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

(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.

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

(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

From right angled ∆ACD,

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,

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

Therefore,

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

---

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.

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

Answer :

(a) From the figure, Sec θ = AB/BD

But in ∆ADC, ∠D = 90^o

AC² =AD² + DC² 

⇒ (13)² =AD² + 25

⇒ AD² = 169 -25

= 144

= (12)²

⇒ AD = 12

AB² = AD² + BD² (in right ∆ABD)

= (12)² + (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

⇒ AB² = (4)² + (3)²

⇒ AB² = 16 + 9

⇒ AB² = 25

⇒ AB² = (5)²

⇒ AB = 5

In right angled triangle ACD

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

(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.

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 = 90^o

tan A = 5/12

⇒ BC/AC =5/12

Let BC = 5x and AC = 12x

From right angled ∆ABC

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

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

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

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,

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

(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.

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,

(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

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

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

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

In right angled ∆ABC

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

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

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

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 = 90^o and ∠Q = 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

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 sin² A + 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

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

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

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

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)

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

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 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

(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,

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

Answer :

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 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 :

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

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.

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 = AB² + 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.

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.

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

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…………

Answer :

1 = sin² θ + cos² θ

= √cos² θ/sin² θ

= cos θ/ sin θ

Here cos θ/sin θ = cot θ

= cot θ

Hence,

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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