# 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.

##### (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.

##### (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,

From right angled ∆ABD,

⇒ AB= 400

⇒ AB= (20)2

⇒ AB = 20

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

= 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)

= 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)

= 12/20

= 3/5

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

= 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:

(i) sin x

(ii) cot x

(iii) cot2 x- cosec2 x

(iv) sec y

(v) tan2 y – 1/cos2 y.

##### (a) From the figure, Sec θ = AB/BD

But in ∆ADC, ∠D = 90o

= 144

= (12)2

= (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

⇒ AB= (4)2 + (3)2

⇒ AB= 16 + 9

⇒ AB2 = 25

⇒ AB2 = (5)2

⇒ AB = 5

In right angled triangle ACD

⇒ 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

= 4/5

(ii) cot x = Base/Perpendicular

= ¾

(iii) cot x = Base/Perpendicular

= 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)

= 12/(8 √2)

= 3/(2 √2)

cot y = Base/Hypotenuse

= 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) 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.

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

= 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

##### (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.

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,

⇒ AD2 = (15)2 – (9)2

⇒ AD2 = 225 – 81

##### (i) cos ∠ABC = Base/Hypotenuse

(In right angled ∆ABD, ∠ABC = ∠ABD)

= BD / AB

= 9/15

= 3/5

##### (ii) sin ∠ACB = sin ∠ACD

= perpendicular/ Hypotenuse

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

D is the mid point of BC

So BD = CD

BD = CD = 6/2 = 3 cm

In right angled ∆ABD

AD2 = 25 – 9 = 16

(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

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,

(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

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

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.

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)+ 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.

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.

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).

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 = 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.

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.

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.

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

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.

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 sinA + 3 cot2 A)/ (tan2 A – cos2 A).

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.

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).

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).

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).

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).

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).

##### (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).

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.

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.

##### (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.

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.

##### (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.

cot α = CD/AD = 3/2

Take CD = 3x then AD = 2x

(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.

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

Hence,

#### 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

##### (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…………

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.

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.

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 :–