tratios, we get
cos45° cos30° + sin45° sin30°

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 3

Question 4.
Solution:
On substituting the value of various Tratios, we get

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 4

Question 5.
Solution:

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 5

Question 6.
Solution:
On substituting the value of various Tratios, we get

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 6

Question 7.
Solution:
On substituting the value of various Tratios, we get

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 7

Question 8.
Solution:
On substituting the value of various Tratios, we get

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 8

Question 9.
Solution:
On substituting the value of various Tratios, we get

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 9

Question 10.
Solution:
(i)

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 10

(ii)

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 10

Question 11.
Solution:
(i)

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 11


R.H.S. = L.H.S.
Hence, sin60° cos30° – cos60° sin30° = sin30°

(ii)
L.H.S. = cos60° cos30° + sin60° sin30°

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 11
RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 11

(iii)

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 11

R.H.S. = L.H.S.
Hence,2sin30° cos30° = sin60°

(iv)

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 11

R.H.S. = sin90° = 1
R.H.S. = L.H.S.
Hence, 2 sin 45° cos45° = sin90°

Question 12.
Solution:
A = 45° 2 A = 90°

(i)Sin 2A = sin90° = 1

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 12

(ii) cos2A = cos90° = 0

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 12

Question 13.
Solution:
A = 30 ⇒ 2A = 60

(i)

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 13

(ii)

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 13

(iii)

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 13

Question 14.
Solution:
(i)

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 14

(ii)

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 15

Question 15.
Solution:
(i)

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 15


(ii)

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 15


(iii)

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 15

Question 16.
Solution:

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 16

Hence, (A + B) = 45

Question 17.
Solution:
Putting A = 30° 2 A = 60°

Question 18.
Solution:
Putting A = 30° 2 A = 60°

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 18
RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 18

Question 19.
Solution:
Putting A = 30° 2 A = 60°

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 19

Question 20.
Solution:
From right angled ∆ABC,

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 20

Question 21.
Solution:
From right angled ∆ABC,

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 21

Question 22.
Solution:
From right angled  ∆ABC,

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 22

(i)

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 22

(ii)
By Pythagoras theorem

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 22

Hence, (i) BC = 3cm and (ii) AB = 3cm.

Question 23.
Solution:
sin (A + B)= 1  sin (A + B) = sin90°

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 23

Adding (1) and (2), we get
2A = 90° ⇒ A = 45°
Putting A = 45° in (1) we get
45° + B = 90° ⇒ B = 45°
Hence, A = 45° and B = 45°.

Question 24.
Solution:

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 24

Solving (1) and (2), we get
2A = 90° ⇒ A = 45°
Putting A = 45° in (1), we get
45° – B = 30° ⇒ B = 45 – 30° = 15°
Hence, A = 45°, B = 15°.

Question 25.
Solution:

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 25

Solving (1) and (2), we get
2A = 90° ⇒ A = 45°
Putting A = 45° in (1), we get
45° – B = 30° ⇒ B = 45° – 30°  = 15°
A = 45°, B = 15°

Question 26.
Solution:

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 26

Question 27.
Solution:

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles Question 27

RS Aggarwal Solutions for Class 10 Maths Chapter 11: Download PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles

Download PDF: RS Aggarwal Solutions for Class 10 Maths Chapter 11–T Ratios Of Some Particular Angles PDF

Chapterwise RS Aggarwal Solutions for Class 10 Maths :

About RS Aggarwal Class 10 Book

Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.

He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.

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