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In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
(i) sin A = 2/3
Solution:
We have,
sin A = 2/3 ……..….. (1)
As we know, by sin definition,
sin A = Perpendicular/ Hypotenuse = 2/3 ….(2)
By comparing eq. (1) and (2), we have
Opposite side = 2 and Hypotenuse = 3
Now, on using Pythagoras theorem in Δ ABC
AC2 = AB2 + BC2
Putting the values of perpendicular side (BC) and hypotenuse (AC) and for the base side as (AB), we get
⇒ 32 = AB2 + 22
AB2 = 32 – 22
AB2 = 9 – 4
AB2 = 5
AB = √5
Hence, Base = √5
By definition,
cos A = Base/Hypotenuse
⇒ cos A = √5/3
Since, cosec A = 1/sin A = Hypotenuse/Perpendicular
⇒ cosec A = 3/2
And, sec A = Hypotenuse/Base
⇒ sec A = 3/√5
And, tan A = Perpendicular/Base
⇒ tan A = 2/√5
And, cot A = 1/ tan A = Base/Perpendicular
⇒ cot A = √5/2
(ii) cos A = 4/5
Solution:
We have,
cos A = 4/5 …….…. (1)
As we know, by cos definition,
cos A = Base/Hypotenuse …. (2)
By comparing eq. (1) and (2), we get
Base = 4 and Hypotenuse = 5
Now, using Pythagoras’ theorem in Δ ABC,
AC2 = AB2 + BC2
Putting the value of base (AB) and hypotenuse (AC) and for the perpendicular (BC), we get
52 = 42 + BC2
BC2 = 52 – 42
BC2 = 25 – 16
BC2 = 9
BC= 3
Hence, Perpendicular = 3
By definition,
sin A = Perpendicular/Hypotenuse
⇒ sin A = 3/5
Then, cosec A = 1/sin A
⇒ cosec A= 1/ (3/5) = 5/3 = Hypotenuse/Perpendicular
And, sec A = 1/cos A
⇒ sec A =Hypotenuse/Base
sec A = 5/4
And, tan A = Perpendicular/Base
⇒ tan A = 3/4
Next, cot A = 1/tan A = Base/Perpendicular
∴ cot A = 4/3
(iii) tan θ = 11/1
Solution:
We have, tan θ = 11…..…. (1)
By definition,
tan θ = Perpendicular/ Base…. (2)
On Comparing eq. (1) and (2), we get;
Base = 1 and Perpendicular = 5
Now, using Pythagoras’ theorem in Δ ABC,
AC2 = AB2 + BC2
Putting the value of base (AB) and perpendicular (BC) to get hypotenuse(AC), we get
AC2 = 12 + 112
AC2 = 1 + 121
AC2= 122
AC= √122
Hence, hypotenuse = √122
By definition,
sin = Perpendicular/Hypotenuse
⇒ sin θ = 11/√122
And, cosec θ = 1/sin θ
⇒ cosec θ = √122/11
Next, cos θ = Base/ Hypotenuse
⇒ cos θ = 1/√122
And, sec θ = 1/cos θ
⇒ sec θ = √122/1 = √122
And, cot θ = 1/tan θ
∴ cot θ = 1/11
(iv) sin θ = 11/15
Solution:
We have, sin θ = 11/15 ………. (1)
By definition,
sin θ = Perpendicular/ Hypotenuse …. (2)
On Comparing eq. (1) and (2), we get,
Perpendicular = 11 and Hypotenuse= 15
Now, using Pythagoras’ theorem in Δ ABC,
AC2 = AB2 + BC2
Putting the value of perpendicular (BC) and hypotenuse (AC) to get the base (AB), we have
152 = AB2 +112
AB2 = 152 – 112
AB2 = 225 – 121
AB2 = 104
AB = √104
AB= √ (2×2×2×13)
AB= 2√(2×13)
AB= 2√26
Hence, Base = 2√26
By definition,
cos θ = Base/Hypotenuse
∴ cosθ = 2√26/ 15
And, cosec θ = 1/sin θ
∴ cosec θ = 15/11
And, secθ = Hypotenuse/Base
∴ secθ =15/ 2√26
And, tan θ = Perpendicular/Base
∴ tanθ =11/ 2√26
And, cot θ = Base/Perpendicular
∴ cotθ =2√26/ 11
(v) tan α = 5/12
Solution:
We have, tan α = 5/12 …. (1)
By definition,
tan α = Perpendicular/Base…. (2)
On Comparing eq. (1) and (2), we get
Base = 12 and Perpendicular side = 5
Now, using Pythagoras’ theorem in Δ ABC,
AC2 = AB2 + BC2
Putting the value of the base (AB) and the perpendicular (BC) to get hypotenuse (AC), we have
AC2 = 122 + 52
AC2 = 144 + 25
AC2= 169
AC = 13 [After taking sq root on both sides]
Hence, Hypotenuse = 13
By definition,
sin α = Perpendicular/Hypotenuse
∴ sin α = 5/13
And, cosec α = Hypotenuse/Perpendicular
∴ cosec α = 13/5
And, cos α = Base/Hypotenuse
∴ cos α = 12/13
And, sec α =1/cos α
∴ sec α = 13/12
And, tan α = sin α/cos α
∴ tan α=5/12
Since, cot α = 1/tan α
∴ cot α =12/5
(vi) sin θ = √3/2
Solution:
We have, sin θ = √3/2 …………. (1)
By definition,
sin θ = Perpendicular/ Hypotenuse….(2)
On Comparing eq. (1) and (2), we get
Perpendicular = √3 and Hypotenuse = 2
Now, using Pythagoras’ theorem in Δ ABC,
AC2 = AB2 + BC2
Putting the value of perpendicular (BC) and hypotenuse (AC) and get the base (AB), we get
22 = AB2 + (√3)2
AB2 = 22 – (√3)2
AB2 = 4 – 3
AB2 = 1
AB = 1
Thus, Base = 1
By definition,
cos θ = Base/Hypotenuse
∴ cos θ = 1/2
And, cosec θ = 1/sin θ
Or cosec θ= Hypotenuse/Perpendicular
∴ cosec θ =2/√3
And, sec θ = Hypotenuse/Base
∴ sec θ = 2/1
And, tan θ = Perpendicular/Base
∴ tan θ = √3/1
And, cot θ = Base/Perpendicular
∴ cot θ = 1/√3
(vii) cos θ = 7/25
Solution:
We have, cos θ = 7/25 ……….. (1)
By definition,
cos θ = Base/Hypotenuse
On Comparing eq. (1) and (2), we get;
Base = 7 and Hypotenuse = 25
Now, using Pythagoras’ theorem in Δ ABC,
AC2= AB2 + BC2
Putting the value of base (AB) and hypotenuse (AC) to get the perpendicular (BC),
252 = 72 +BC2
BC2 = 252 – 72
BC2 = 625 – 49
BC2 = 576
BC= √576
BC= 24
Hence, Perpendicular side = 24
By definition,
sin θ = perpendicular/Hypotenuse
∴ sin θ = 24/25
Since, cosec θ = 1/sin θ
Also, cosec θ= Hypotenuse/Perpendicular
∴ cosec θ = 25/24
Since, sec θ = 1/cosec θ
Also, sec θ = Hypotenuse/Base
∴ sec θ = 25/7
Since, tan θ = Perpendicular/Base
∴ tan θ = 24/7
Now, cot = 1/tan θ
So, cot θ = Base/Perpendicular
∴ cot θ = 7/24
(viii) tan θ = 8/15
Solution:
We have, tan θ = 8/15 …………. (1)
By definition,
tan θ = Perpendicular/Base …. (2)
On Comparing eq. (1) and (2), we get
Base = 15 and Perpendicular = 8
Now, using Pythagoras’ theorem in Δ ABC,
AC2 = 152 + 82
AC2 = 225 + 64
AC2 = 289
AC = √289
AC = 17
Hence, Hypotenuse = 17
By definition,
Since, sin θ = perpendicular/Hypotenuse
∴ sin θ = 8/17
Since, cosec θ = 1/sin θ
Also, cosec θ = Hypotenuse/Perpendicular
∴ cosec θ = 17/8
Since, cos θ = Base/Hypotenuse
∴ cos θ = 15/17
Since, sec θ = 1/cos θ
Also, sec θ = Hypotenuse/Base
∴ sec θ = 17/15
Since, cot θ = 1/tan θ
Also, cot θ = Base/Perpendicular
∴ cot θ = 15/8
(ix) cot θ = 12/5
Solution:
We have, cot θ = 12/5 …………. (1)
By definition,
cot θ = 1/tan θ
cot θ = Base/Perpendicular ……. (2)
On Comparing eq. (1) and (2), we have
Base = 12 and Perpendicular side = 5
Now, using Pythagoras’ theorem in Δ ABC,
AC2= AB2 + BC2
Putting the value of base (AB) and perpendicular (BC) to get the hypotenuse (AC),
AC2 = 122 + 52
AC2= 144 + 25
AC2 = 169
AC = √169
AC = 13
Hence, Hypotenuse = 13
By definition,
Since, sin θ = perpendicular/Hypotenuse
∴ sin θ= 5/13
Since, cosec θ = 1/sin θ
Also, cosec θ= Hypotenuse/Perpendicular
∴ cosec θ = 13/5
Since, cos θ = Base/Hypotenuse
∴ cos θ = 12/13
Since, sec θ = 1/cosθ
Also, sec θ = Hypotenuse/Base
∴ sec θ = 13/12
Since, tanθ = 1/cot θ
Also, tan θ = Perpendicular/Base
∴ tan θ = 5/12
(x) sec θ = 13/5
Solution:
We have, sec θ = 13/5…….… (1)
By definition,
sec θ = Hypotenuse/Base…………. (2)
On Comparing eq. (1) and (2), we get
Base = 5 and Hypotenuse = 13
Now, using Pythagoras’ theorem in Δ ABC,
AC2 = AB2 + BC2
And putting the value of the base side (AB) and hypotenuse (AC) to get the perpendicular side (BC),
132 = 52 + BC2
BC2 = 132 – 52
BC2=169 – 25
BC2= 144
BC= √144
BC = 12
Hence, Perpendicular = 12
By definition,
Since, sin θ = perpendicular/Hypotenuse
∴ sin θ= 12/13
Since, cosec θ= 1/ sin θ
Also, cosec θ= Hypotenuse/Perpendicular
∴ cosec θ = 13/12
Since cos θ= 1/sec θ
Also, cos θ = Base/Hypotenuse
∴ cos θ = 5/13
Since, tan θ = Perpendicular/Base
∴ tan θ = 12/5
Since, cot θ = 1/tan θ
Also, cot θ = Base/Perpendicular
∴ cot θ = 5/12
(xi) cosec θ = √10
Solution:
We have, cosec θ = √10/1 ……..… (1)
By definition,
cosec θ = Hypotenuse/ Perpendicular …….….(2)
And, cosecθ = 1/sin θ
On comparing eq.(1) and(2), we get
Perpendicular side = 1 and Hypotenuse = √10
Now, using Pythagoras’ theorem in Δ ABC,
AC2 = AB2 + BC2
Putting the value of perpendicular (BC) and hypotenuse (AC) to get the base side (AB),
(√10)2 = AB2 + 12
AB2= (√10)2 – 12
AB2= 10 – 1
AB = √9
AB = 3
So, Base side = 3
By definition,
Since, sin θ = Perpendicular/Hypotenuse
∴ sin θ = 1/√10
Since, cos θ = Base/Hypotenuse
∴ cos θ = 3/√10
Since, sec θ = 1/cos θ
Also, sec θ = Hypotenuse/Base
∴ sec θ = √10/3
Since, tan θ = Perpendicular/Base
∴ tan θ = 1/3
Since, cot θ = 1/tan θ
∴ cot θ = 3/1
(xii) cos θ =12/15
Solution:
We have; cos θ = 12/15 ………. (1)
By definition,
cos θ = Base/Hypotenuse……… (2)
By comparing eq. (1) and (2), we get;
Base =12 and Hypotenuse = 15
Now, using Pythagoras’ theorem in Δ ABC, we get
AC2 = AB2+ BC2
Putting the value of base (AB) and hypotenuse (AC) to get the perpendicular (BC),
152 = 122 + BC2
BC2 = 152 – 122
BC2 = 225 – 144
BC 2= 81
BC = √81
BC = 9
So, Perpendicular = 9
By definition,
Since, sin θ = perpendicular/Hypotenuse
∴ sin θ = 9/15 = 3/5
Since, cosec θ = 1/sin θ
Also, cosec θ = Hypotenuse/Perpendicular
∴ cosec θ= 15/9 = 5/3
Since, sec θ = 1/cos θ
Also, sec θ = Hypotenuse/Base
∴ sec θ = 15/12 = 5/4
Since, tan θ = Perpendicular/Base
∴ tan θ = 9/12 = 3/4
Since, cot θ = 1/tan θ
Also, cot θ = Base/Perpendicular
∴ cot θ = 12/9 = 4/3
Exercise-Wise RD Sharma Class 10 Maths Chapter 5 – Trigonometric Ratios
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What are the most important topics covered in Chapter 5 of RD Sharma Solutions for Class 10 Maths?
Trigonometry is the science of measuring triangles, and it is covered in the RD Sharma Solutions for Class 10 Maths Chapter 5 books. In addition, the trigonometric ratios and their relationships will be covered in this chapter. Also discussed are the trigonometric ratios of some specific angles.
In RD Sharma Solutions for Class 10 Maths Chapter 5, how many exercises are there?
In RD Sharma Solutions for Class 10 Maths Chapter 5, there are three exercises. Based on the notions of trigonometric ratios, the first exercise comprises 18 questions with sub-questions, the second exercise contains 20 questions with sub-questions, and the third exercise contains 8 questions with sub-questions.
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