Frank Solutions Class 9 Maths Chapter 26 Trigonometrical Ratios

Frank Solutions Class 9 Maths Chapter 26 Trigonometrical Ratios provide solutions in a stepwise manner as per students’ intelligence quotient. The main aim of developing these solutions is to make the learning process smooth and hassle-free among students. Those who aspire to excel in solving difficult problems easily, can make use of Frank Solutions. Here, students can find the Frank Solutions Class 9 Maths Chapter 26 Trigonometrical Ratios PDF, from the links provided below.

Chapter 26 Trigonometrical Ratios has problems pertaining to the measurement of the sides and angles of a triangle. The branch of Mathematics, which deals with the study of sides and angles of a right-angled triangle is known as trigonometry. It is most essential to understand this chapter in-depth, as it plays a vital role in further classes also.

Frank Solutions for Class 9 Maths Chapter 26 Trigonometrical Ratios Download PDF

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Access Frank Solutions for Class 9 Maths Chapter 26 Trigonometrical Ratios

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:

FRANK Solutions Class 9 Maths Chapter 26 - 1

In △ABC,

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:

FRANK Solutions Class 9 Maths Chapter 26 - 2

In △ABC,

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:

FRANK Solutions Class 9 Maths Chapter 26 - 3

△ABC 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)

FRANK Solutions Class 9 Maths Chapter 26 - 4

Solution:

FRANK Solutions Class 9 Maths Chapter 26 - 5

Since, AD is the median on BC,

We have,

BD = DC = (1/2) x BC

= (1/2) x 12

= 6 cm

△ADB is a right angled triangle

Therefore,

AB2 = AD2 + BD2

AB2 = 82 + 62

AB2 = 64 + 36

AB2 = 100

We get,

AB = 10 cm

△ADC 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:

FRANK Solutions Class 9 Maths Chapter 26 - 6

tan R = 5 / 12

PQ / QR = 5 / 12

Hence,

PQ = 5 and QR = 12

In right-angled △PQR,

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, △ABC is right angled at B. AD divides BC in the ratio 1: 2.

Find

(i) tan ∠BAC / tan ∠BAD

(ii) cot ∠BAC / cot ∠BAD

FRANK Solutions Class 9 Maths Chapter 26 - 7

Solution:

Given

△ABC 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:

FRANK Solutions Class 9 Maths Chapter 26 - 8

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 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:

FRANK Solutions Class 9 Maths Chapter 26 - 9

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)

FRANK Solutions Class 9 Maths Chapter 26 - 10

Solution:

△ABC 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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