ML Aggarwal Trigonometrical Ratios Exe-17 Class 9 ICSE Maths Solutions

ML Aggarwal Trigonometrical Ratios Exe-17 Class 9 ICSE Maths Solutions Ch-17. Step by Step Solutions of Exe-17 questions on Trigonometrical Ratios of ML Aggarwal for ICSE Class 9th Mathematics. Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Trigonometrical Ratios Exe-17 Class 9 ICSE Maths Solutions

Board ICSE
Subject Maths
Class 9th
Chapter-17 Trigonometrical Ratios
Topics Solution of Exe-17 Questions
Academic Session 2024-2025

Exe-17 Solutions of ML Aggarwal for ICSE Class-9 Ch-17, Trigonometrical Ratios

Question 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

 (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

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

(i) sin

(ii) cos A

(iii) sin2 A + cos2 A

(iv) sec2 A – tan2 A.

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

Answer :

(a) From right angled triangle OMP,

By Pythagoras theorem,

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,

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

(ML Aggarwal Trigonometrical Ratios Exe-17 Class 9 ICSE Maths)

Question 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

(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.

(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.

(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.

Answer :

(a) 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

(b) From Figure

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

BD = BC – CD

= 21 – 5

= 16

(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.

From right angled ∆ACD,

AC = AD2 + CD2

⇒ AD= AC2 – CD2

⇒ AD= (13)2 – (5)2

⇒ AD= 169 – 25

⇒ AD= 144)

⇒ AD= (12)2

⇒ AD = 12

From right angled ∆ABD,

AB= AD2 + BD2

⇒ AB= 400

⇒ AB= (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 – coty = (5/3)2 – (4/3)2

= (25/9) – (16/9)

= (25-16)/9

= 9/9

= 1

Therefore,

cosec2 y – coty = 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


Trigonometrical Ratios Exercise-17

ML Aggarwal Class 9 ICSE Maths Solutions

Page 413

Question 3.

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

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

(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.

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

Answer :

(a) From the figure, Sec θ = AB/BD

But in ∆ADC, ∠D = 90o

AC2 =AD+ DC

⇒ (13)2 =AD+ 25

⇒ AD= 169 -25

= 144

= (12)2

⇒ AD = 12

AB= AD+ BD(in right ∆ABD)

= (12)+ (16)2

= 144 + 256

= 400

= (20)2

⇒ AB = 20

Sec θ = AB/BD

= 20/16

= 5/4

(b) let given ∆ABC

BD = 3, AC = 12, AD = 4

In right angled ∆ABD

AB2 =AD2 + BD2

⇒ AB= (4)2 + (3)2

⇒ AB= 16 + 9

⇒ AB2 = 25

⇒ AB2 = (5)2

⇒ AB = 5

In right angled triangle ACD

AC2 = AD+ CD2

⇒ CD2 = AC2 – AD2

⇒ CD2 = (12)– (4)2

⇒ CD= 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/cosy

= (1/2√2)2 – 1/(2√2/3)2

= ¼ × – ¼ × 2

= (1/8) – (9/8)

= (1-9)/8

= -8/8

= -1

tany – 1/cosy = –1

(ML Aggarwal Trigonometrical Ratios Exe-17 Class 9 ICSE Maths)

Question  4.

(a) From the figure (1) given below, find the values of:
(i) 2 sin y- cos y

(ii)2sinx-cosx
(iii) 1- sin x + cos y
(iv) 2 cos x-3 sin y +4 tan z.

(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 x°

(ii) y.

(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 x° (ii) y.

Answer :

(a) In a right angled ∆BCD,

BC2 = BD2 + CD2

BC2 = 92 + 122

BC2 = 81 + 144 = 225

⇒ BC2 = 152

⇒ BC = 15

In a right angled ∆ABC,

AC2 = AB2 + BC2

AB2 = AC2 – BC2

AB2 = 252 – 152

AB2 = 625 – 225 = 400

So we get

AB2 = 202

AB = 20

(i)

In right angled ∆BCD

sin y = perpendicular/ hypotenuse

⇒ sin y = BD/ BC

sin y = 9/15 = 3/5

In right angled ∆BCD

cos y = base/hypotenuse

⇒ cos y = CD/BC

cos y = 12/15 = 4/5

2sin y – cos y = 2× 3/5 – 4/5

= 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

sin x = 15/25 = 3/5

In right angled ∆ABC

cos x = base/hypotenuse

⇒ cos x = AB/AC

cos x = 20/25 = 4/5

2 sin x – cos x = 2× 3/5 – 4/5

= 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

sin x = 12/25 = 3/5

In right angled ∆BCD

cos y = base/hypotenuse

⇒ cos y = CD/BC

cos y = 12/15 = 4/5

1 – sin x + cos y

= 1 – 3/5 + 4/5

= (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

cos x = 20/25 = 4/5

In right angled ∆BCD

sin y = perpendicular/hypotenuse

⇒ sin y = BD/BC

sin y = 9/15 = 3/5

In right angled ∆ABC

tan x = perpendicular/base

⇒ tan x = BC/AB

tan x = 15/20 = ¾

Here,

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

= 8/5 – 9/5 3/1

Taking LCM

= (8 – 9 + 15)/5

= (23 – 9)/ 5

= 14/5

(b)  AB = y units, BC = 3 units, CA = 5 units

(i) In right angled ∆ABC

sin x = perpendicular/hypotenuse

⇒ sin x = BC/AC

sin x = 3/5

(ii) In right angled ∆ABC

AC2 = BC2 + AB2

We can write it as

AB2 = AC2 – BC2

AB2 = 52 – 32

AB2 = 25 – 9 = 16

So we get

AB2 = 42

⇒ AB = 4

y = 4 units

Hence,

y = 4 units.

Question  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.

Answer :

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

AB2 = (5x)+ (12x)2

⇒ AB= 25x2 + 144x2

⇒ AB= 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

Question  6

(a) In ..ABC, A = 90°. If AB 7 cm and BC- AC 1 cm, find :
(1) sin C
(i) tan B
(b) In …PQR,Q = 90°. If PQ = 40 cm and PR + QR = 50 cm, find :
(1) sin P
(ii) cos P
(ti) tan R

Answer :

(a) In right ∆ABC

∠A = 90°

AB = 7 cm

BC – AC = 1 cm

⇒ BC = 1 + AC

(a) In ∆ABC, ∠A = 90°. If AB = 7 cm and BC – AC = 1 cm, find:

BC2 = AB2 + AC2

Substituting the value of BC

(1 + AC)2 = AB2 + AC2

⇒ 1 + AC2 + 2AC = 72 + AC2

1 + AC2 + 2AC = 49 AC2

⇒ 2AC = 49 – 1 – 48

AC = 48/2 = 24 cm

Here,

BC = 1 + AC

BC = 1 + 24 = 25 cm

(i) sin C = AB/BC = 7/25

(ii) tan B = AC/AB = 24/7

(b) In right ∆PQR

∠Q = 90°

PQ = 40 cm

⇒ PQ + QR = 50 cm

We can write it as

PQ = 50 – QR

(b) In ∆PQR, ∠Q = 90°. If PQ = 40 cm and PR + QR = 50 cm, find:

PR2 = PQ2 + QR2

⇒ (50 – QR)2 = (40)2 + QR2

2500 + QR2 – 100QR = 1600 + QR2

So we get

2500 – 1600 = 100QR

⇒ 100QR = 900

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


Trigonometrical Ratios Exercise-17

ML Aggarwal Class 9 ICSE Maths Solutions

Page 414

Question  7.

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

(i) cos ∠ABC

(ii) sin ∠ACB.

Answer :

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,

AB= AD+ BD2

⇒ AD= AB2 – BD2

⇒ AD2 = (15)2 – (9)2

⇒ AD2 = 225 – 81

⇒ AD= 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

(ML Aggarwal Trigonometrical Ratios Exe-17 Class 9 ICSE Maths)

Question  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.

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.

Answer :

(a) ∆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

BD = CD = 6/2 = 3 cm

In right angled ∆ABD

AB2 = AD2 + BD2

AD2 = AB2 – BD2

AD2 = 52 – 32

AD2 = 25 – 9 = 16

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

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

AC2 = AB2 + BC2

AC2 = 22 + 12

⇒ AC2 = 4 + 1 = 5

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

sin2 θ + tan2 θ = (2/√5)2 + (2/1)2

= 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

In right angled ∆ABD,

AB2 = AD2 + BD2

BD2 = AB2 – AD2

(15)2 = (5x)2 – (4x)2

⇒ 225 = 25x2 – 16x2

225 = 9x2

⇒ x2 = 225/9 = 25

x = √25 = 5

AD = 4 × 5 = 20

AB = 5 × 5 = 25

In right angled ∆ACD

tan C = perpendicular/base

tan C = AD/CD = 1/1

Consider AD = X then CD = x

In right angled ∆ADC

AC2 = AD2 + CD2

AC2 = x2 + x2 …(1)

So the equation becomes

AC2 = 202 + 202

⇒ AC2 = 400 + 400 = 800

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

tan B = AD/BD

Substituting the values

tan B = 20/15 = 4/3

In right angled ∆ABD,

cos B = base/hypotenuse

cos B = BD/AB

Substituting the values

cos B = 15/25 = 3/5

Here,

LHS = tan2 B – 1/cos2 B

= (4/3)2 – 1/(3/5)2

= (4)2/(3)2 – (5)2/(3)2

= 16/9 – 25/9

So we get

= (16 – 25)/9

= -9/9

= – 1

= RHS

Hence, proved.

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

(i) cos θ

(ii) tan θ.

Answer :

Let ∆ ABC be a right angled at B

Let ∠ACB = θ

sin θ = 3/5

AB/AC = 3/5

Let AB = 3x

then AC = 5x

In right angled ∆ ABC,

ML aggarwal class-10 Trigonimetric Ratiochapter 17 img 11

(5x)2 = (3x)+ BC2

⇒ BC2 = (5x)– (3x)2

⇒ BC= (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

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

Answer :

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

tan θ = 5/12

AB/BC = 5/12

Consider AB = 5x and BC = 12x

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

AC2 = AB2 + BC2

AC2 = (5x)2 + (12x)2

AC2 = 25x2 + 144x2 = 169x2

AC2 = (13x)2

⇒ AC = 13x

In right angled ∆ ABC

sin θ = perpendicular/hypotenuse

sin θ = AB/AC = 5x/13x = 5/13

In right angled ∆ ABC

cos θ = base/hypotenuse

So we get

cos θ = BC/AC

cos θ = 12x/13x

= 12/13

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

Answer :

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 class-10 Trigonimetric Ratiochapter 17 img 13

In right angled ∆ ABC

AC2 = AB2 + BC2

(10x)2 = (6x)2 + BC2

BC2 = 100x2 – 36x2 = 64x2

BC2 = (8x)2

⇒ BC = 8x

In right angled ∆ ABC

cos θ = base/hypotenuse

⇒ cos θ = BC/AC

cos θ = 8x/10x = 4/5

In right angled ∆ ABC

tan θ = perpendicular/base

⇒ tan θ = AB/BC

tan θ = 6x/8x = ¾

Here

cos θ + tan θ = 4/5 + ¾

Taking LCM

= (16 + 15)/ 20

= 31/20

= 1 (11/20)

(ML Aggarwal Trigonometrical Ratios Exe-17 Class 9 ICSE Maths)

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

Answer :

Let ∆ABC be a right angled

∠ACB = θ

Given that, tan θ = 4/3

(AB/BC = 4/3)

tan θ = 4/3

(AB/BC = 4/3)

Let AB = 4x,

then BC = 3x

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

In right angled ∆ABC

AC2 = AB+ BC2

⇒ AC= AB+ BC2

⇒ AC2 = AB2 + BC2

⇒ AC= AB+ BC2

⇒ (AC= (4x)2 + (3x)2

⇒ AC= 16x+ 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

Question 13. If cosec = √5 and θ is less than 90°, find the value of cot θ – cos θ.

Answer :

Given cosec θ = √5/1 = OP/PM

OP = √5 and PM = 1

f cosec = √5 and θ is less than 90°, find the value of cot θ – cos θ.

Now OP2 = OM2 + PM2

(√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

cot θ – cos θ = 2 – (2/√5)

= 2 (√5 – 1)/ √5

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

Answer :

Given that sin θ = p/q

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

AB/AC = p/q

Let AB = px

And then AC = qx

In right angled triangle ABC

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

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

Answer :

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

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

tanθ = PM/OL = 8/15

PM = 8, OM = 15

But OP2 = OM2 + PM2 using Pythagoras theorem,

= 152 + 82

= 225 + 64

= 289

= 172

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)

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

(5 sin A – 2 cos A)/ tan A.

Answer :

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

Given A is an acute angle and 13 sin A = 5, Evaluate: (5 sin A – 2 cos A)/ tan A.

Let AB = 5x

AC = 13 x

In right angled triangle ABC,

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

(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

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

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

Answer :

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

Given that cosec A = √2

Given A is an acute angle and cosec A = √2, find the value of (2 sin2 A + 3 cot2 A)/ (tan2 A – cos2 A).

AC/BC = √2/1

Let AC = √2x

Then BC = x

In right angled triangle ABC

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

2 sin2A + 3 cot2A/(tan2A – cos2A) = 8


Trigonometrical Ratios Exercise-17

ML Aggarwal Class 9 ICSE Maths Solutions

Page 415

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

Answer :

Diagonals AC and BD of rhombus ABCD meet at O

AC = 8 cm and BD = 6 cm

O is the mid point of AC

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

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

CD2 = 42 + 32

CD2 = 16 + 9 = 25

⇒ CD2 = 52

⇒ CD = 5 cm

In right angled ∆COD

sin ∠OCD = perpendicular/ hypotenuse

sin ∠OCD = OD/CD = 3/5

(ML Aggarwal Trigonometrical Ratios Exe-17 Class 9 ICSE Maths)

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

Answer :

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

tan θ = AB/BC = 5/12

Take AB = 5x then BC = 12x

In right angled ∆ABC,

AC2 = AB2 + BC2

AC2 = (5x)2 + (12x)2

AC2 = 25x2 + 144x2 = 169x2

AC2 = (13x)2

⇒ AC = 13x

In right angled ∆ABC

cos θ = base/hypotenuse

cos θ = BC/AC

cos θ = 12x/13x = 12/13

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

⇒ sin θ = AB/AC

sin θ = 5x/13x = 5/13

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

Taking LCM

= [(12 + 5)/13]/[(12 – 5)/13]

= (17/13)/(7/13)

= 17/13 × 13/7

= 17/7

Hence,

(cos θ + sin θ)/(cos θ – sin θ) = 17/7

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

Answer :

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

sin A/ cos A = tan A

tan A = 5/12

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

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

tan A = BC/AB = 5/12

Take BC = 5x then AB = 12x

In right angled ∆ABC

AC2 = BC2 + AB2

AC2 = (5x)2 + (12x)2

⇒ AC2 = 25x2 + 144x2 = 169x2

AC2 = (13x)2

⇒ AC = 13x

In right angled ∆ABC

sin A = perpendicular/hypotenuse

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

In right angled ∆ABC

cos A = base/hypotenuse

⇒ cos A = AB/AC = 12x/13x = 12/13

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

= [(5+12)/13]/[24/13 – 5/13]

= [(5+12)/13]/[(24 – 5)/13]

= (17/13)/(19/13)

= 17/13 × 13/19

= 17/19

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

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

Answer :

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

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

AC2 = BC2 + AB2

AC2 = (px)2 + (qx)2

⇒ AC2 = p2x2 + q2x2

⇒ AC2 = x2 (p2 + q2)

AC = √x2 (p2 + q2)

⇒ AC = x(√p2 + q2)

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

⇒ sin θ = BC/AC

sin θ = px/x(√p2 + q2)

sin θ = p/(√p2 + q2)

In right angled ∆ABC

cos θ = base/hypotenuse

⇒ cos θ = AB/AC

cos θ = qx/x(√p2 + q2)

cos θ = q/(√p2 + q2)

ML aggarwal class-10 Trigonimetric Ratiochapter 17 img 23

Question 22. If 3 cot θ = 4, find the value of (5 sinθ – 3 cosθ)/(5 sinθ + 3 cosθ).

Answer :

3 cot θ = 4

⇒ cot θ = 4/3

ML aggarwal class-10 Trigonimetric Ratiochapter 17 img 24

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

AC2 = AB2 + BC2

AC2 = (3x)2 + (4x)2

⇒ AC2 = 9x2 + 16x2 = 25x2

AC2 = (5x)2

⇒ AC = 5x

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

⇒ sin θ = AB/AC

sin θ = 3x/5x = 3/5

In right angled ∆ABC

cos θ = base/hypotenuse

⇒ cos θ = BC/AC

cos θ = 4x/5x

= 4/5

ML aggarwal class-10 Trigonimetric Ratiochapter 17 img 25

(ML Aggarwal Trigonometrical Ratios Exe-17 Class 9 ICSE Maths)

Question 23.

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

(ii) If cosecθ = 13/12, find the value of (2 sinθ – 3 cosθ)/(4 sinθ – 9 cosθ).

Answer :

(i) 5 cosθ – 12 sinθ = 0

5 cosθ = 12 sinθ

⇒ sin θ/cos θ = 5/12

⇒ tan θ = 5/12

Dividing both numerator and denominator by cos θ

ML aggarwal class-10 Trigonimetric Ratiochapter 17 img 26

(ii) cosec θ = 13/12

cosec θ = 1/sin θ

1/sin θ = 13/12

⇒ sin θ = 12/13

Here cos2 θ = 1 – sin2 θ

= 1 – (12/13)2

= 1 – 144/169

Taking LCM

= (169 – 144)/ 169

= 25/169

= (5/13)2

cos θ = 5/13

Here,

ML aggarwal class-10 Trigonimetric Ratiochapter 17 img 27

Question 24. If 5 sin θ = 3, find the value of (secθ – tanθ)/(secθ + tanθ).

Answer :

If 5 sin θ = 3, find the value of (secθ – tanθ)/(secθ + tanθ).

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

5 sin θ = 3

sin θ = AB/AC = 3/5

Take AB = 3x then AC = 5x

In right angled ∆ABC

AC2 = AB2 + BC2

⇒ BC2 = AC2 – AB2

BC2 = (5x)2 – (3x)2

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 class-10 Trigonimetric Ratiochapter 17 img 29

(ML Aggarwal Trigonometrical Ratios Exe-17 Class 9 ICSE Maths)

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

Answer :

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

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

sin θ = cos θ

⇒ sin θ/cos θ = 1

tan θ = AB/BC = 1

Take AB = x then BC = x

In right angled ∆ABC,

AC2 = AB2 + BC2

⇒ AC2 = x2 + x2 = 2x2

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

= (2×1) + ½ – 1

= 2 + ½ – 1

= 1+ ½

Taking LCM

= (2 + 1)/2

= 3/2

Hence, 2 tan2 θ + sin2 θ – 1 = 3/2.

Question 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.

Answer :

(i) cos θ tan θ = sin θ

LHS = cos θ tan θ

tan θ = sin θ/cos θ

= cos θ (sin θ/cos θ)

= 1× sin θ/1

= sin θ

= RHS

Hence, LHS = RHS.

(ii) sin θ cot θ = cos θ

LHS = sin θ cot θ

cot θ = cos θ/sin θ

= sin θ (cos θ/sin θ)

= 1× cos θ/1

= cos θ

= RHS

Hence, LHS = RHS.

(iii) sin2θ/cosθ + cosθ = 1/cosθ

LHS = sin2θ/cosθ + cosθ/1

Taking LCM

= (sin2θ + cos2θ)/cosθ

sin2θ + cos2θ = 1

= 1/cos θ

= RHS

Hence,

LHS = RHS.

Question 27. If in ∆ABC, ∠C = 90° and tan A = ¾, prove that sin A cos B + cos A sin B = 1.

Answer :

tan A = BC/AC = ¾

 If in ∆ABC, ∠C = 90° and tan A = ¾, prove that sin A cos B + cos A sin B = 1.

AB2 = AC2 + BC2

= 42 + 32

= 16 + 9

= 25

= 52

AB = 5

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

= (3/5 × 3/5) + (4/5 × 4/5)

= 9/25 + 16/25

= (9 + 16)/ 25

= 25/25

= 1

= RHS

Hence, LHS = RHS.

(ML Aggarwal Trigonometrical Ratios Exe-17 Class 9 ICSE Maths)

Question 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 = x° and ∠SAB = y°, then find:

(i) tan x°

(ii) sin y°.

(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.

(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 = x° and ∠SAB = y°, then find: (i) tan x° (ii) sin y°. (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.

Answer :

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

AB = 18 cm, BC = 7.5 cm, RS = 5 cm, ∠BSR = x° and ∠SAB = y°

By Geometry ∆ARS and ∆ABC are similar

AR/AB = RS/BC

AR/18 = 5/7.5

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

RB = AB – AR

⇒ RB = 18 – 12 = 6

In right angled ∆ABC

AC2 = AB2 + BC2

AC2 = 182 + 7.52

AC2 = 324 + 56.25 = 380.25

⇒ AC = √380.25 = 19.5 cm

(i) In right angled ∆BSR

tan x° = perpendicular/base

tan x° = RB/RS = 6/5

(ii) In right angled ∆ASR

sin y° = perpendicular/hypotenuse

AS2 = 122 + 52

AS2 = 144 + 25 = 169

⇒ AS = √169 = 13 cm

sin y° = RS/AS = 5/13

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

In right angled ∆ABC

AC2 = AB+ BC2

AC2 = 122 + 52

AC2 = 144 + 25 = 169

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


Trigonometrical Ratios Exercise-17

ML Aggarwal Class 9 ICSE Maths Solutions

Page 416

Question 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.

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.

Answer :

In right ∆ADC

cot α = CD/AD = 3/2

Take CD = 3x then AD = 2x

AC2 = CD2 + AD2

(15)2 = (3x)2 + (2x)2

13x2 = 225

⇒ x2 = 225/13

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)

= 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

= 1350/13

= 103 (11/13) cm2

Question 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.

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.

Answer :

BC = 12, BD = 13

In right angled ∆BCD

BD2 = BC2 + CD2

CD2 = BD2 – BC2

CD2 = (13)2 – (12)2

⇒ CD2 = 169 – 144 = 25

CD = √25 = 5

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.

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

sin θ = ED/AD or cos θ = AE/AD

AD = ED/sin θ or AD = AE/cos θ

AD = 12/sin θ or AD = 9/cos θ

Hence,

AD = 12/ sin θ or AD = 9/cos θ.

Question 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

Answer :

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

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

(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]

= 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

sin2 A + cos2 A = 1

= 2 (sin2 A + cos2 A)

= 2 (1)

= 2

= RHS

Hence, LHS = RHS.

(ii) cot2 A – 1/sin2 A + 1 = 0

LHS = cot2 A – 1/sin2 A + 1

1/sin A = cosec A

= cot2 A – cosec2 A + 1

= (1 + cot2 A) – cosec2 A

1 + cot2 A = cosec2 A

= cosec2 A – cosec2 A

= 0

= RHS

Hence, LHS = RHS.

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

LHS = 1/(1 + tan2 A) + 1/(1 + cot2 A)

sec2 A – tan2 A = 1

⇒ sec2 A = 1 + tan2 A

⇒ cosec2 A – cot2 A = 1

⇒ cosec2 A = 1 + cot2 A

= 1/sec2 A + 1/cosec2 A

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

= cos2 A + sin2 A

= 1

= RHS

Hence,

LHS = RHS.

Question  32. Simplify…………

ML aggarwal class-10 Trigonimetric Ratiochapter 17 img 36

Answer :

ML aggarwal class-10 Trigonimetric Ratiochapter 17 img 37

1 = sin2 θ + cos2 θ

= √cos2 θ/sin2 θ

= cos θ/ sin θ

Here cos θ/sin θ = cot θ

= cot θ

Hence,

ML aggarwal class-10 Trigonimetric Ratiochapter 17 img 38

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

Answer :

sin θ + cosec θ = 2

⇒ sin θ + 1/sin θ = 2

sin2 θ + 1 = 2 sin θ

⇒ sin2 θ – 2 sin θ + 1 = 0

(sin θ – 1)2 = 0

⇒ sin θ – 1 = 0

⇒ sin θ = 1

sin2 θ + cosec2 θ = sin2 θ + 1/sin2 θ

= 12 + 1/12

= 1 + 1/1

= 1 + 1

= 2

(ML Aggarwal Trigonometrical Ratios Exe-17 Class 9 ICSE Maths)

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

Answer :

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

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

= [(a cosθ)2 + (b sinθ)2 + 2 (a cosθ) (b sinθ)] + [(a sinθ)2 + (b cosθ)2 – 2 (a sinθ) (b cosθ)]

= a2 cos2θ + b2 sin2θ + 2 ab sinθ cosθ + a2 sin2θ + b2 cos2θ – 2 ab sinθ cosθ

= a2 cos2θ + b2 sin2θ + a2 sin2θ + b2 cos2θ

= a2 (cos2θ + sin2θ) + b2 (sin2θ + cos2θ)

Here sin2θ + cos2θ = 1

= a2 (1) + b2 (1)

= a2 + b2

Hence, x2 + y2 = a2 + b2.

—  : End of ML Aggarwal Trigonometrical Ratios Exe-17 Class 9 ICSE Maths Solutions :–

Return to :-  ML Aggarawal Maths Solutions for ICSE  Class-9

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