Frank Solutions for Class 9 Maths Chapter 26 Trigonometrical Ratios provide solutions in a stepwise manner as per students’ intelligence levels. The main aim of developing these solutions is to make the learning process smooth and hassle-free for students. Those who aspire to excel in solving difficult problems easily can practise Frank Solutions regularly.
Chapter 26, Trigonometrical Ratios, contains problems pertaining to the measurement of the sides and angles of a triangle. The branch of Mathematics which deals with the study of the sides and angles of a right-angled triangle is known as Trigonometry. It is essential to understand this chapter in-depth, as it plays a vital role in higher classes also. Students can download the Frank Solutions for Class 9 Maths Chapter 26 Trigonometrical Ratios PDF from the links provided below.
Frank Solutions for Class 9 Maths Chapter 26 Trigonometrical Ratios Download PDF
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1. In each of the following, one trigonometric ratio is given. Find the values of the other trigonometric ratios.
(i) sin A = 12 / 13
(ii) cos B = 4 / 5
(iii) cot A = 1 / 11
(iv) cosec C = 15 / 11
(v) tan C = 5 / 12
(vi) sin B = √3 / 2
(vii) cos A = 7 / 25
(viii) tan B = 8 / 15
(ix) sec B = 15 / 12
(x) cosec C = √10
Solution:
(i) Sin A = 12 / 13
Sin A = Perpendicular / Hypotenuse
Sin A = 12 / 13
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Base = √(Hypotenuse)2 – (Perpendicular)2
Base = √(13)2 – (12)2
Base = √169 – 144
Base = √25
We get,
Base = 5
Cos A = Base / Hypotenuse
Cos A = 5 / 13
Sec A = (1 / cos A)
Sec A = 13 / 5
Cot A = (1 / tan A)
Cot A = 5 / 12
Cosec A = (1 / sin A)
Cosec A = 13 / 12
(ii) Cos B = 4 / 5
Cos B = Base / Hypotenuse
Cos B = 4 / 5
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Perpendicular = √(Hypotenuse)2 – (Base)2
Perpendicular = √(5)2 – (4)2
Perpendicular = √25 – 16
Perpendicular = √9
We get,
Perpendicular = 3
Sin B = Perpendicular / Hypotenuse
Sin B = 3 / 5
Tan B = Perpendicular / Base
Tan B = 3 / 4
Sec B = (1 / cos B)
Sec B = 5 / 4
Cot B = (1 / tan B)
Cot B = 4 / 3
Cosec B = (1 / sin B)
Cosec B = 5 / 3
(iii) Cot A = 1 / 11
Cot A = (1 / tan A)
Cot A = Base / Perpendicular
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
(Hypotenuse) = √(Perpendicular)2 + (Base)2
(Hypotenuse) = √(11)2 + (1)2
(Hypotenuse) = √(121 + 1)
(Hypotenuse) = √122
Cos A = Base / Hypotenuse
Cos A = 1 / √122
Tan A = Perpendicular / Base
Tan A = 11
Sec A = (1 / cos A)
Sec A = √122
Sin A = Perpendicular / Hypotenuse
Sin A = 11 / √122
Cosec A = (1 / sin A)
Cosec A = √122 / 11
(iv) Cosec C = 15 / 11
Cosec C = (1 / sin C)
Cosec C = (Hypotenuse / Perpendicular)
Cosec C = 15 / 11
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Base = √(Hypotenuse)2 – (Perpendicular)2
Base = √(15)2 – (11)2
Base = √225 – 121
Base = √104
Sin C = Perpendicular / Hypotenuse
Sin C = 11 / 15
Cos C = Base / Hypotenuse
Cos C = √104 / 15
Tan C = Perpendicular / Base
Tan C = 11 / √104
Sec C = (1 / Cos C)
Sec C = 15 / √104
Cot C = (1 / tan C)
Cot C = √104 / 11
(v) Tan C = 5 / 12
Tan C = Perpendicular / Base
Tan C = 5 / 12
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
(Hypotenuse) = √(Perpendicular)2 + (Base)2
(Hypotenuse) = √(5)2 + (12)2
(Hypotenuse) = √25 + 144
(Hypotenuse) = √169
We get,
(Hypotenuse) = 13
Cot C = (1 / tan C)
Cot C = 12 / 5
Sin C = Perpendicular / Hypotenuse
Sin C = 5 / 13
Cos C = Base / Hypotenuse
Cos C = 12 / 13
Sec C = (1 / Cos C)
Sec C = 13 / 12
Cosec C = (1 / Sin C)
Cosec C = 13 / 5
(vi) Sin B = √3 / 2
Sin B = Perpendicular / Hypotenuse
Sin B = √3 / 2
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Base = √(Hypotenuse)2 – (Perpendicular)2
Base = √(2)2 – (√3)2
Base = √4 – 3
Base = √1
We get,
Base = 1
Cos B = Base / Hypotenuse
Cos B = 1 / 2
Tan B = Perpendicular / Base
Tan B = √3 / 1
Tan B = √3
Sec B = (1 / Cos B)
Sec B = 2
Cot B = (1 / tan B)
Cot B = 1 / √3
Cosec B = 1 / Sin A
Cosec B = 2 / √3
(vii) Cos A = 7 / 25
Cos A = Base / Hypotenuse
Cos A = 7 / 25
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Perpendicular = √(Hypotenuse)2 – (Base)2
Perpendicular = √(25)2 – (7)2
Perpendicular = √625 – 49
Perpendicular = √576
We get,
Perpendicular = 24
Sin A = Perpendicular / Hypotenuse
Sin A = 24 / 25
Tan A = Perpendicular / Base
Tan A = 24 / 7
Sec A = (1 / cos A)
Sec A = 25 / 7
Cot A = (1 / tan A)
Cot A = 7 / 24
Cosec A = (1 / sin A)
Cosec A = 25 / 24
(viii) Tan B = 8 / 15
Tan B = Perpendicular / Base
Tan B = 8 / 15
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
(Hypotenuse) = √(Perpendicular)2 + (Base)2
(Hypotenuse) = √(8)2 + (15)2
(Hypotenuse) = √64 + 225
(Hypotenuse) = √289
We get,
Hypotenuse = 17
Cot B = 1 / tan B
Cot B = 15 / 8
Sin B = Perpendicular / Hypotenuse
Sin B = 8 / 17
Cos B = Base / Hypotenuse
Cos B = 15 / 17
Sec B = (1 / cos B)
Sec B = 17 / 15
Cosec B = (1 / sin B)
Cosec B = 17 / 8
(ix) Sec B = 15 / 12
Sec B = (1 / cos B)
Sec B = Hypotenuse / Base
Sec B = 15 / 12
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Perpendicular = √(Hypotenuse)2 – (Base)2
Perpendicular = √(15)2 – (12)2
Perpendicular = √225 – 144
Perpendicular = √81
We get,
Perpendicular = 9
Sin B = Perpendicular / Hypotenuse
Sin B = 9 / 15
Tan B = Perpendicular / Base
Tan B = 9 / 12
Cot B = (1 / tan B)
Cot B = 12 / 9
Cosec B = (1 / sin B)
Cosec B = 15 / 9
Cos B = Base / Hypotenuse
Cos B = 12 / 15
(x) Cosec C = √10
Cosec C = (1 / sin C)
Cosec C = Hypotenuse / Perpendicular
Cosec C = √10 / 1
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Base = √(Hypotenuse)2 – (Perpendicular)2
Base = √(√10)2 – (1)2
Base = √10 – 1
Base = √9
We get,
Base = 3
Sin C = Perpendicular / Hypotenuse
Sin C = 1 / √10
Cos C = Base / Hypotenuse
Cos C = 3 / √10
Tan C = Perpendicular / Base
Tan C = 1 / 3
Sec C = (1 / cos C)
Sec C = √10 / 3
Cot C = (1 / tan C)
Cot C = 3
2. In △ABC, ∠A = 900. If AB = 5 units and AC = 12 units, find:
(i) sin B
(ii) cos C
(iii) tan B
Solution:
In â–³ABC,
BC2 = AB2 + AC2
BC = √AB2 + AC2
BC = √52 + 122
BC = √25 + 144
BC = √169
We get,
BC = 13
AC = 12 units
BC = 13 units
AB = 5 units
(i) sin B = Perpendicular / Hypotenuse
sin B = AC / BC
sin B = 12 / 13
(ii) cos C = Base / Hypotenuse
cos C = AC / BC
cos C = 12 / 13
(iii) tan B = Perpendicular / Base
tan B = AC / AB
tan B = 12 / 5
3. In △ABC, ∠B = 900. If AB = 12 units and BC = 5 units, find:
(i) sin A
(ii) tan A
(iii) cos C
(iv) cot C
Solution:
In â–³ABC,
AC2 = AB2 + BC2
AC = √122 + 52
AC = √144 + 25
AC = √169
We get,
AC = 13
AB = 12 units
BC = 5 units
AC = 13 units
(i) sin A = Perpendicular / Hypotenuse
sin A = BC / AC
sin A = 5 / 13
(ii) tan A = Perpendicular / Base
tan A = BC / AB
tan A = 5 / 12
(iii) cos C = Base / Hypotenuse
cos C = BC / AC
cos C = 5 / 13
(iv) cot C = Base / Perpendicular
cot C = BC / AB
cot C = 5 / 12
4. If sin A = 3/ 5, find cos A and tan A.
Solution:
Given
sin A = 3 / 5
sin A = Perpendicular / Hypotenuse
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
(Base)2 = (Hypotenuse)2 – (Perpendicular)2
Base = √(Hypotenuse)2 – (Perpendicular)2
Base = √52 – 32
Base = √25 – 9
Base = √16
We get,
Base = 4
cos A = Base / Hypotenuse
cos A = 4 / 5
tan A = Perpendicular / Base
tan A = 3 / 4
5. If sin θ = 8 / 17, find the other five trigonometric ratios.
Solution:
Given
sin θ = 8 / 17
sin θ = Perpendicular / Hypotenuse
Base = √(Hypotenuse)2 – (Perpendicular)2
Base = √172 – 82
Base = √289 – 64
Base = √225
We get,
Base = 15
cos θ = Base / Hypotenuse
cos θ = 15 / 17
tan θ = Perpendicular / Base
tan θ = 8 / 15
cosec θ = 1/ sin θ
cosec θ = 17 / 8
sec θ = 1/cos θ
sec θ = 17 / 15
cot θ = 1/ tan θ
cot θ = 15 / 8
6. If tan A = 0.75, find the other trigonometric ratios for A.
Solution:
Given
tan A = 0.75
tan A = 75 / 100
We get,
tan A = 3 / 4
tan A = Perpendicular / Base
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Hypotenuse = √(Perpendicular)2 + (Base)2
Hypotenuse = √32 + 42
Hypotenuse = √9 + 16
Hypotenuse = √25
We get,
Hypotenuse = 5
sin A = Perpendicular / Hypotenuse
sin A = 3 / 5
sin A = 0.6
cos A = Base / Hypotenuse
cos A = 4 / 5
cos A = 0.8
cosec A = 1 / sin A
cosec A = 5 / 3
cosec A = 1.66
sec A = 1 / cos A
sec A = 5 / 4
sec A = 1.25
cot A = 1 / tan A
cot A = 4 / 3
cot A = 1.33
7. If sin A = 0.8, find the other trigonometric ratios for A.
Solution:
Given
sin A = 0.8
sin A = 8 / 10
sin A = 4 / 5
sin A = Perpendicular / Hypotenuse
By Pythagoras theorem,
We have
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Base = √(Hypotenuse)2 – (Perpendicular)2
Base = √52 – 42
Base = √25 – 16
Base = √9
We get,
Base = 3
cos A = Base / Hypotenuse
cos A = 3 / 5
cos A = 0.6
tan A = Perpendicular / Base
tan A = 4 / 3
tan A = 1.33
cosec A = 1/sin A
cosec A = 5 / 4
cosec A = 1.25
sec A = 1/ cos A
sec A = 5 / 3
sec A = 1.66
cot A = 1/ tan A
cot A = 3 / 4
cot A = 0.75
8. If 8 tan θ = 15, find
(i) sin θ
(ii) cot θ
(iii) sin2 θ – cot2 θ
Solution:
Given
8 tan θ = 15
tan θ = 15 / 8
tan θ = Perpendicular / Base
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Hypotenuse = √(Perpendicular)2 + (Base)2
Hypotenuse = √152 + 82
Hypotenuse = √225 + 64
Hypotenuse = √289
We get,
Hypotenuse = 17
(i) sin θ = Perpendicular / Hypotenuse
sin θ = 15 / 17
(ii) cot θ = 1 / tan θ
cot θ = 8 / 15
(iii) sin2 θ – cot2 θ = (sin θ + cot θ) (sin θ – cot θ)
sin2 θ – cot2 θ = {(15 / 17) + (8 / 15)} {(15 / 17) – (8 / 15)}
sin2 θ – cot2 θ = {(225 + 136) / 255}{(225 – 136) / 255}
sin2 θ – cot2 θ = (361 / 255) (89 / 255)
On calculation, we get,
sin2 θ – cot2 θ = 32129 / 65025
9. In an isosceles triangle ABC, AB = BC = 6 cm and ∠B = 900. Find the values of
(a) cos C
(b) cosec C
(c) cos2 C + cosec2 C
Solution:
â–³ABC is an isosceles right-angled triangle
Therefore,
AC2 = AB2 + BC2
AC2 = 62 + 62
AC2 = 36 + 36
AC2 = 72
We get,
AC = 6√2 cm
(a) cos C = BC / AC
cos C = (6 / 6√2)
cos C = (1 / √2)
(b) cosec C = AC / AB
cosec C = (6√2 / 6)
cosec C = √2
(c) cos2 C + cosec2 C = (1/√2)2 + (√2)2
cos2 C + cosec2 C = (1/2) + 2
On further calculation, we get,
cos2 C + cosec2 C = 5 / 2
10. In the given figure, AD is the median on BC from A. If AD = 8 cm and BC = 12 cm, find the value of
(a) sin x
(b) cos y
(c) tan x. cot y
(d) (1/sin2 x) – (1/ tan2 x)
Solution:
Since AD is the median on BC,
We have,
BD = DC = (1/2) x BC
= (1/2) x 12
= 6 cm
â–³ADB is a right-angled triangle
Therefore,
AB2 = AD2 + BD2
AB2 = 82 + 62
AB2 = 64 + 36
AB2 = 100
We get,
AB = 10 cm
â–³ADC is a right-angled triangle
Therefore,
AC2 = AD2 + DC2
AC2 = 82 + 62
AC2 = 64 + 36
AC2 = 100
We get,
AC = 10 cm
(a) sin x = AD / AB
sin x = 8 / 10
sin x = 4 / 5
(b) cos y = AD / AC
cos y = 8 / 10
cos y = 4 / 5
(c) cos x = BD / AB
cos x = 6 / 10
cos x = 3 / 5
And,
sin y = DC / AC
sin y = 6 / 10
sin y = 3 / 5
Hence,
tan x = sin x / cos x
tan x = {(4/5) / (3/5)}
We get,
tan x = 4 / 3
cot y = cos y / sin y
cot y = {(4/5) / (3/5)}
We get,
cot y = 4 / 3
Therefore,
tan x. cot y = (4/3) x (4/3)
tan x . cot y = 16 / 9
(d) (1/ sin2 x) – (1/ tan2 x) = {1/ (4/5)2} – {1/ (4/3)2}
On calculating further, we get,
(1/ sin2 x ) – (1/ tan2 x) = (25 / 16) – (9 / 16)
(1/ sin2 x) – (1/ tan2 x) = (16 / 16)
(1/ sin2 x) – (1/ tan2 x) = 1
11. In a right-angled triangle PQR, ∠PQR = 900, QS ⊥PR and tan R = (5/ 12), find the value of
(a) sin ∠PQS
(b) tan ∠SQR
Solution:
tan R = 5 / 12
PQ / QR = 5 / 12
Hence,
PQ = 5 and QR = 12
In right-angled â–³PQR,
PR2 = PQ2 + QR2
PR2 = 52 + 122
PR2 = 25 + 144
PR2 = 169
PR = √169
We get,
PR = 13
(a) ∠PQS + ∠P = 900 and ∠P + ∠R = 900
Hence,
∠PQS + ∠P = ∠P + ∠R
∠PQS = ∠R
Therefore,
sin ∠PQS = sin R = PQ / PR = 5 / 13
(b) ∠SQR + ∠R = 900 and ∠R + ∠P = 900
Hence,
∠SQR + ∠R = ∠R + ∠P
∠SQR = ∠P
Therefore,
tan ∠SQR = tan P = QR / PQ = 12 / 5
12. In the given figure, â–³ABC is right-angled at B. AD divides BC in the ratio 1: 2.
Find
(i) tan ∠BAC / tan ∠BAD
(ii) cot ∠BAC / cot ∠BAD
Solution:
Given
â–³ABC is right angled at B
BD: DC = 1: 2 as AD divides BC in the ratio 1: 2
i.e BD = x, DC = 2x
Hence,
BC = 3x
(i) tan ∠BAC / tan ∠BAD = (BC / AB) / (BD / AB)
tan ∠BAC / tan ∠BAD = BC / BD
tan ∠BAC / tan ∠BAD = 3x / x
tan ∠BAC / tan ∠BAD = 3
(ii) cot ∠BAC / cot ∠BAD = (AB / BC) / (AB / BD)
cot ∠BAC / cot ∠BAD = BD / BC
cot ∠BAC / cot ∠BAD = x / 3x
cot ∠BAC / cot ∠BAD = 1 / 3
13. If sin A = 7/25, find the value of:
(a) (2 tan A) / (cot A – sin A)
(b) cos A + (1/cot A)
(c) cot2 A – cosec2 A
Solution:
Consider △ABC, where ∠B = 900
sin A = Perpendicular / Hypotenuse
sin A = BC / AC
sin A = 7/25
cosec A = 1/sin A
cosec A = 25 / 7
By the Pythagoras theorem,
We have,
AC2 = AB2 + BC2
AB2 = AC2 – BC2
AB2 = 252 – 72
AB2 = 625 – 49
AB2 = 576
AB = √576
We get,
AB = 24
Now,
cos A = Base / Hypotenuse
cos A = AB / AC
cos A = 24 / 25
tan A = Perpendicular / Base
tan A = BC / AB
tan A = 7 / 24
cot A = (1/ tan A)
cot A = 24 / 7
(a) (2 tan A) / cot A – sin A = {2 x (7/24)} / {(24/7) – (7/25)}
On further calculation, we get,
= (7/12) / (551/175)
= (7/12) x (175 / 551)
We get,
= 1225 / 6612
(b) cos A + 1/cot A = cos A + tan A
cos A + 1/cot A = (24 / 25) + (7 / 24)
On calculating further, we get,
cos A + 1/cot A = (576 + 175) / 600
cos A + 1/cot A = 751 / 600
(c) cot2 A – cosec2 A = (24 / 7)2 – (25 / 7)2
cot2 A – cosec2 A = (576 / 49) – (625 / 49)
cot2 A – cosec2 A = (576 – 625) / 49
We get,
cot2 A – cosec2 A = – 49 / 49
cot2 A – cosec2 A = – 1
14. If cosec θ = 29 / 20, find the value of:
(a) cosec θ – (1/ cot θ)
(b) sec θ / (tan θ – cosec θ)
Solution:
Consider △ABC, where ∠A = 900
Cosec θ = Hypotenuse / Perpendicular
Cosec θ = BC / AB
Cosec θ = 29 / 20
By Pythagoras theorem,
We have,
BC2 = AB2 + AC2
AC2 = BC2 – AB2
AC2 = 292 – 202
AC2 = 841 – 400
AC2 = 441
AC = √441
We get,
AC = 21
Now,
sec θ = Hypotenuse / Base
sec θ = BC / AC
sec θ = 29 / 21
tan θ = Perpendicular / Base
tan θ = AB / AC
tan θ = 20 / 21
cot θ = 1/ tan θ
cot θ = 21 / 20
(a) cosec θ – (1/ cot θ) = (29 / 20) – {1 / (21/20)}
cosec θ – (1/ cot θ) = (29 / 20) – (20 / 21)
cosec θ – (1/ cot θ) = (609 – 400) / 420
We get,
cosec θ – (1/ cot θ) = 209 / 420
(b) sec θ / (tan θ – cosec θ) = (29 / 21) / {(20 / 21) – (29 / 20)}
sec θ / (tan θ – cosec θ) = (29 / 21) / {(400 – 609) / 420}
sec θ / (tan θ – cosec θ) = (29 / 21) / {(-209 / 420)}
sec θ / (tan θ – cosec θ) = (29 / 21) x (-420 / 209)
sec θ / (tan θ – cosec θ) = – 580 / 209
15. In the given figure, AC = 13 cm, BC = 12 cm and ∠B = 900. Without using tables, find the values of:
(a) sin A cos A
(b) (cos A – sin A) / (cos A + sin A)
Solution:
â–³ABC is a right-angled triangle.
By Pythagoras theorem,
We have,
AC2 = AB2 + BC2
AB2 = AC2 – BC2
AB2 = 132 – 122
AB2 = 169 – 144
AB2 = 25
AB = √25
We get,
AB = 5 cm
sin A = BC / AC
sin A = 12 / 13
cos A = AB / AC
cos A = 5 / 13
(a) sin A cos A = (12 / 13) x (5 / 13)
sin A cos A = 60 / 169
(b) (cos A – sin A) / (cos A + sin A) = {(5/ 13) – (12 / 13)} / {(5/ 13) + (12 / 13)}
(cos A – sin A) / (cos A + sin A) = (-7 / 13) / (17 / 13)
(cos A – sin A) / (cos A + sin A) = (-7 / 13) x (13 / 17)
(cos A – sin A) / (cos A + sin A) = -7 / 17
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