RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 6 T-Ratios of Some Particular Angles Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 6 T-Ratios of Some Particular Angles Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 6
Chapter NameT-Ratios of Some Particular Angles
Exercise6
CategoryRS Aggarwal Solutions

RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles

Question 1.
Solution:
On substituting the value of various T-ratios, we get
sin60° cos30° + cos60° sin30°
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 1.1

Question 2.
Solution:
On substituting the value of various T-ratios, we get
cos60° cos30° – sin60° sin30°
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 2.1

Question 3.
Solution:
On substituting the value of various Tratios, we get
cos45° cos30° + sin45° sin30°
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 3.1

Question 4.
Solution:
On substituting the value of various Tratios, we get
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 4.1

Question 5.
Solution:
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 5.1

Question 6.
Solution:
On substituting the value of various Tratios, we get
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 6.1

Question 7.
Solution:
On substituting the value of various Tratios, we get
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 7.1

Question 8.
Solution:
On substituting the value of various Tratios, we get
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 8.1

Question 9.
Solution:
On substituting the value of various Tratios, we get
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 9.1

Question 10.
Solution:
(i)
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 10.1

(ii)
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 10.2

Question 11.
Solution:
(i)
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 11.1
R.H.S. = L.H.S.
Hence, sin60° cos30° – cos60° sin30° = sin30°

(ii)
L.H.S. = cos60° cos30° + sin60° sin30°
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 11.2
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 11.3

(iii)
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 11.4
R.H.S. = L.H.S.
Hence,2sin30° cos30° = sin60°

(iv)
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 11.5
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 Class 10 Chapter 6 T-Ratios of Some Particular Angles 12.1

(ii) cos2A = cos90° = 0
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 12.2

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

(i)
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 13.1

(ii)
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 13.2

(iii)
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 13.3

Question 14.
Solution:
(i)
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 14.1

(ii)
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 14.2

Question 15.
Solution:
(i)
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 15.1
(ii)
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 15.2
(iii)
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 15.3

Question 16.
Solution:
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 16.1
Hence, (A + B) = 45

Question 17.
Solution:
Putting A = 30° 2 A = 60°
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 17.1

Question 18.
Solution:
Putting A = 30° 2 A = 60°
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 18.1

Question 19.
Solution:
Putting A = 30° 2 A = 60°
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 19.1

Question 20.
Solution:
From right angled ∆ABC,
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 20.1

Question 21.
Solution:
From right angled ∆ABC,
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 21.1

Question 22.
Solution:
From right angled  ∆ABC,
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 22.1

(i)
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 22.2

(ii)
By Pythagoras theorem
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 22.3
Hence, (i) BC = 3cm and (ii) AB = 3cm.

Question 23.
Solution:
sin (A + B)= 1  sin (A + B) = sin90°
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 23.1
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 Class 10 Chapter 6 T-Ratios of Some Particular Angles 24.1
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 Class 10 Chapter 6 T-Ratios of Some Particular Angles 25.1
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 Class 10 Chapter 6 T-Ratios of Some Particular Angles 26.1

Question 27.
Solution:
RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles 27.1

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