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.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 6 |

Chapter Name | T-Ratios of Some Particular Angles |

Exercise | 6 |

Category | RS 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°

**Question 2.****Solution:**

On substituting the value of various T-ratios, we get

cos60° cos30° – sin60° sin30°

**Question 3.****Solution:**

On substituting the value of various Tratios, we get

cos45° cos30° + sin45° sin30°

**Question 4.****Solution:**

On substituting the value of various Tratios, we get

**Question 5.****Solution:**

**Question 6.****Solution:**

On substituting the value of various Tratios, we get

**Question 7.****Solution:**

On substituting the value of various Tratios, we get

**Question 8.****Solution:**

On substituting the value of various Tratios, we get

**Question 9.****Solution:**

On substituting the value of various Tratios, we get

**Question 10.****Solution:**

(i)

(ii)

**Question 11.****Solution:**

(i)

R.H.S. = L.H.S.

Hence, sin60° cos30° – cos60° sin30° = sin30°

(ii)

L.H.S. = cos60° cos30° + sin60° sin30°

(iii)

R.H.S. = L.H.S.

Hence,2sin30° cos30° = sin60°

(iv)

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

(ii) cos2A = cos90° = 0

**Question 13.****Solution:**

A = 30 ⇒ 2A = 60

(i)

(ii)

(iii)

**Question 14.****Solution:**

(i)

(ii)

**Question 15.****Solution:**

(i)

(ii)

(iii)

**Question 16.****Solution:**

Hence, (A + B) = 45

**Question 17.****Solution:**

Putting A = 30° 2 A = 60°

**Question 18.****Solution:**

Putting A = 30° 2 A = 60°

**Question 19.****Solution:**

Putting A = 30° 2 A = 60°

**Question 20.****Solution:**

From right angled ∆ABC,

**Question 21.****Solution:**

From right angled ∆ABC,

**Question 22.****Solution:**

From right angled ∆ABC,

(i)

(ii)

By Pythagoras theorem

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

**Question 23.****Solution:**

sin (A + B)= 1 sin (A + B) = sin90°

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:**

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:**

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:**

**Question 27.****Solution:**

**All Chapter RS Aggarwal Solutions For Class 10 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class10**

**All Subject NCERT Solutions For Class 10**

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rsaggarwalsolutions.in to your friends.