RS Aggarwal Solutions Class 10 Chapter 5 Trigonometric Ratios


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RS Aggarwal Solutions Class 10 Chapter 5 Trigonometric Ratios





Exercise 5





Question 1:
Solution:





Given function: sin θ = √3/2





Let us first draw a right ∆ABC, ∠B = 90 degrees and ∠A = 𝜃





rs aggarwal class 10 chapter 10 img 1




(where k is a positive)





We know that sin 𝜃 = BC/AC = (Perpendicular)/Hypotenuse = √3/2





By Pythagoras Theorem:





AC2 = AB2 + BC2





Or AB2 = AC2 – BC2 = 4k2 – 3k2 = k2





AB = k





Find other T-rations using their definitions:





Cos = AB/AC = 1/2





Tan 𝜃 = BC/𝐴𝐵 = √3





𝑐𝑜𝑠𝑒𝑐 𝜃 = 1/sin𝜃 = 2/√3





sec 𝜃 = 1/cos 𝜃 = 2





cot 𝜃 = 1/tan 𝜃 = 1/√3





Question 2:
Solution:





Given function: cos θ = 7/25





Draw a right ∆ABC, ∠B = 90 degrees and ∠A = θ





rs aggarwal class 10 chapter 10 img 2




(where k is a positive)





We know that cos θ = AB/AC = Base/Hypotenuse = 7/25





By Pythagoras Theorem:





AC2 = AB2 + BC2





Or BC2 = AC2 – AB2 = 625k2 – 49k2 = 576k2





AB = 24k





Find other T-rations using their definitions:





Sin 𝜃 = BC/AC = 24/25





Tan 𝜃 = BC/𝐴𝐵 = 24/7





𝑐𝑜𝑠𝑒𝑐 𝜃 = 1/sin𝜃 = 25/24





sec 𝜃 = 1/cos 𝜃 = 25/7





cot 𝜃 = 1/tan 𝜃 = 7/24





Question 3:
Solution:





Given function: tan θ=15/8





Draw a right ∆ABC, ∠B = 90 degrees and ∠A = θ





rs aggarwal class 10 chapter 10 img 3




We know that tan θ = BC/AB = perpendicular/base = 15/8





(where k is a positive)





By Pythagoras Theorem:





AC2 = AB2 + BC2





= 64k2 + 225k2





= 289k2





AC = 17k





Find other T-rations using their definitions:





Sin 𝜃 = BC/AC = 15/17





cos 𝜃 = AB/AC = 8/17





𝑐𝑜𝑠𝑒𝑐 𝜃 = 1/sin𝜃 = 17/15





sec 𝜃 = 1/cos 𝜃 = 17/8





cot 𝜃 = 1/tan 𝜃 = 8/15





Question 4:
Solution:





Given function: cot θ= 2





Draw a right ∆ABC, ∠C = 90 degrees and ∠A = θ





rs aggarwal class 10 chapter 10 img 4




We know that cot θ = AC/BC = base/perpendicular = 2/1





(where k is a positive)





By Pythagoras Theorem:





AB2 = BC2 + AC2





= k2 + 4k2





= 5k2





AB = k√5





Find other T-rations using their definitions:





Sin 𝜃 = BC/AB = k/(k√5) = 1/√5





cos 𝜃 = AC/AB = (2k)/(k√5) = 2/√5





tan θ = BC/AC = sinθ /cosθ = k/(2k) = 1/2





𝑐𝑜𝑠𝑒𝑐 𝜃 = 1/sin𝜃 = √5





sec 𝜃 = 1/cos 𝜃 = √5/2





Question 5:
Solution:





Given function: cosec θ = √10





Draw a right ∆ABC, ∠C = 90 degrees and ∠A = θ





rs aggarwal class 10 chapter 10 img 5




We know that, cosecθ = AB/BC = hypotenuse/perpendicular = (k√10)/k





(where k is a positive)





By Pythagoras Theorem:





AB2 = AC2 + BC2





AB2 = AC2 – BC2





= 10k2 – k2





= 9k2





AC = 3k





Find other T-rations using their definitions:





Sin 𝜃 = BC/AB = 1/√10





cos 𝜃 = AC/AB = (3k)/(k√10) = 3/√10





tan θ = BC/AC = sinθ /cosθ = 1/3





secθ = AB/AC = 1/cosθ = √10/3





cotθ = AC/BC = 1/tanθ = 3





Question 6:
Solution:





Draw a triangle, ΔABC, Let ∠ACB = θ and ∠B = 90 degrees





rs aggarwal class 10 chapter 10 img 6




Sin θ = (a2 – b2) / (a2 + b2)





AB = (a2 – b2)





AC = (a2 + b2)





By Pythagoras theorem:





BC = √[(a2 + b2)2 – (a2 – b2)2]





BC = √(4a2 b2)





or BC = 2ab





Find other T-rations using their definitions:





cos 𝜃 = base/hypotenuse = 2ab / (a2 + b2)





tan θ = perpendicular/base = (a2 – b2) / 2ab





cosec 𝜃 = 1/sin 𝜃 = (a2 + b2)/(a2 – b2)





sec θ = 1/cos 𝜃 = (a2 + b2)/2ab





cotθ = 1/tanθ = 2ab/(a2 – b2)





Question 7:
Solution:





Given: 15 cot A = 8





cotA = (8k)/(15k) = 1/tanA = AC/BC





rs aggarwal class 10 chapter 10 img 7




Where k is any positive.





By Pythagoras theorem:





AB2 = BC2 + AC2





= (15k)2 + (8k)2





= 289k2





AB = 17k





Find other T-rations using their definitions:





sin A = perpendicular/ hypotenuse = (15k)/(17k) = 15/17





sec A = hypotenuse /base = 17/8





Question 8:
Solution:





Draw a triangle, ΔABC, Let ∠ACB = θ and ∠B = 90 degress





Sin A = perpendicular/ hypotenuse = 9/41





rs aggarwal class 10 chapter 10 img 8




By Pythagoras theorem:





AC2 = AB2 + BC2





AB2 = AC2 – BC2





= 412 – 92





= 1600





AB = 40





Find other T-ratios using their definitions:





cos A = base/ hypotenuse = 40/41





tan A = perpendicular /base = 9/40





Question 9:
Solution:





cos θ = 0.6 = (6k)/(10k) = AC/AB





rs aggarwal class 10 chapter 10 img 9




Where k is any positive.





By Pythagoras theorem:





AB2 = BC2 + AC2





BC2 = AB2 – AC2





= (10k)2 + (6k)2





= 64k2





BC = 8k





Find other T-rations using their definitions:





sin θ = perpendicular/ hypotenuse = 8/10





tan θ = perpendicular/base = 8/6





Now,





LHS = 5sinθ – 3tanθ





= 5(8/10) – 3(8/6)





= 4 – 3(4/3)





= 4(3) – 3(4)





= 12 – 12





= 0





=RHS





Hence proved.





Question 10:
Solution:





cosec θ = 2





or 1/sinθ = 2





(cosecθ is reciprocal of sin θ)





sin θ = 1/2





which implies θ = 30 degrees.





Find the values of cos θ and cot θ at θ = 30 degrees.





cos 30^0 = √3/2 and cot 30^0 = √3





Now,





LHS = cot 0 + sin0/(1 +cos0)





= √3 + 1/2 / (1 + √3/2)





rs aggarwal class 10 chapter 10 img 11




=2





=RHS





Hence proved.





Question 11:
Solution:





Given: tan θ = 1/√7





tanθ = k/(k√7) = BC/AC





rs aggarwal class 10 chapter 10 img 13




Where k is any positive.





By Pythagoras theorem:





AB2 = BC2 + AC2





= k2 + 7k2





AB = 2k√2





Find cosec θ and sec θ using their definitions:





cosec θ = AB/BC = 2k√2/k = 2√2





secθ = AB/AC = 2√2/√7





Now,





LHS =





rs aggarwal class 10 chapter 10 img 14




= 48/64





=3/4





= RHS





Hence proved





Question 12.
Solution:





Given: tan θ = 20/21





tanθ = 20k/(21k)





rs aggarwal class 10 chapter 10 img 16




Where k is any positive.





By Pythagoras theorem:





AC2 = AB2 + BC2





= 441k2 + 400k2





AC = 29k





Find sin θ and cos θ using their definitions:





sin θ = perpendicular/ hypotenuse = 20/29





cos 𝜃 = base/hypotenuse = 21/29





Now,





LHS =





rs aggarwal class 10 chapter 10 img 17




rs aggarwal class 10 chapter 10 img 18




=30/70





= 3/7





=RHS





Hence proved





Question 13:
Solution:





Given: secθ=5/4





sec θ = 5k/(4k) = 5/4 and cos θ = 4/5





rs aggarwal class 10 chapter 10 img 20




Where k is any positive.





By Pythagoras theorem:





AC2 = AB2 + BC2





BC2 = AC2 – AB2





= 25k2 – 16k2





BC = 9k2 = 3k





Find sin θ, tan θ and cot θ using their definitions:





sin θ = perpendicular/ hypotenuse = 3/5





tan θ = perpendicular/base = 3/4





cotθ = 1/tanθ = 4/3





Now,





LHS =





rs aggarwal class 10 chapter 10 img 21




=RHS





Question 14:
Solution:





Given: cotθ=3/4





or cotθ = 3k/4k





rs aggarwal class 10 chapter 10 img 23




Where k is any positive.





By Pythagoras theorem:





AC2 = AB2 + BC2





= 9k2 +16k2





= 25k2





AC = 5k





Find sec θ and cosec θ using their definitions:





sec θ = hypotenuse/base = 5/3





cosec 𝜃 = hypotenuse/perpendicular = 5/4





Now,





rs aggarwal class 10 chapter 10 img 24




Question 15:
Solution:





Given: sinθ=3/4





or sinθ=3k/4k





rs aggarwal class 10 chapter 10 img 26




Where k is any positive.





By Pythagoras theorem:





AC2 = AB2 + BC2





16k2 = AB2 + 9k2





AB = √7





Find sec θ and cot θ using their definitions:





sec θ = hypotenuse/base = 4/√7





cotθ = base/perpendicular = √7/3





Now,





rs aggarwal class 10 chapter 10 img 27




Question 16:
Solution:





Given: sinθ = a/b





or sinθ= ak/bk





rs aggarwal class 10 chapter 10 img 29




Where k is any positive.





By Pythagoras theorem:





AC2 = AB2 + BC2





AB2 = AC2 – BC2





= b2 – a2b2−a2−−−−−−√





Find sec θ and tan θ using their definitions:





sec θ = hypotenuse/base = b/√(b2 – a2)





tan θ = perpendicular/base = a/√(b2 – a2)





Now,





LHS = sec θ + tan θ





rs aggarwal class 10 chapter 10 img 30




= RHS





Question 17:
Solution:





Given: cosθ=3/5





cosθ = (3k)/(5k) = AC/AB





rs aggarwal class 10 chapter 10 img 32




Where k is any positive.





By Pythagoras theorem:





AB2 = BC2 + AC2





BC = 4k





Find sin θ, tan θ and cot θ using their definitions:





sin θ = perpendicular/ hypotenuse = 4/5





tan θ = perpendicular/base = 4/3





cot θ = 1/tan θ = 3/4





Now,





LHS =





rs aggarwal class 10 chapter 10 img 33




= 3/160





= RHS





Question 18:
Solution:





Given: tan θ = 4/3





tanθ = (4k)/(3k) = BC/AC





rs aggarwal class 10 chapter 10 img 34




Where k is any positive.





By Pythagoras theorem:





AB2 = BC2 + AC2





AB2 = 16k2 + 9k2





AB = 5k





Find sin θ and cos θ using their definitions:





sin θ = perpendicular/ hypotenuse = 4/5





cos 𝜃 = base/hypotenuse = 3/5





Now,





LHS = sin θ + cos θ





= 4/5 + 3/5





= 7/5





= RHS





Question 19:
Solution:





Given: tan θ = a/b





tanθ = (ak)/(bk) = BC/AB





rs aggarwal class 10 chapter 10 img 36




Where k is any positive.





By Pythagoras theorem:





AC2 = AB2 + BC2AC=a2+b2−−−−−−√





Find sin θ and cos θ using their definitions:





sin θ = perpendicular/ hypotenuse = a/(√(a2+b2)





cos 𝜃 = base/hypotenuse = b/(√(a2+b2)





Now,





LHS:





rs aggarwal class 10 chapter 10 img 37




Question 20:
Solution:





Given: 3 tan θ = 4





or tan θ = 4/3





rs aggarwal class 10 chapter 10 img 39




Question 21:
Solution:





Given: 3 cot θ = 2





or cot θ = 2/3





Where k is any positive.





rs aggarwal class 10 chapter 10 img 40




By Pythagoras theorem:





AC2 = AB2 + BC2





AC2 = (2k)2 + (3k)2





AC2 = 4k2 + 9k2 = 13k2





AC = √13 k





Find sin θ and cos θ using their definitions:





sin θ = perpendicular/ hypotenuse = 3/√13





cos 𝜃 = base/hypotenuse =2/√13





Now,





rs aggarwal class 10 chapter 10 img 41




Question 22:
Solution:





Given: 3cotθ=4





or cot θ = 4/3





rs aggarwal class 10 chapter 10 img 42




Where k is any positive.





By Pythagoras theorem:





AC2 = AB2 + BC2





= 16 + 9





AC = 5 k





Find sin θ and cos θ using their definitions:





sin θ = perpendicular/ hypotenuse = 3/5





cos 𝜃 = base/hypotenuse = 4/5





Now,





rs aggarwal class 10 chapter 10 img 43




Question 23:
Solution:





Given: sec θ = 17/8





or sec θ = 17k/8k





rs aggarwal class 10 chapter 10 img 45




Where k is any positive.





By Pythagoras theorem:





AC2 = AB2 + BC2





BC2 = AC2 – AB2





= 289k2 – 64k2





= 225 k2





BC = 15 k





rs aggarwal class 10 chapter 10 img 46




Question 24:
Solution:





Draw a triangle using given instructions:





rs aggarwal class 10 chapter 10 img 47




From figure: Δ ABC and Δ ABD are right angled triangles





where AD = 10cm BC = CD = 4cm





BD = BC + CD = 8cm





By Pythagoras theorem:





AD2 = BD2 + AB2





(10)2 = (8)2 + AB2





100 = 64 + AB2





AB2 = 36 = (6)2





or AB = 6cm





Again,





AC2 = BC2 + AB2





AC2 = (4)2 + (6)2





AC2 = 16 + 36 = 52





or AC = √52 = 2√13 cm





(i) Find sin θ





sin θ = BC/AC = 4/2√13 = 2√13/13





(ii) Find cos θ





cosθ = AB/AC = 6/2√13 = 3/√13 = 3√13/13





Question 25:





Solution:





Draw a triangle using given instructions:





rs aggarwal class 10 chapter 10 img 48




From figure: Δ ABC is a right angled triangle





By Pythagoras theorem:





AC2 = BC2 + AB2





AC = 25





(i) Find sin A





sin A = BC/AC = 7/25





(ii) Find cos A





cos A = AB/AC = 24/25





(iii) sin C = AB/AC = 24/25





(iv) cosC = BC/AC = 7/25





Question 26:
Solution:





Draw a triangle using given instructions:





rs aggarwal class 10 chapter 10 img 49




From figure: Δ ABC is a right angled triangle





AB2 = AC2 + BC2





(29)2 = AC2 + (21)2





841 = AC2 + 441





AC2 = 400





or AC = 20





Find sin θ and cos θ:





sinθ = AC/AB = 20/29





cosθ = BC/AB = 21/29





Now:





LHS = cos2θ – sin2θ





= (21/29)2 – (20/29)2





= 41/841





= RHS





Hence proved.





Question 27:
Solution:





rs aggarwal class 10 chapter 10 img 50




From figure: Δ ABC is a right angled triangle





AC2 = BC2 + AB2





AC2 = (5)2 + (12)2





AC2 = 25 + 144





AC2 = 169 = (13)2





or AC = 13





Now from figure,





i. cos A = AB/AC = 12/13





ii. cosec A = AC/BC = 13/5





iii. cos C = BC/AC = 5/13





iv. cosec C = AC/AB = 13/12





Hence proved.





Question 28:
Solution:





Given: sinα = k/(2k) = BC/AB





rs aggarwal class 10 chapter 10 img 51




Where k is any positive.





By Pythagoras theorem:





AB2 = BC2 + AC2





AC2 = AB2 – BC2





= (2k)2 – (k)2





= 3k2





or AC= k√3





Find cos α:





cos α = base/hypotenuse = √3/2





Now,





rs aggarwal class 10 chapter 10 img 52




Question 29:
Solution:





Given: tan A = BC/AB = k/(k√3)





rs aggarwal class 10 chapter 10 img 53




Where k is any positive.





By Pythagoras theorem:





AC2 = BC2 + AB2





= (k)2 – (√3 k)2





= k2 + 3k2





= 4k2





or AC= 2k





Find sin A, cos A, sin C and cos C





sin A = BC/AC = k/(2k) = 1/2





cos A = AB/AC = (k√3)/(2k) = √3/2





sin C = AB/AC = (k√3)/(2k) = √3/2





cos C = BC/AC = k/(2k) = ½





(i) sinA cosC + cosA sinC = 1





LHS = sinA cosC + cosA sinC = (1/2)(1/2) + ( √3/2)( √3/2)





= 1/4 + 3/4





= 4/4





= 1





RHS = LHS





(ii) cosA cosC – sinA sinC = 0





LHS = cosA cosC – sinA sinC





= (1/2)(√3/2) – (1/2)(√3/2)





= (√3/4) – (√3/4)





= 0





=RHS





Question 30:
Solution:





Consider two right triangles XAY and WBZ such that sin A = sin B





rs aggarwal class 10 chapter 10 img 54




To Prove: ∠A = ∠B





From figures:





sin A = XY/XA and sin B = WZ / WB





XY/XA = WZ / WB = k (say)





or XY/WZ = XA/ WB …(1)





sin A = sin B (Given)





We have,





XY = WZ k and XA = WB k …(2)





By Pythagoras: Apply on both the triangles





WB2 = WZ2 + BZ2





BZ2 = WB2 – WZ2





And,





XA2 = XY2 + AY2





AY2 = XA2 – XY2





rs aggarwal class 10 chapter 10 img 55




Question 31:
Solution:





Consider ΔABC to be a right angled triangle.





angle C = 90 degree





tan A = BC/AC and





tan B = AC/BC





Given: tanA = tanB





So, BC/AC = AC/BC





BC2 = AC2





BC = AC





Which implies, ∠ A = ∠ B (using triangle opposite and equal angles property)





Question 32:
Solution:





Consider ΔABC to be a right angled triangle at B.





angle C = 90 degree





Given: tan A = 1 …(1)





tan A = 1 = BC/AB





AB = BC





Again, tan A = sin A/cos A





sin A = cos A …using (1)





By Pythagoras theorem:





AC2 = BC2 + AB2





AC2 = 2BC2





(AC/BC)2 = 2





Or AC/BC = √2





cosecA = √2





or sin A = 1/√2





and cosA = 1/√2





Now,





2 sinA cosA = 2(1/√2)( 1/√2)





= 2(1/2)





= 1





= RHS





Question 33:





Solution:





rs aggarwal class 10 chapter 10 img 56




Δ PQR is a right angled triangle.





By Pythagoras theorem:





PR2 = RQ2 + PQ2





(x + 2)2 = x2 + PQ2





PQ2 = 4 + 4x





or PQ = 2√(x+1)





Now,





cot ϕ = QR/PQ = x/2(√(x+1)





tan θ = QR/PQ = x/2(√(x+1)





(i) √(x+1) cot ϕ = √(x+1) {x/2(√(x+1)} = x/2





(ii) √(x^3 + x^ 2) tan θ = √(x^3 + x^ 2) {x/2(√(x+1)} = x2/2





(iii) cos θ = PQ/PR = 2(√(x+1) / (x+2)





Question 34:





rs aggarwal class 10 chapter 10 img 57




(Using trig property: sin2 + cos2 A = 1)





= 1 – 1





= 0





Question 35:





rs aggarwal class 10 chapter 10 img 58




Complete RS Aggarwal Solutions Class 10





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