ML Aggarwal Solutions for Class 9 Maths Chapter 17 Trigonometric Ratios

ML Aggarwal Solutions For Class 9 Maths Chapter 17 Trigonometric Ratios consists of accurate solutions, which help the students to quickly complete their homework and prepare well for the exams. It ensures that you get all the necessary information of all concepts included in the chapter. Class 9 is a critical level as it forms the base for students for the forthcoming academic years. So, students who aspire to attain good marks in Maths practice ML Aggarwal Solutions for Class 9 Maths. Chapter 17 of ML Aggarwal Solutions deals with trigonometric ratios. In trigonometry, trigonometric ratios are derived from the sides of a right-angled triangle. There are six 6 ratios such as sine, cosine, tangent, cotangent, cosecant, and secant. This chapter of ML Aggarwal Solutions for Class 9 Maths contains one exercise with chapter test. These solutions provided by BYJU’S cover all these concepts, with detailed explanations.

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ML Aggarwal Solutions for Class 9 Maths Chapter 17 Trigonometric Ratios
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Access answers to ML Aggarwal Solutions for Class 9 Maths Chapter 17 – Trigonometric Ratios

Exercise 17

1. (a) From the figure (1) given below, find the values of:

(i) sin θ

(ii) cos θ

(iii) tan θ

(iv) cot θ

(v) sec θ

(vi) cosec θ

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 1

(b) From the figure (2) given below, find the values of:

(i) sin A

(ii) cos A

(iii) sin2 A + cos2 A

(iv) sec2 A – tan2 A.

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 2

Solution:

(a) From right angled triangle OMP,

By Pythagoras theorem, we get

OP2 = OM2 +MP2

MP2 = OP2 + OM2

MP2 = (15)2 – (12)2

MP2 = 225 – 144

MP2 = 81

MP2 = 92

MP = 9

(i) sin θ = MP/OP

= 9/15

= 3/5

(ii) cos θ = OM/OP

= 12/15

= 4/5

(iii) tan θ = MP/OP

= 9/12

= ¾

(iv) cot θ = OM/MP

= 12/9

= 4/3

(v) sec θ = OP/OM

= 15/12

= 5/4

(vi) cosec θ = OP/MP

= 15/9

= /3

(b) From right angled triangle ABC,

By Pythagoras theorem, we get

AB2 = AC2 + BC2

AB2 = (12)2 + (5)2

AB2 = 144 + 25

AB2 = 169

AB2 = 132

AB = 13

(i) sin A = BC/AB

= 5/13

(ii) cos A = AC/AB

= 12/13

(iii)Sin2 A + cos2 A = (BC/AB)2 + (AC/AB)2

= (5/13)2 + (12/13)2

= (25/169) + (144/169)

= (25 + 144)/ 169

= 169/169

= 1

Sin2 A + cos2 A = 1

(iv) Sec2 A – tan2 A = (AB/AC)2 – (BC/AC)2

= (13/12)2 – (5/12)2

= (169/144) – (25/144)

= (169 – 25)/ 144

= 144/144

= 1

Sec2 A – tan2 A = 1

2. (a) From the figure (1) given below, find the values of:
(i) sin B

(i) cos C

(iii) sin B + sin C

(iv) sin B cos C + sin C cos B.

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 3

(b) From the figure (2) given below, find the values of:

(i) tan x

(ii) cos y

(iii) cosec2 y – cot2 y

(iv) 5/sin x + 3/sin y – 3 cot y.

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 4

Solution:

From right angled triangle ABC,

By Pythagoras theorem, we get

BC2 = AC2 + AB2

AC2 = BC2 – AB2

AC2 = (10)2 – (6)2

AC2 = 100 – 36

AC2 = 64

AC2 = 82

AC = 8

(i) sin B = perpendicular/ hypotenuse

= AC/BC

= 8/10

= 4/5

(ii) cos C = Base/hypotenuse

= AC/BC

= 8/10

= 4/5

(iii) sin B = Perpendicular/hypotenuse

= AC/BC

= 8/10

= 4/5

Sin C = perpendicular/hypotenuse

= AB/BC

= 6/10

= 3/5

Now,

Sin B + sin C = (4/5) + (3/5)

= (4 + 3)/5

= 7/5

(iv) sin B = 4/5

Cos C = 4/5

Sin C = perpendicular/ hypotenuse

= AB/BC

= 6/10

= 3/5

Cos B = Base/Hypotenuse

= AB/BC

= 6/10

= 3/5

sin B cos C + sin C cos B

= (4/5) × (4/5) × (3/5) × (3/5)

= (26/25) × (9/25)

= (16+9)/25

= 25/25

= 1

From Figure

AC = 13, CD = 5, BC =21,

BD = BC – CD

= 21 – 5

= 16

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 5

From right angled ∆ACD,

By Pythagoras theorem we get

AC = AD2 + CD2

AD2 = AC2 – CD2

AD2 = (13)2 – (5)2

AD2 = 169 – 25

AD2 = 144)

AD2 = (12)2

AD = 12

From right angled ∆ABD,

By Pythagoras angled ∆ABD

By Pythagoras theorem we get

AB2 = AD2 + BD2

AB2 = 400

AB2 = (20)2

AB = 20

(i) tan x = perpendicular/Base (in right angled ∆ACD)

=CD/AD

= 5/12

(ii) cos y = Base/Hypotenuse (in right angled ∆ABD)

= BD/AB

= (20)/12 – (5/3)

Cot y = Base/Perpendicular (in right ∆ABD)

=BD/AB

= 16/20 = 4/5

(iii) cos y = Hypotenuse/ perpendicular (in right angled ∆ABD)

BD/AB

= 20/12

= 5/3

Cot y = Base/Perpendicular (in right ∆ABD)

AB/AD

= 16/12

= 4/3

Cosec2 y – cot2 y = (5/3)2 – (4/3)2

= (25/9) – (16/9)

= (25-16)/9

= 9/9

= 1

Hence, cosec2 y – cot2 y = 1

(iv) sin x = Perpendicular/Hypotenuse (in right angled ∆ACD)

= AD/AB

= 12/20

= 3/5

Cot y = Base/Perpendicular (in right angled ∆ABD)

= BD/AD

= 16/12

= 4/3

(5/sin x) + (3/sin y) – 3cot y

= 5/(5/13) + 3/(3/5) – 3 × 4/3

= 5 × 13/5 + 3 × 5/3 – 3 × 4/3

= 1 × 13/1 + 1 × 5/1 – 1 × 4/1

= 13 + 5 – 4 = 18 – 4

= 14

Hence 5/sin x + 3/sin y – 3cot y = 14

3. (a) From the figure (1) given below, find the value of sec θ.

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 6
(b) From the figure (2) given below, find the values of:
(i) sin x

(ii) cot x

(iii) cot2 x- cosec2 x

(iv) sec y

(v) tan2 y – 1/cos2 y.

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 7

Solution:

(a) From the figure, Sec θ = AB / BD

But in ∆ADC, ∠D = 90o

AC2 =AD2 + DC2 (Pythagoras Theorem)

(13)2 =AD2 + 25

AD2 = 169 -25

= 144

= (12)2

AD = 12

(in right ∆ABD)

AB2 = AD2 + BD2

= (12)2 + (16)2

= 144 + 256

= 400

= (20)2

AB = 20

Now, Sec θ = AB / BD

= 20/16

= 5/4

(b) let given ∆ABC

BD = 3, AC = 12, AD = 4

In right angled ∆ABD

By Pythagoras theorem

AB2 =AD2 + BD2

AB2 = (4)2 + (3)2

AB2 = 16 + 9

AB2 = 25

AB2 = (5)2

AB = 5

In right angled triangle ACD

By Pythagoras theorem,

AC2 = AD2 + CD2

CD2 = AC2 – AD2

CD2 = (12)2 – (4)2

CD2 = 144 – 16

CD2 = 128

CD = √128

CD = √64 × 2 CD

= 8√2

(i) sin x = perpendicular/Hypotenuse

= AD/AB

= 4/5

(ii) cot x = Base/Perpendicular

= BD/AD

= ¾

(iii) cot x = Base/ Perpendicular

BD/AD

= 3/4

(iv) cosec x = Hypotenuse / Perpendicular

AB/BD

= 5/4

Cot2 x – cosec2 x

= (3/4)2 – (5/4)2

= 9/16 – 25/16

(9 -25)/16

= -16/16

= -1

Perpendicular = Hypotenuse/Base (in right angled ∆ACD)

= AD/CD

= 12/(8 √2)

= 3/(2 √2)

Cot y = Base/ Hypotenuse

= AD/CD

= 4/8 √ 2

= 1/2 √2

Cot y = Base / Hypotenuse (in right angled ∆ACD)

= CD/AC

= 8√2/12

= 2√/3

Now tan2 y = 1/cos2 y

= (1/2√2)2 – 1/(2√2/3)2

= ¼ × – ¼ × 2

= (1/8) – (9/8)

= (1-9)/8

= -8/8

= -1

tan2 y – 1/cos2 y = – 1.

4. (a) From the figure (1) given below, find the values of:

(i) 2 sin y – cos y

(ii) 2 sin x – cos x

(iii) 1 – sin x + cos y

(iv) 2 cos x – 3 sin y + 4 tan x

(b) In the figure (2) given below, ∆ABC is right-angled at B. If AB = y units, BC = 3 units and CA = 5 units, find

(i) sin x0

(ii) y.

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 8

Solution:

(a) In a right angled ∆BCD,

Using Pythagoras theorem

BC2 = BD2 + CD2

Substituting the values

BC2 = 92 + 122

By further calculation

BC2 = 81 + 144 = 225

BC2 = 152

BC = 15

In a right angled ∆ABC,

Using Pythagoras theorem

AC2 = AB2 + BC2

We can write it as

AB2 = AC2 – BC2

Substituting the values

AB2 = 252 – 152

By further calculation

AB2 = 625 – 225 = 400

So we get

AB2 = 202

AB = 20

(i) We know that

In right angled ∆BCD

sin y = perpendicular/ hypotenuse

sin y = BD/ BC

Substituting the values

sin y = 9/15 = 3/5

In right angled ∆BCD

cos y = base/hypotenuse

cos y = CD/BC

Substituting the values

cos y = 12/15 = 4/5

Here

2sin y – cos y = 2 × 3/5 – 4/5

We can write it as

= 6/5 – 4/5

= 2/5

Therefore, 2 sin y – cos y = 2/5

(ii) In right angled ∆ABC

sin x = perpendicular/ hypotenuse

sin x = BC/AC

Substituting the values

sin x = 15/25 = 3/5

In right angled ∆ABC

cos x = base/hypotenuse

cos x = AB/AC

Substituting the values

cos x = 20/25 = 4/5

Here

2 sin x – cos x = 2 × 3/5 – 4/5

We can write it as

= 6/5 – 4/5

= 2/5

Therefore, 2 sin x – cos x = 2/5.

(iii) In right angled ∆ABC

sin x = perpendicular/hypotenuse

sin x = BC/AC

Substituting the values

sin x = 12/25 = 3/5

In right angled ∆BCD

cos y = base/hypotenuse

cos y = CD/BC

Substituting the values

cos y = 12/15 = 4/5

Here

1 – sin x + cos y = 1 – 3/5 + 4/5

By further calculation

= (5 – 3 + 4)/ 5

So we get

= (9 – 3)/ 5

= 6/5

Therefore, 1 – sin x + cos y = 6/5.

(iv) In right angled ∆BCD

cos x = base/hypotenuse

cos x = AB/AC

Substituting the values

cos x = 20/25 = 4/5

In right angled ∆BCD

sin y = perpendicular/hypotenuse

sin y = BD/BC

Substituting the values

sin y = 9/15 = 3/5

In right angled ∆ABC

tan x = perpendicular/base

tan x = BC/AB

Substituting the values

tan x = 15/20 = ¾

Here

2 cos x – 3 sin y + 4 tan x = 2 × 4/5 – 3 × 3/5 + 4 × ¾

By further calculation

= 8/5 – 9/5 3/1

Taking LCM

= (8 – 9 + 15)/5

So we get

= (23 – 9)/ 5

= 14/5

(b) It is given that

AB = y units, BC = 3 units, CA = 5 units

(i) In right angled ∆ABC

sin x = perpendicular/hypotenuse

sin x = BC/AC

Substituting the values

sin x = 3/5

(ii) In right angled ∆ABC

Using Pythagoras theorem

AC2 = BC2 + AB2

We can write it as

AB2 = AC2 – BC2

Substituting the values

AB2 = 52 – 32

By further calculation

AB2 = 25 – 9 = 16

So we get

AB2 = 42

AB = 4

y = 4 units

Therefore, y = 4 units.

5. In a right-angled triangle, it is given that angle A is an acute angle and that
tan A=5/12. Find the values of:
(i) cos A
(ii) cosec A- cot A.

Solution:

Here, ABC is right angled triangle

∠A is an acute angle and ∠C = 90o

tan A = 5/12

BC/AC =5/12

Let BC = 5x and AC = 12x

From right angled ∆ABC

By Pythagoras theorem, we get

AB2 = (5x)2 + (12x)2

AB2 = 25x2 + 144x2

AB2 = 169x2

(i) cos A = Base/ Hypotenuse

= AC / AB

= 12x/13x

=12/13

(ii) cosec A = Hypotenuse/perpendicular

= AC / BC

= 13x /5x

= 13/5

cosec A – cot A = 13/5 – 12/5

= (13-12)/5

= 1/5

6. (a) In ∆ABC, ∠A = 900. If AB = 7 cm and BC – AC = 1 cm, find:

(i) sin C

(ii) tan B

(b) In ∆PQR, ∠Q = 900. If PQ = 40 cm and PR + QR = 50 cm, find:

(i) sin P

(ii) cos P

(iii) tan R.

Solution:

(a) In right ∆ABC

∠A = 900

AB = 7 cm

BC – AC = 1 cm

BC = 1 + AC

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 9

We know that

BC2 = AB2 + AC2

Substituting the value of BC

(1 + AC)2 = AB2 + AC2

1 + AC2 + 2AC = 72 + AC2

By further calculation

1 + AC2 + 2AC = 49 AC2

2AC = 49 – 1 – 48

So we get

AC = 48/2 = 24 cm

Here

BC = 1 + AC

Substituting the value

BC = 1 + 24 = 25 cm

(i) sin C = AB/BC = 7/25

(ii) tan B = AC/AB = 24/7

(b) In right ∆PQR

∠Q = 900

PQ = 40 cm

PQ + QR = 50 cm

We can write it as

PQ = 50 – QR

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 10

Using Pythagoras theorem

PR2 = PQ2 + QR2

(50 – QR)2 = (40)2 + QR2

By further calculation

2500 + QR2 – 100QR = 1600 + QR2

So we get

2500 – 1600 = 100QR

100QR = 900

By division

QR = 900/100 = 9

We get

PR = 50 – 9 = 41

(i) sin P = QR/PR = 9/41

(ii) cos P = PQ/PR = 40/41

(iii) tan R = PQ/QR = 40/9

7. In triangle ABC, AB = 15 cm, AC = 15 cm and BC = 18 cm. Find

(i) cos ∠ABC

(ii) sin ∠ACB.

Solution:

Here ABC is a triangle in which

AB = 15 cm, AC = 15 cm and BC = 18 cm

Draw AD perpendicular to BC , D is mid-point of BC.

Then, BD – DC = 9 cm

in right angled triangle ABD,

By Pythagoras theorem, we get

AB2 = AD2 + BD2

AD2 = AB2 – BD2

AD2 = (15)2 – (9)2

AD2 = 225 – 81

AD2 = 144

AD – 12 cm

(i) cos ∠ABC = Base/ / Hypotenuse

(In right angled ∆ABD, ∠ABC = ∠ABD)

= BD / AB

= 9/15

= 3/5

(ii) sin ∠ACB = sin ∠ACD

= perpendicular/ Hypotenuse

= AD/AC

= 12/15

= 4/5

8. (a) In the figure (1) given below, ∆ABC is isosceles with AB = AC = 5 cm and BC = 6 cm. Find

(i) sin C

(ii) tan B

(iii) tan C – cot B.

(b) In the figure (2) given below, ∆ABC is right-angled at B. Given that ∠ACB = θ, side AB = 2 units and side BC = 1 unit, find the value of sin2 θ + tan2 θ.

(c) In the figure (3) given below, AD is perpendicular to BC, BD = 15 cm, sin B = 4/5 and tan C = 1.

(i) Calculate the lengths of AD, AB, DC and AC.

(ii) Show that tan2 B – 1/cos2 B = – 1.

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 11

Solution:

(a) It is given that

∆ABC is isosceles with AB = AC = 5 cm and BC = 6 cm

Construct AD perpendicular to BC

D is the mid point of BC

So BD = CD

Here

BD = CD = 6/2 = 3 cm

In right angled ∆ABD

Using Pythagoras theorem

AB2 = AD2 + BD2

We can write it as

AD2 = AB2 – BD2

Substituting the values

AD2 = 52 – 32

By further calculation

AD2 = 25 – 9 = 16

So we get

AD2 = 42

AD = 4 cm

(i) In right angled ∆ACD

sin C = perpendicular/hypotenuse

sin C = AD/AC = 4/5

(ii) In right angled ∆ABD

tan B = perpendicular/base

tan B = AD/BD = 4/3

(iii) In right angled ∆ACD

tan C = perpendicular/base

tan C = AD/CD = 4/3

In right angled ∆ABD

cot B = base/perpendicular

cot B = BD/AD = ¾

Here

tan C – cot B = 4/3 – ¾

Taking LCM

tan C – cot B = (16 – 9)/ 12 = 7/12

(b) It is given that

∆ABC is right-angled at B

AB = 2 units and BC = 1 unit

In right angled ∆ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = 22 + 12

AC2 = 4 + 1 = 5

So we get

AC2 = 5

AC = √5 units

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

sin θ = AB/AC = 2/√5

In right angled ∆ABC

tan θ = perpendicular/base

tan θ = AB/BC = 2/1

We know that

sin2 θ + tan2 θ = (2/√5)2 + (2/1)2

By further calculation

= 4/5 + 4/1

Taking LCM

= (4 + 20)/ 5

= 24/5

= 4 4/5

(c) (i) In ∆ABC

AD is perpendicular to BC

BD = 15 cm

sin B = 4/5

tan C = 1

In ∆ABD

sin B = perpendicular/hypotenuse

sin B = AD/AB = 4/5

Consider AD = 4x and AB = 5x

Using Pythagoras theorem

In right angled ∆ABD

AB2 = AD2 + BD2

We can write it as

BD2 = AB2 – AD2

Substituting the values

(15)2 = (5x)2 – (4x)2

225 = 25x2 – 16x2

By further calculation

225 = 9x2

x2 = 225/9 = 25

So we get

x = √25 = 5

Here

AD = 4 × 5 = 20

AB = 5 × 5 = 25

In right angled ∆ACD

tan C = perpendicular/base

So we get

tan C = AD/CD = 1/1

Consider AD = X then CD = x

In right angled ∆ADC

Using Pythagoras theorem

AC2 = AD2 + CD2

Substituting the values

AC2 = x2 + x2 …..(1)

So the equation becomes

AC2 = 202 + 202

AC2 = 400 + 400 = 800

So we get

AC = √800 = 20√2

Length of AD = 20 cm

Length of AB = 25 cm

Length of DC = 20 cm

Length of AC = 20√2 cm

(ii) In right angled ∆ABD

tan B = perpendicular/base

So we get

tan B = AD/BD

Substituting the values

tan B = 20/15 = 4/3

In right angled ∆ABD

cos B = base/hypotenuse

So we get

cos B = BD/AB

Substituting the values

cos B = 15/25 = 3/5

Here

LHS = tan2 B – 1/cos2 B

Substituting the values

= (4/3)2 – 1/(3/5)2

By further calculation

= (4)2/(3)2 – (5)2/(3)2

= 16/9 – 25/9

So we get

= (16 – 25)/9

= -9/9

= – 1

= RHS

Hence, proved.

9. If sin θ =3/5 and θ is acute angle, find

(i) cos θ

(ii) tan θ.

Solution:

Let ∆ ABC be a right angled at B

Let ∠ACB = θ

Given that, sin θ = 3/5

AB/AC = 3/5

Let AB = 3x

then AC = 5x

In right angled ∆ ABC,

By Pythagoras theorem,

We get

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 12

(5x)2 = (3x)2 + BC2

BC2 = (5x)2 – (3x)2

BC2 = (2x)2

BC = 4x

(i) cos θ = Base/ Hypotenuse

= BC / AC

= 4x /5x

= 4/5

(ii) tan θ = perpendicular/Base

= AB/BC

= 3x/4x

= ¾

10. Given that tan θ = 5/12 and θ is an acute angle, find sin θ and cos θ.

Solution:

Consider ∆ ABC be right angled at B and ∠ACB = θ

It is given that

tan θ = 5/12

AB/BC = 5/12

Consider AB = 5x and BC = 12x

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 13

In right angled ∆ ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = (5x)2 + (12x)2

By further calculation

AC2 = 25x2 + 144x2 = 169x2

So we get

AC2 = (13x)2

AC = 13x

In right angled ∆ ABC

sin θ = perpendicular/hypotenuse

So we get

sin θ = AB/AC = 5x/13x = 5/13

In right angled ∆ ABC

cos θ = base/hypotenuse

So we get

cos θ = BC/AC

Substituting the values

cos θ = 12x/13x = 12/13

11. If sin θ = 6/10, find the value of cos θ + tan θ.

Solution:

Consider ∆ ABC be right angled at B and ∠ACB = θ

It is given that

sin θ = AB/AC

sin θ = 6/10

Take AB = 6x then AC = 10x

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 14

In right angled ∆ ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

(10x)2 = (6x)2 + BC2

By further calculation

BC2 = 100x2 – 36x2 = 64x2

So we get

BC2 = (8x)2

BC = 8x

In right angled ∆ ABC

cos θ = base/hypotenuse

cos θ = BC/AC

Substituting the values

cos θ = 8x/10x = 4/5

In right angled ∆ ABC

tan θ = perpendicular/base

tan θ = AB/BC

Substituting the values

tan θ = 6x/8x = ¾

Here

cos θ + tan θ = 4/5 + ¾

Taking LCM

= (16 + 15)/ 20

= 31/20

= 1 11/20

12. If tan = 4/3, find the value of sin θ + cos θ (both sin θ and cos θ are positive).

Solution:

Let ∆ABC be a right angled

∠ACB = θ

Given that, tan θ = 4/3

(AB/BC = 4/3)

Give that, tan θ = 4/3

(AB/BC = 4/3)

Let AB = 4x,

then BC = 3x

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 15

In right angled ∆ABC

By Pythagoras theorem, we get

AC2 = AB2 + BC2

AC2 = AB2 + BC2

AC2 = AB2 + BC2

AC2 = AB2 + BC2

(AC2 = (4x)2 + (3x)2

AC2 = 16x2 + 9x2

AC2 = 25x2

AC2 = (5x)2

AC = 5x

Sin θ = perpendicular/Hypotenuse

= AB/AC

= 4x/5x

= 4/5

Cos θ = Base/Hypotenuse

= BC/AC

= 3x/5x

= 3/5

Sin θ + cos θ

= 4/5 + 3/5

= (4 + 3)/5

= 7/5

Hence, Sin θ + cos θ = 7/5 = 1 2/5

13. 1f cosec = √5 and θ is less than 900, find the value of cot θ – cos θ.

Solution:

Given cosec θ = √5/1 = OP/PM

OP = √5 and PM = 1

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 16

Now OP2 = OM2 + PM2 using Pythagoras theorem

(√5)2 = OM2 + 12

5 = OM2 + 1

OM2 = 5 – 1

OM2 = 4

OM = 2

Now cot θ = OM/PM

= 2/1

= 2

Cos θ = OM/OP

= 2/√5

Now cot θ – Cos θ = 2 – (2/√5)

= 2 (√5 – 1)/ √5

14. Given sin θ = p/q, find cos θ + sin θ in terms of p and q.

Solution:

Given that sin θ = p/q

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 17

Which implies,

AB/AC = p/q

Let AB = px

And then AC = qx

In right angled triangle ABC

By Pythagoras theorem,

We get

AC2 = AB2 + BC2

BC2 = AC2 – AB2

BC2 = q2x2 – p2x2

BC2 = (q2 – p2)x2

BC = √( q2 – p2)x

In right angled triangle ABC,

Cos θ = base/ hypotenuse

= BC/AC

= √( q2 – p2)x/qx

= √( q2 – p2)/ q

Now,

Sin θ + cos θ = p/q + √( q2 – p2)/ q

= [p + √( q2 – p2)]/ q

15. If θ is an acute angle and tan = 8/15, find the value of sec θ + cosec θ.

Solution:

Given tan θ = 8/15

θ is an acute angle

in the figure triangle OMP is a right angled triangle,

∠M = 90o and ∠Q = θ

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 18

Tanθ = PM/OL = 8/15

Therefore, PM = 8, OM = 15

But OP2 = OM2 + PM2 using Pythagoras theorem,

= 152 + 82

= 225 + 64

= 289

= 172

Therefore, OP = 17

Sec θ = OP/OM

= 17/15

Cosec θ = OP/PM

= 17/8

Now,

Sec θ + cosec θ = (17/15) + (17/8)

= (136 + 255)/ 120

= 391/120

= 3 31/120

16. Given A is an acute angle and 13 sin A = 5, evaluate:

(5 sin A – 2 cos A)/ tan A.

Solution:

Let triangle ABC be a right angled triangle at B and A is an acute angle

Given that 13 sin A = 5

Sin A = 5/13

AB/Ac = 5/13

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 19

Let AB = 5x

AC = 13 x

In right angled triangle ABC,

Using Pythagoras theorem,

We get

AC2 = AB2 + BC2

BC2 = AC2 – BC2

BC2 = (13x)2 – (5x)2

BC2 = 169x2 – 25x2

BC2 = 144x2

BC = 12x

Sin A = 5/13

Cos A = base/ hypotenuse

= BC/AC

= 12x/ 13x

= 12/13

Tan A = perpendicular/ base

= AB/BC

= 5x/ 12x

= 5/ 12

Now,

(5 sin A – 2 cos A)/ tan A = [(5) (5/13) – (2) (12/13)]/ (5/12)

= (1/13)/ (5/12)

= 12/65

Hence (5 sin A – 2 cos A)/ tan A = 12/65

17. Given A is an acute angle and cosec A = √2, find the value of

(2 sin2 A + 3 cot2 A)/ (tan2 A – cos2 A).

Solution:

Let triangle ABC be a right angled at B and A is a acute angle.

Given that cosec A = √2

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 20

Which implies,

AC/BC = √2/1

Let AC = √2x

Then BC = x

In right angled triangle ABC

By using Pythagoras theorem,

We get

AC2 = AB2 + BC2

(√2x)2 = AB2 + x2

AB2 = 2x2 – x2

AB = x

Sin A = perpendicular/ hypotenuse

= BC/AC

= 1/ √2

Cot A = base/ perpendicular

= x/x

= 1

Tan A = perpendicular/ base

= BC/AB

= x/x

= 1

Cos A = base/ hypotenuse

= AB/AC

= x/ √2x

= 1/√2

Substituting these values we get

2 sin2 A + 3 cot2 A/ (tan2 A – cos2 A) = 8

18. The diagonals AC and BD of a rhombus ABCD meet at O. If AC = 8 cm and BD = 6 cm, find sin ∠OCD.

Solution:

It is given that

Diagonals AC and BD of rhombus ABCD meet at O

AC = 8 cm and BD = 6 cm

O is the mid point of AC

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 21

We know that

AO = OC = AC/2 = 8/2 = 4 cm

O is the mid point of BD

BO = OD = BD/2 = 6/2 = 3 cm

In right angled ∆COD

CD2 = OC2 + OD2

Substituting the values

CD2 = 42 + 32

So we get

CD2 = 16 + 9 = 25

CD2 = 52

CD = 5 cm

In right angled ∆COD

sin ∠OCD = perpendicular/ hypotenuse

So we get

sin ∠OCD = OD/CD = 3/5

19. If tan θ = 5/12, find the value of (cos θ + sin θ)/ (cos θ – sin θ).

Solution:

Consider ∆ABC be right angled at B and ∠ACB = θ

It is given that

tan θ = AB/BC = 5/12

Take AB = 5x then BC = 12x

In right angled ∆ABC,

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = (5x)2 + (12x)2

By further calculation

AC2 = 25x2 + 144x2 = 169x2

So we get

AC2 = (13x)2

AC = 13x

In right angled ∆ABC

cos θ = base/hypotenuse

cos θ = BC/AC

Substituting the values

cos θ = 12x/13x = 12/13

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

sin θ = AB/AC

Substituting the values

sin θ = 5x/13x = 5/13

Here

(cos θ + sin θ)/ (cos θ – sin θ) = [12/13 + 5/13]/ [12/13 – 5/13]

Taking LCM

= [(12 + 5)/ 13]/[(12 – 5)/ 13]

So we get

= 17/13/ 7/13

= 17/13 × 13/7

= 17/7

Therefore, (cos θ + sin θ)/ (cos θ – sin θ) = 17/7 = 2 3/7.

20. Given 5 cos A – 12 sin A = 0, find the value of (sin A + cos A)/ (2 cos A – sin A).

Solution:

It is given that

5 cos A – 12 sin A = 0

We can write it as

5 cos A = 12 sin A

So we get

sin A/ cos A = 5/12

We know that sin A/ cos A = tan A

tan A = 5/12

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 22

Consider ∆ABC right angled at B and ∠A is acute angle

Here

tan A = BC/AB = 5/12

Take BC = 5x then AB = 12x

In right angled ∆ABC

Using Pythagoras theorem

AC2 = BC2 + AB2

Substituting the values

AC2 = (5x)2 + (12x)2

AC2 = 25x2 + 144x2 = 169x2

So we get

AC2 = (13x)2

AC = 13x

In right angled ∆ABC

sin A = perpendicular/hypotenuse

So we get

sin A = BC/AC = 5x/13x = 5/13

In right angled ∆ABC

cos A = base/hypotenuse

So we get

cos A = AB/AC = 12x/13x = 12/13

Here

(sin A + cos A)/ (2 cos A – sin A) = [5/13 + 12/13]/ [2 × 12/13 – 5/13]

By further calculation

= [(5 + 12)/13]/ [24/13 – 5/13]

So we get

= [(5 + 12)/13]/[(24 – 5)/13]

= 17/13/ 19/13

= 17/13 × 13/19

= 17/19

Therefore, (sin A + cos A)/ (2 cos A – sin A) = 17/19

21. If tan θ = p/q, find the value of (p sin θ – q cos θ)/ (p sin θ + q cos θ).

Solution:

It is given that

tan θ = p/q

Consider ∆ABC be right angled at B and ∠BCA = θ

tan θ = BC/AB = p/q

BC = px then AB = qx

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 23

In right angled ∆ABC

Using Pythagoras theorem

AC2 = BC2 + AB2

Substituting the values

AC2 = (px)2 + (qx)2

AC2 = p2x2 + q2x2

AC2 = x2 (p2 + q2)

So we get

AC = √x2 (p2 + q2)

AC = x(√p2 + q2)

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

sin θ = BC/AC

Substituting the values

sin θ = px/ x(√p2 + q2)

So we get

sin θ = p/ (√p2 + q2)

In right angled ∆ABC

cos θ = base/hypotenuse

cos θ = AB/AC

Substituting the values

cos θ = qx/ x(√p2 + q2)

So we get

cos θ = q/ (√p2 + q2)

Here

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 24

22. If 3 cot θ = 4, find the value of (5 sin θ – 3 cos θ)/ (5 sin θ + 3 cos θ).

Solution:

It is given that

3 cot θ = 4

cot θ = 4/3

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 25

Consider ∆ABC be right angled at B and ∠ACB = θ

cot θ = BC/AB = 4/3

Take BC = 4x then AB = 3x

In right angled ∆ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = (3x)2 + (4x)2

AC2 = 9x2 + 16x2 = 25x2

So we get

AC2 = (5x)2

AC = 5x

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

sin θ = AB/AC

Substituting the values

sin θ = 3x/5x = 3/5

In right angled ∆ABC

cos θ = base/hypotenuse

cos θ = BC/AC

Substituting the values

cos θ = 4x/5x = 4/5

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 26

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 27

23. (i) If 5 cos θ – 12 sin θ = 0, find the value of (sin θ + cos θ)/(2 cos θ – sin θ).

(ii) If cosec θ = 13/12, find the value of (2 sin θ – 3 cos θ)/(4 sin θ – 9 cos θ).

Solution:

(i) It is given that

5 cos θ – 12 sin θ = 0

We can write it as

5 cos θ = 12 sin θ

sin θ/ cos θ = 5/12

tan θ = 5/12

Dividing both numerator and denominator by cos θ

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 28

(ii) It is given that

cosec θ = 13/12

We know that cosec θ = 1/ sin θ

1/sin θ = 13/12

sin θ = 12/13

Here cos2 θ = 1 – sin2 θ

Substituting the values

= 1 – (12/13)2

By further calculation

= 1 – 144/169

Taking LCM

= (169 – 144)/ 169

= 25/169

So we get

= (5/13)2

cos θ = 5/13

Here

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 29

24. If 5 sin θ = 3, find the value of (sec θ – tan θ)/ (sec θ + tan θ).

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 30

Consider ∆ABC be right angled at B and ∠ACB = θ

It is given that

5 sin θ = 3

sin θ = AB/AC = 3/5

Take AB = 3x then AC = 5x

In right angled ∆ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

BC2 = AC2 – AB2

Substituting the values

BC2 = (5x)2 – (3x)2

So we get

BC2 = 25x2 – 9x2 = 16x2

BC2 = (4x)2

BC = 4x

In right angled ∆ABC

sec θ = hypotenuse/base

sec θ = AC/BC = 5x/4x = 5/4

In right angled ∆ABC

tan θ = perpendicular/base

tan θ = AB/BC = 3x/4x = ¾

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 31

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 32

25. If θ is an acute angle and sin θ = cos θ, find the value of 2 tan2 θ + sin2 θ – 1.

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 33

Consider ∆ABC be right angled at B and ∠ACB = θ

It is given that

sin θ = cos θ

sin θ/cos θ = 1

tan θ = AB/BC = 1

Take AB = x then BC = x

In right angled ∆ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

AC2 = x2 + x2 = 2x2

So we get

AC = √2x2

AC = (√2)x

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

sin θ = AB/AC = x/√2x = 1/√2

Here

2 tan2 θ + sin2 θ – 1 = 2 × (1)2 + (1/√2)2 – 1

By further calculation

= 2 × 1 + ½ – 1

= 2 + ½ – 1

= 1+ ½

Taking LCM

= (2 + 1)/2

= 3/2

Therefore, 2 tan2 θ + sin2 θ – 1 = 3/2.

26. Prove the following:

(i) cos θ tan θ = sin θ

(ii) sin θ cot θ = cos θ

(iii) sin2 θ/ cos θ + cos θ = 1/ cos θ.

Solution:

(i) cos θ tan θ = sin θ

LHS = cos θ tan θ

We know that tan θ = sin θ/cos θ

= cos θ (sin θ/cos θ)

So we get

= 1 × sin θ/1

= sin θ

= RHS

Therefore, LHS = RHS.

(ii) sin θ cot θ = cos θ

LHS = sin θ cot θ

We know that cot θ = cos θ/sin θ

= sin θ (cos θ/sin θ)

= 1 × cos θ/1

= cos θ

= RHS

Therefore, LHS = RHS.

(iii) sin2 θ/cos θ + cos θ = 1/cos θ

LHS = sin2 θ/cos θ + cos θ/1

Taking LCM

= (sin2 θ + cos2 θ)/cos θ

We know that sin2 θ + cos2 θ = 1

= 1/cos θ

= RHS

Therefore, LHS = RHS.

27. If in ∆ABC, ∠C = 900 and tan A = ¾, prove that sin A cos B + cos A sin B = 1.

Solution:

It is given that

tan A = BC/AC = ¾

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 34

Using Pythagoras theorem

AB2 = AC2 + BC2

Substituting the values

= 42 + 32

= 16 + 9

= 25

= 52

So we get AB = 5

Here

sin A = BC/AC = 3/5

cos A = AC/AB = 4/5

cos B = BC/AB = 3/5

sin B = AC/AB = 4/5

LHS = sin A cos B + cos A sin B

Substituting the values

= 3/5 × 3/5 + 4/5 × 4/5

By further calculation

= 9/25+ 16/25

= (9 + 16)/ 25

= 25/25

= 1

= RHS

Therefore, LHS = RHS.

28. (a) In figure (1) given below, ∆ABC is right-angled at B and ∆BRS is right-angled at R. If AB = 18 cm, BC = 7.5 cm, RS = 5 cm, ∠BSR = x0 and ∠SAB = y0, then find:

(i) tan x0

(ii) sin y0.

(b) In the figure (2) given below, ∆ABC is right angled at B and BD is perpendicular to AC. Find

(i) cos ∠CBD

(ii) cot ∠ABD.

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 35

Solution:

(a) ∆ABC is right-angled at B, ∆BSC is right-angled at S and ∆BRS is right-angled at R

It is given that

AB = 18 cm, BC = 7.5 cm, RS = 5 cm, ∠BSR = x0 and ∠SAB = y0

By Geometry ∆ARS and ∆ABC are similar

AR/AB = RS/BC

Substituting the values

AR/18 = 5/7.5

By further calculation

AR = (5 × 18)/7.5 = (1 × 18)/1.5

Multiply both numerator and denominator by 10

AR = (18 × 10)/15

AR = (10 × 6)/5

AR = (2 × 6)/1 = 12

So we get

RB = AB – AR

RB = 18 – 12 = 6

In right angled ∆ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = 182 + 7.52

By further calculation

AC2 = 324 + 56.25 = 380.25

AC = √380.25 = 19.5 cm

(i) In right angled ∆BSR

tan x0 = perpendicular/base

tan x0 = RB/RS = 6/5

(ii) In right angled ∆ASR

sin y0 = perpendicular/hypotenuse

Using Pythagoras theorem

AS2 = 122 + 52

By further calculation

AS2 = 144 + 25 = 169

AS = √169 = 13 cm

So we get

sin y0 = RS/AS = 5/13

(b) We know that

∆ABC is right angled at B and BD is perpendicular to AC

In right angled ∆ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = 122 + 52

By further calculation

AC2 = 144 + 25 = 169

So we get

AC2 = (13)2

AC = 13

By Geometry ∠CBD = ∠A and ∠ABD = ∠C

(i) cos ∠CBD = cos ∠A = base/hypotenuse

In right angled ∆ABC

cos ∠CBD = cos ∠A = AB/AC = 12/13

(ii) cos ∠ABD = cos ∠C = base/perpendicular

In right angled ∆ABC

cos ∠ABD = cos ∠C = BC/AB = 5/12

29. In the adjoining figure, ABCD is a rectangle. Its diagonal AC = 15 cm and ∠ACD = α. If cot α = 3/2, find the perimeter and the area of the rectangle.

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 36

Solution:

In right ∆ADC

cot α = CD/AD = 3/2

Take CD = 3x then AD = 2x

Using Pythagoras theorem

AC2 = CD2 + AD2

Substituting the values

(15)2 = (3x)2 + (2x)2

By further calculation

13x2 = 225

x2 = 225/13

So we get

x = √225/13 = 15/√13

Length of rectangle (l) = 3x = (3 × 15)/ √13 = 45/√13 cm

Breadth of rectangle (b) = 2x = (2 × 15)/ √13 = 30/√13 cm

(i) Perimeter of rectangle = 2 (l + b)

Substituting the values of l and b

= 2 (45/√13 + 30/√13)

So we get

= 2 × 75/√13

= 150/√13 cm

(ii) Area of rectangle = l × b

Substituting the values of l and b

= 45/√13 × 30/√13

So we get

= 1350/13

= 103 11/13 cm2

30. Using the measurements given in the figure alongside,

(a) Find the values of:

(i) sin ϕ

(ii) tan θ.

(b) Write an expression for AD in terms of θ.

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 37

Solution:

From the figure

BC = 12, BD = 13

In right angled ∆BCD

Using Pythagoras theorem

BD2 = BC2 + CD2

It can be written as

CD2 = BD2 – BC2

Substituting the values

CD2 = (13)2 – (12)2

CD2 = 169 – 144 = 25

So we get

CD = √25 = 5

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 38

Construct BE perpendicular to AB

CD = BE = 5 and EA = AE = 14 – 5 = 9

(a) (i) sin ϕ = perpendicular/hypotenuse

In right angled ∆BCD

sin ϕ = CD/BD = 5/13

(ii) tan θ = perpendicular/hypotenuse

In right angled ∆AED

tan θ = ED/AE = BC/AE = 12/9 = 4/3 (Since ED = BC)

(b) In right angled ∆AED

sin θ = perpendicular/hypotenuse

cos θ = base/perpendicular

We can write it as

sin θ = ED/AD or cos θ = AE/AD

AD = ED/sin θ or AD = AE/ cos θ

Substituting the values

AD = 12/sin θ or AD = 9/cos θ

Therefore, AD = 12/ sin θ or AD = 9/cos θ.

31. Prove the following:

(i) (sin A + cos A)2 + (sin A – cos A)2 = 2

(ii) cot2 A – 1/sin2 A + 1 = 0

(iii) 1/(1 + tan2 A) + 1/(1 + cot2 A) = 1

Solution:

(i) (sin A + cos A)2 + (sin A – cos A)2 = 2

LHS = (sin A + cos A)2 + (sin A – cos A)2

Using the formula

(a + b)2 = a2 + b2 + 2ab and (a – b)2 = a2 + b2 – 2ab

= [(sin A)2 + (cos A)2 + 2 sin A cos A] + [(sin A)2 + (cos A)2 – 2 sin A cos A]

By further calculation

= sin2 A + cos2 A + 2 sin A cos A + sin2 A + cos2 A – 2 sin A cos A

= sin2 A + cos2 A + sin2 A + cos2 A

= 2 sin2 A + 2 cos2 A

We know that sin2 A + cos2 A = 1

= 2 (sin2 A + cos2 A)

= 2 (1)

= 2

= RHS

Therefore, LHS = RHS.

(ii) cot2 A – 1/sin2 A + 1 = 0

LHS = cot2 A – 1/sin2 A + 1

We know that

1/sin A = cosec A

= cot2 A – cosec2 A + 1

= (1 + cot2 A) – cosec2 A

We know that 1 + cot2 A = cosec2 A

= cosec2 A – cosec2 A

= 0

= RHS

Therefore, LHS = RHS.

(iii) 1/(1 + tan2 A) + 1/(1 + cot2 A) = 1

LHS = 1/(1 + tan2 A) + 1/(1 + cot2 A)

We know that

sec2 A – tan2 A = 1

sec2 A = 1 + tan2 A

cosec2 A – cot2 A = 1

cosec2 A = 1 + cot2 A

So we get

= 1/sec2 A + 1/cosec2 A

Here 1/sec A = cos A and 1/cosec A = sin A

= cos2 A + sin2 A

= 1

= RHS

Therefore, LHS = RHS.

32. Simplify

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 39

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 40

We know that 1 = sin2 θ + cos2 θ

= √cos2 θ/sin2 θ

= cos θ/ sin θ

Here cos θ/sin θ = cot θ

= cot θ

Therefore,
ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 41= cot θ.

33. If sin θ + cosec θ = 2, find the value of sin2 θ + cosec2 θ.

Solution:

It is given that

sin θ + cosec θ = 2

sin θ + 1/sin θ = 2

By further calculation

sin2 θ + 1 = 2 sin θ

sin2 θ – 2 sin θ + 1 = 0

So we get

(sin θ – 1)2 = 0

sin θ – 1 = 0

sin θ = 1

Here

sin2 θ + cosec2 θ = sin2 θ + 1/sin2 θ

Substituting the values

= 12 + 1/12

= 1 + 1/1

= 1 + 1

= 2

34. If x = a cos θ + b sin θ and y = a sin θ – b cos θ, prove that x2 + y2 = a2 + b2.

Solution:

It is given that

x = a cos θ + b sin θ …. (1)

y = a sin θ – b cos θ …. (2)

By squaring and adding both the equations

x2 + y2 = (a cos θ + b sin θ)2 + (a sin θ – b cos θ)2

Using the formula

(a + b)2 = a2 + b2 + 2ab and (a – b)2 = a2 + b2 – 2ab

= [(a cos θ)2 + (b sin θ)2 + 2 (a cos θ) (b sin θ)] + [(a sin θ)2 + (b cos θ)2 – 2 (a sin θ) (b cos θ)]

By further calculation

= a2 cos2 θ + b2 sin2 θ + 2 ab sin θ cos θ + a2 sin2 θ + b2 cos2 θ – 2 ab sin θ cos θ

= a2 cos2 θ + b2 sin2 θ + a2 sin2 θ + b2 cos2 θ

So we get

= a2 (cos2 θ + sin2 θ) + b2 (sin2 θ + cos2 θ)

Here sin2 θ + cos2 θ = 1

= a2 (1) + b2 (1)

= a2 + b2

Therefore, x2 + y2 = a2 + b2.

Chapter test

1. (a)From the figure (i) given below, calculate all the six t-ratios for both acute………

(b)From the figure (ii) given below, find the values of x and y in terms of t-ratios

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 42
ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 43

Solution:

(a) From right angled triangle ABC,

By Pythagoras theorem, we get

AC2 = AB2 + BC2

AB2 = AC2 – BC2

AB2 = (3)2 – (2)2

AB2 = 9 – 4

AB2 = 5

AB = √5

(i) sin A = perpendicular/ hypotenuse

= BC/AC

= 2/3

(ii) cos A = base/ hypotenuse

= AB/AC

= √5/3

(iii) tan A = perpendicular/ base

= BC/AB

= 2/ √5

(iv) cot A = base/perpendicular

= AB/ BC

= √5/2

(v) sec A = hypotenuse/ base

= AC/AB

= 3/ √5

(vi) cosec A = hypotenuse/perpendicular

= AC/BC

= 3/2

(b) From right angled triangle ABC,

∠BAC = θ

Then we know that,

Cot θ = base/ perpendicular

= AB/ BC

= x/ 10

x = 10 cot θ

also, cosec θ = hypotenuse/ perpendicular

= AC/ BC

= y/ 10

y = 10 cosec θ

Therefore, x = 10 cot θ and y = 10 cosec θ.

2. (a) From the figure (1) given below, find the values of:

(i) sin ∠ABC

(ii) tan x – cos x + 3 sin x.

(b) From the figure (2) given below, find the values of:

(i) 5 sin x

(ii) 7 tan x

(iii) 5 cos x – 17 sin y – tan x.

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 44

Solution:

(a) From the figure

BC = 12, CD = 9 and BC = 20

In right angled ∆ABC,

Using Pythagoras theorem

AB2 = AC2 + BC2

It can be written as

AC2 = AB2 – BC2

Substituting the values

AC2 = (20)2 – (12)2

By further calculation

AC2 = 400 – 144 = 256

So we get

AC2 = (16)2

AC = 16

In right angled ∆BCD

Using Pythagoras theorem

BD2 = BC2 + CD2

Substituting the values

BD2 = 122 + 92

By further calculation

BD2 = 144 + 81 = 225

So we get

BD2 = (15)2

BD = 15

(i) In right angled ∆BCD

sin ∠ABC = perpendicular/hypotenuse

So we get

sin ∠ABC = AC/AB = 16/20 = 4/5

(ii) In right angled ∆BCD

tan x = perpendicular/base

So we get

tan x = BC/CD = 12/9 = 4/3

In right angled ∆BCD

cos x = base/hypotenuse

So we get

cos x = CD/BD = 9/15 = 3/5

In right angled ∆BCD

sin x = perpendicular/hypotenuse

So we get

sin x = BC/BD = 12/15 = 4/5

tan x – cos x + 3 sin x = 4/3 – 3/5 + 3 × 4/5

By further calculation

= 4/3 – 3/5 + 12/5

Taking LCM

= (4 × 5 – 3 × 3 + 12 × 3)/ 15

So we get

= (20 – 9 + 36)/ 15

= (56 – 9)/ 15

= 27/15

= 3 2/15

Therefore, tan x – cos x + 3 sin x = 3 2/15.

(b) In the figure

AC = 17, AB = 25, AD = 15

In right angled ∆ACD

Using Pythagoras theorem

AC2 = AD2 + CD2

Substituting the values

(17)2 = (15)2 + (CD)2

By further calculation

CD2 = (17)2 – (15)2

CD2 = 289 – 225 = 64

So we get

CD2 = 82

CD = 8

In right angled ∆ABD

Using Pythagoras theorem

AB2 = AD2 + BD2

Substituting the values

(25)2 = (15)2 + BD2

By further calculation

BD2 = (25)2 – (15)2

BD2 = 625 – 225 = 400

So we get

BD2 = (20)2

BD = 20

(i) In right angled ∆ABD

5 sin x = 5 (perpendicular/hypotenuse)

So we get

= 5 (AD/AB)

= 5 × 15/25

= 15/5

= 3

(ii) In right angled ∆ABD

7 tan x = 7 (perpendicular/base)

So we get

= 7 (AD/AB)

= 7 × 15/20

= 7 × ¾

= 21/4

= 5 ¼

(iii) In right angled ∆ABD

cos x = base/hypotenuse

So we get

cos x = BD/AB = 20/25 = 4/5

In right angled ∆ACD

sin y = perpendicular/hypotenuse

So we get

sin y = CD/AC = 8/17

In right angled ∆ABD

tan x = perpendicular/base

So we get

tan x = AD/BD = 15/20 = ¾

5 cos x – 17 sin y – tan x = 5 × 4/5 – 17 × 8/17 – ¾

It can be written as

= 4/1 – 8/1 – ¾

Taking LCM

= (16 – 32 – 3)/4

= (16 – 35)/4

So we get

= – 19/4

= – 4 ¾

Therefore, 5 cos x – 17 sin y – tan x = – 4 ¾.

3. If q cos θ = p, find tan θ – cot θ in terms of p and q.

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 45

Consider ABC as a triangle right angled at B and ∠ACB = θ

It is given that

q cos θ = p

cos θ = BC/AC = p/q

Take BC = px then AC = qx

In right angled ∆ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

It can be written as

AB2 = AC2 – BC2

Substituting the values

AB2 = (qx)2 – (px)2

AB2 = q2x2 – p2x2

Taking out the common terms

AB2 = (q2 – p2)x2

So we get

AB = √(q2 – p2) x2

AB = (√q2 – p2)x

In right angled ∆ABC

tan θ = perpendicular/base

So we get

tan θ = AB/BC = [(√q2 – p2)x]/px

tan θ = (√q2 – p2)/p

In right angled ∆ABC

cot θ = base/perpendicular

So we get

cot θ = BC/AB = px/[(√q2 – p2)x]

cot [(√q2 – p2)x] = p/(√q2 – p2)

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 46

4. Given 4 sin θ = 3 cos θ, find the values of:

(i) sin θ

(ii) cos θ

(iii) cot2 θ – cosec2 θ.

Solution:

It is given that

4 sin θ = 3 cos θ

sin θ/cos θ = ¾

tan θ = ¾

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 47

Consider ∆ABC right angled at B and ∠ACB = θ

tan θ = perpendicular/base

Substituting the values

¾ = AB/BC

AB/BC = ¾

Take AB = 3x then BC = 4x

In right angled ∆ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = (3x)2 + (4x)2

By further calculation

AC2 = 9x2 + 16x2 = 25x2

So we get

AC2 = (5x)2

AC = 5x

(i) In right angled ∆ABC

sin θ = perpendicular/hypotenuse

So we get

sin θ = AB/AC = 3x/5x = 3/5

(ii) In right angled ∆ABC

cos θ = base/hypotenuse

So we get

cos θ = BC/AC = 4x/5x = 4/5

(iii) In right angled ∆ABC

cot θ = base/perpendicular

So we get

cot θ = BC/AB = 4x/3x = 4/3

In right angled ∆ABC

cosec θ = hypotenuse/perpendicular

So we get

cosec θ = AC/AB = 5x/3x = 5/3

Here

cot2 θ – cosec2 θ = (4/3)2 – (5/3)2

By further calculation

= 16/9 – 25/9

= (16 – 25)/9

= -9/9

= – 1

Therefore, cot2 θ – cosec2 θ = -1.

5. If 2 cos θ = √3, prove that 3 sin θ – 4 sin3 θ = 1.

Solution:

It is given that

2 cos θ = √3

cos θ = √3/2

We know that

sin2 θ = 1 – cos2 θ

Substituting the values

= 1 – (√3/2)2

= 1 – ¾

= ¼

sin θ = √ ¼ = ½

Consider

LHS = 3 sin θ – 4 sin3 θ

It can be written as

= sin θ (3 – 4 sin2 θ)

Substituting the values

= ½ (3 – 4 × ¼)

= ½ (3 – 1)

= ½ × 1

= 1

= RHS

Therefore, proved.

6. If (sec θ – tan θ)/ (sec θ + tan θ) = ¼, find sin θ.

Solution:

We know that

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 48

By cross multiplication

4 – 4 sin θ = 1 + sin θ

We get

4 – 1 = sin θ + 4 sin θ

3 = 5 sin θ

sin θ = 3/5

7. If sin θ + cosec θ = 3 1/3, find the value of sin2 θ + cosec2 θ.

Solution:

It is given that

sin θ + cosec θ = 3 1/3 = 10/3

By squaring on both sides

(sin θ + cosec θ)2 = (10/3)2

Expanding using formula (a + b)2 = a2 + b2 + 2ab

sin2 θ + cosec2 θ + 2 sin θ cosec θ = 100/9

We know that sin θ = 1/cosec θ

sin2 θ + cosec2 θ + 2 sin θ × 1/ sin θ = 100/9

By further calculation

sin2 θ + cosec2 θ + 2 = 100/9

sin2 θ + cosec2 θ = 100/9 – 2

Taking LCM

sin2 θ + cosec2 θ = (100 – 18)/9 = 82/9

So we get

sin2 θ + cosec2 θ = 9 1/9

8. In the adjoining figure, AB = 4 m and ED = 3 m.

If sin α = 3/5 and cos β = 12/13, find the length of BD.

ML Aggarwal Solutions for Class 9 Maths Chapter 17 - Image 49

Solution:

It is given that

sin α = AB/AC = 3/5

AB = 3 and AC = 5

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

52 = 32 + BC2

By further calculation

25 = 9 + BC2

BC2 = 25 – 9 = 16

So we get

BC2 = 42

BC = 4

We know that

tan α = AB/BC = 4/5

cos β = CD/CE = 12/13

CD = 12 and CE = 13

Using Pythagoras theorem

CE2 = CD2 + ED2

Substituting the values

132 = 122 + ED2

By further calculation

ED2 = 132 – 122

ED2 = 169 – 144 = 25

So we get

ED2 = (5)2

ED = 5

tan β = ED/CD = 5/12

From the figure

tan α = AB/BC = 4/BC

So we get

¾ = 4/BC

BC = (4 × 4)/3 = 16/3 m

tan β = ED/CD = 3/CD

5/12 = 3/CD

So we get

CD = (12 × 3)/5 = 36/5 m

Here

BD = BC + CD

Substituting the values

= 16/3 + 36/5

Taking LCM

= (80 + 108)/15

= 188/15 m

= 12 8/15 m