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


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





Exercise 6





(Question 1 to Question 9)





Question 1:
Solution:





We know that,





sin 60° = √3/2 = cos 30°





and sin 30° = 1/2 = cos 60°





Now,





sin 60° cos 30° + cos 60° sin 30° = (√3/2)( √3/2) + (1/2)(1/2)





= (3/4) + (1/4)





= 4/4





= 1





Question 2:
Solution:





We know that,





cos 60° = 1/2 = sin 30°





and cos 30° = √3/2 = sin 60°





Now,





cos 60° cos 30° — sin 60° sin 30° = (1/2) × (√3/2) – (√3/2) × (1/2)





= (√3/4) – (√3/4)





= 0





Question 3:
Solution:





We know that,





cos 45° = 1/√2 = sin 45°





cos 30° = √3/2 and





sin 30° = 1/2





Now,





cos 45° cos 30° + sin 45° sin 30° = (1/√2)×(√3/2) + (1/√2)(1/2)





= (√3 / 2√2) + (1 / 2√2)





= (√3 + 1) / (2√2)





Question 4:
Solution:





We know that,





sin 30° = 1/2 , sin 60° = √3/2 , sin 90° = 1





cos 30° = √3/2 , cos 45° = 1/√2 , cos 60° = 1/2





sec 60° = 2, tan 45° = 1 and cot 45° = 1





Now,





rs aggarwal class 10 chapter 11 img 2




Question 5:





Solution:





We know that,





cos 30° = √3/2 ⇨ cos2 30° = 3/4





cos 60° = 1/2 ⇨ cos2 60° = 1/4





sec 30° =(2/√3) ⇨ sec2 30° = 4/3





tan 45° = 1 ⇨ tan2 45° = 1





sin 30° = 1/2 ⇨ sin2 30° = 1/4





Now,





(5cos260°+4sec230°-tan245°)/(sin230°+cos230°)





rs aggarwal class 10 chapter 11 img 3




= 67/12





Question 6:
Solution:





We know that,





sin 45° = 1/√2 , cos 60° = 1/2





sin 30° = 1/2 and cos 90° = 0





Now,





2cos2 60° + 3sin2 45° − 3sin2 30° + 2cos2 90°





rs aggarwal class 10 chapter 11 img 4




Question 7:
Solution:





We know that,





Cot 30° = √3, cos 30° = √3/2





Sec 45° = √2 , cosec 30° = 2





Now,





cot230° − 2cos230° − ¾ sec245° + ¼ cosec230°





rs aggarwal class 10 chapter 11 img 5




= 1





Question 8:
Solution:





sin 30° = 1/2 ⇒ sin2 30° = 1/4





cos 45° = 1/√2 = sin 45°





cot 45° = 1 ⇒ cot2 45° = 1





cos 60° = 1/2 ⇒ sec 60° = 2 ⇒ sec2 60° = 4





cos 30° = √3/2 ⇒ sec 30° = 2/√3 ⇒ sec2 30° = 4/3





cosec 45° = 1/sin 45° = √2 ⇒ cosec2 45° = 2





(sin230° + 4cot245° − sec260°)(cosec245° sec230°)





rs aggarwal class 10 chapter 11 img 6




=1/4 x 8/3





= 2/3





Question 9:
Solution:





4/cot230°+1/sin230° – 2cos245°-sin2





rs aggarwal class 10 chapter 11 img 7




= 4/3 + 4/1 – 1 – 0





= 26/6





= 13/3





Question 10:
Solution:





(i)





rs aggarwal class 10 chapter 11 img 9




(ii)





rs aggarwal class 10 chapter 11 img 10




Question 11:
Solution:





(i) L.H.S. = sin 60° cos 30° — cos 60° sin 30°





= (√3/2) × (√3/2) – (1/2)(1/2)





= (3/4) – (1/4)





= 2/4





= 1/2





R.H.S.:





sin 30° = 1/2





LHS = RHS





(ii)





L.H.S. = cos 60° cos 30° + sin 60° sin 30°





= (1/2) × (√3/2) + (√3/2)(1/2)





= (√3/4) + (√3/4)





= √3/2





R.H.S.





cos 30° = √3/2





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





(iii)





L.H.S. = 2 sin 30° cos 30°





= 2 × (1/2) × (√3/2)





= √3/2





R.H.S.





sin 60° = √3/2





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





(iv)L.H.S. = 2 sin 45° cos 45°





= 2 × (1/√2) × (1/√2)





= (2 × 1/2)





= 1





R.H.S. = sin 90° = 1





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





Question 12:
Solution:





A = 45° then 2 A = 90°





(i)Sin 2A = sin90°





RHS:





2 sin 45° cos 45° = 2 × (1/√2) × (1/√2)





= 1





LHS:





sin 90° = 1





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





(ii) cos 2A = cos90° = 0





rs aggarwal class 10 chapter 11 img 11




Question 13.
Solution:





A = 30 ⇒ 2A = 60





(i)





rs aggarwal class 10 chapter 11 img 13




(ii)





rs aggarwal class 10 chapter 11 img 14




(iii)





rs aggarwal class 10 chapter 11 img 15




Question 14:
Solution:





(i) sin (A + B) = sin A cos B + cos A sin B





If A = 60° and B = 30°, then





To verify: sin 90° = sin 60° cos 30° + cos 60° sin 30°





RHS: sin 60° cos 30° + cos 60° sin 30°





= (√3/2) × (√3/2) + (1/2)(1/2)





= (3/4) + (1/4)





= 4/4





= 1





= sin 90°





= LHS





(ii) cos (A + B) = cos A cos B — sin A sin B





If A = 60° and B = 30°





Verify: cos (90°) = cos 60° cos 30° — sin60° sin 30°





R.H.S. = cos 60° cos 30° – sin 60° sin 30°





= (1/2) × (√3/2) – (√3/2)(1/2)





= (√3/4) – (√3/4)





= 0





= cos 90°





= L.H.S.





Question 15:
Solution:





(i) sin (A – B) = sin A cos B – cos A sin B





If A = 60° and B = 30°, then





LHS :





= sin(60°-30°)





= sin 30°





= 1/2





R.H.S. = sin 60° cos 30° – cos 60° sin 30°





= (√3/2) × (√3/2) – (1/2) (1/2)





= (3/4) – (1/4)





= 2/4





= 1/2





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





(ii) cos (A – B) = cos A cos B + sin A sin B





If A = 60° and B = 30°, then





Verify: cos (30°) = cos 60° cos 30° + sin 60° sin 30°





R.H.S. = cos 60° cos 30° + sin 60° sin 30°





= (1/2) × (√3/2) + (√3/2)(1/2)





= (√3/4) + (√3/4)





= √3/2





= cos 30°





= L.H.S.





rs aggarwal class 10 chapter 11 img 16




Verified.





Question 16:
Solution:





rs aggarwal class 10 chapter 11 img 17




=( 5/6) /( 5/6)





= 1





This implies, tan(A + B) = 1





= tan 450





Or A + B = 450 . Proved





Question 17:
Solution:





Put A = 30° ⇨ 2A = 60°





rs aggarwal class 10 chapter 11 img 18




The value of tan 60is √3.





Question 18:
Solution:





Put A = 30° then 2 A = 60°cosA=1+cos2A2−−−−−−√





rs aggarwal class 10 chapter 11 img 19




The value of cos 30° is √3/2.





Question 19:
Solution:





Put A = 30° then 2 A = 60°





sinA=1–cos2A2−−−−−−√





Squaring both side, we get





sin2A=1–cos2A2





And,





rs aggarwal class 10 chapter 11 img 20




Sin 300 = 1/2





Question 20:
Solution:





Draw a right angled ∆ABC using given instructions:





rs aggarwal class 10 chapter 11 img 21




Here sin 30 = BC/AC





½ = BC/20





Or BC = 10 cm





By Pythagoras theorem:





(AB)2 = (AC)2 – (BC)2





=(20)2 – (10)2





= 300





AB = 10 √3 cm





Question 21:
Solution:





Draw a right angled ∆ABC using given instructions:





rs aggarwal class 10 chapter 11 img 22




Here sin 30 = BC/AC





½ = 6/AC





Or AC = 12cm





By Pythagoras theorem:





(AB)2 = (AC)2 – (BC)2





=(12)2 – (6)2





= 108





AB = 6 √3 cm





Question 22:
Solution:





From right angled ∆ABC,





rs aggarwal class 10 chapter 11 img 23




  1. BC/AC = sin45




BC32√=12√





Or BC = 3





  1. By Pythagoras theorem




(AB)2 = (AC)2 – (BC)2





= (3√2)2 – (3)2





= 18 – 9





= 9





AB = 3 cm





Question 23:
Solution:





sin (A + B)= 1 or sin (A + B) = sin90° [As sin90° = 1)





A + B = 90° …(1)





Again, Cos(A-B) = 1





= cos 0°





A – B = 0 …(2)





Adding (1) and (2), we get





2A = 90° or A = 45°





Putting A = 45° in (1) we get





45° + B = 90° or B = 45°





Hence, A = 45° and B = 45°.





Question 24:
Solution:





sin (A – B)= 1/2





or sin (A – B) = sin 30°





A – B = 30° …(1)





Again, Cos(A+B) = 1/2





= cos 60°





A + B = 60° …(2)





Solving (1) and (2), we get





2A = 90° or A = 45°





Putting A = 45° in (1), we get





45° – B = 30° or B = 45 – 30° = 15°





Therefore, A = 45°, B = 15°.





Question 25:





Solution:





tan (A – B)= 1/√3





or tan(A – B) = tan 30°





A – B = 30° …(1)





Again, tan(A+B) = √3





= tan 60°





A + B = 60° …(2)





Solving (1) and (2), we get





2A = 90° or A = 45°





Putting A = 45° in (1), we get





45° – B = 30° or B = 45° – 30° = 15°





Therefore, A = 45°, B = 15°





Question 26:





Solution:





Given: 3x = cosec θ and 3/x = cot θ





We know that: cose2 θ – cot2 θ = 1





Substituting the values, we get





(3x)2 – (3/x)2 = 1





9( x2 – 1/ x2) = 1





( x2 – 1/ x2) = 1/9





3 ( x2 – 1/ x2) = 1/3





Question 27:
Solution:





Given: sin (A + B) = sin A cos B + cos A sin B and





cos (A – B) = cos A cos B + sin A sin B





(i) To find: sin 75°





Put A = 30° and B = 45°, then





sin 75° = sin 30° cos 45° + cos 30° sin 45°





= (1/2) × (1/√2) + (√3/2) × (1/√2)





= (1/2√2) + (√3/2√2)





= (1+√3)/2√2





(ii) Find cos 75°





Put A = 45° and B = 30°, then





cos 15° = cos 45° cos 30° + sin 45° sin 30°





= (1/√2) × (√3/2) + (1/√2) × (1/2)





= (√3 / 2√2) + (1/2√2)





= (1+√3)/2√2





Complete RS Aggarwal Solutions Class 10





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