# ML Aggarwal Trigonometrical Ratios Chapter Test Class 9 ICSE Maths Solutions

ML Aggarwal Trigonometrical Ratios Chapter Test Class 9 ICSE Maths Solutions Ch-17. Step by Step Solutions of Ch-Test 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 Chapter Test Class 9 ICSE Maths Solutions

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

### Solution of Ch-Test Questions on Trigonometrical Ratios

#### Question 1.

(a)From the figure (i0) 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

##### (a) From right angled triangle ABC,

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

cotθ = base/ perpendicular

= AB/ BC

= x/ 10

x = 10 cotθ

Also,

cosecθ = hypotenuse/ perpendicular

= AC/ BC

= y/ 10

y = 10 cosec θ

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

#### Question 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.

##### (a) BC = 12, CD = 9 and BC = 20

In right angled ∆ABC,

AB2 = AC2 + BC2

AC2 = AB2 – BC2

AC2 = (20)2 – (12)2

AC2 = 400 – 144 = 256

So we get

AC2 = (16)2

⇒ AC = 16

In right angled ∆BCD

BD2 = BC2 + CD2

BD2 = 122 + 92

BD2 = 144 + 81 = 225

BD2 = (15)2

⇒ BD = 15

(i) In right angled ∆BCD

sin ∠ABC = perpendicular/hypotenuse

sin ∠ABC = AC/AB = 16/20 = 4/5

(ii) In right angled ∆BCD

tan x = perpendicular/base

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

In right angled ∆BCD

cos x = base/hypotenuse

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

In right angled ∆BCD

sin x = perpendicular/hypotenuse

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

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

= 4/3 – 3/5 + 12/5

Taking LCM

= [(4×5) – (3×3) + (12×3)]/15

= (20 – 9 + 36)/15

= (56 – 9)/15

= 27/15

= 3 (2/15)

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

##### (b) AC = 17, AB = 25, AD = 15

In right angled ∆ACD

(17)2 = (15)2 + (CD)2

CD2 = (17)2 – (15)2

⇒ CD2 = 289 – 225 = 64

CD2 = 82

⇒ CD = 8

In right angled ∆ABD

(25)2 = (15)2 + BD2

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)

= 5 × 15/25

= 15/5

= 3

(ii) In right angled ∆ABD

7 tan x = 7 (perpendicular/base)

= 7× 15/20

= 7× ¾

= 21/4

= 5 ¼

(iii) In right angled ∆ABD

cos x = base/hypotenuse

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

In right angled ∆ACD

sin y = perpendicular/hypotenuse

sin y = CD/AC = 8/17

In right angled ∆ABD

tan x = perpendicular/base

tan x = AD/BD = 15/20 = ¾

5 cosx – 17 siny – tanx = (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

= –19/4

= –4 ¾

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

#### Question 3. If q cosθ = p, find tanθ – cotθ in terms of p and q.

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

q cos θ = p

cos θ = BC/AC = p/q

Take BC = px then AC = qx

In right angled ∆ABC

AC2 = AB2 + BC2

AB2 = AC2 – BC2

AB2 = (qx)2 – (px)2

⇒ AB2 = q2x2 – p2x2

AB2 = (q2 – p2)x2

AB = √(q2 – p2) x2

⇒ AB = (√q2 – p2)x

In right angled ∆ABC

tan θ = perpendicular/base

tan θ = AB/BC = [(√q2 – p2)x]/px

⇒ tan θ = (√q2 – p2)/p

In right angled ∆ABC

cot θ = base/perpendicular

cot θ = BC/AB = px/[(√q2 – p2)x]

⇒ cot [(√q2 – p2)x] = p/(√q2 – p2)

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

(i) sin θ

(ii) cos θ

(iii) cot2 θ – cosec2 θ.

4 sin θ = 3 cos θ

⇒ sin θ/cos θ = ¾

⇒ tan θ = ¾

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

tan θ = perpendicular/base

¾ = AB/BC

⇒ AB/BC = ¾

Take AB = 3x then BC = 4x

In right angled ∆ABC

AC2 = AB2 + BC2

AC2 = (3x)2 + (4x)2

AC2 = 9x2 + 16x2 = 25x2

AC2 = (5x)2

⇒ AC = 5x

(i) In right angled ∆ABC

sin θ = perpendicular/hypotenuse

sin θ = AB/AC = 3x/5x = 3/5

(ii) In right angled ∆ABC

cos θ = base/hypotenuse

cos θ = BC/AC = 4x/5x = 4/5

(iii) In right angled ∆ABC

cot θ = base/perpendicular

cot θ = BC/AB = 4x/3x = 4/3

In right angled ∆ABC

cosec θ = hypotenuse/perpendicular

cosec θ = AC/AB = 5x/3x = 5/3

cot2 θ – cosec2 θ = (4/3)2 – (5/3)2

= 16/9 – 25/9

= (16 – 25)/9

= -9/9

= -1

Hence, cot2 θ – cosec2 θ = -1.

#### Question 5. If 2 cos 0 = 3, prove that 3 sin 0 – 4 sin3 0 = 1.

2 cos θ = √3

⇒ cos θ = √3/2

sin2 θ = 1 – cos2 θ

= 1 – (√3/2)2

= 1 – ¾

= ¼

sin θ = √ ¼ = ½

LHS = 3 sin θ – 4 sin3 θ

= sinθ (3 – 4sin2 θ)

= ½ [3 – (4 × ¼)]

= ½ (3 – 1)

= ½ × 2

= 1

= RHS

Hence proved.

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

4 – 4 sin θ = 1 + sin θ

4 – 1 = sin θ + 4 sin θ

⇒ 3 = 5 sin θ

⇒ sin θ = 3/5

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

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 sinθ× 1/sin θ) = 100/9

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

sin2 θ + cosec2 θ = 9 1/9

#### Question 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.

sin α = AB/AC = 3/5

AB = 3 and AC = 5

AC2 = AB2 + BC2

52 = 32 + BC2

25 = 9 + BC2

⇒ BC2 = 25 – 9 = 16

BC2 = 42

⇒ BC = 4

tan α = AB/BC = 4/5

cos β = CD/CE = 12/13

CD = 12 and CE = 13

CE2 = CD2 + ED2

132 = 122 + ED2

ED2 = 132 – 122

⇒ ED2 = 169 – 144 = 25

ED2 = (5)2

⇒ ED = 5

⇒ tan β = ED/CD = 5/12

tan α = AB/BC = 4/BC

¾ = 4/BC

⇒ BC = (4×4)/3 = 16/3 m

tan β = ED/CD = 3/CD

⇒ 5/12 = 3/CD

CD = (12 × 3)/5 = 36/5 m

Here,

BD = BC + CD

= 16/3 + 36/5

Taking LCM

= (80 + 108)/15

= 188/15 m

= 12 8/15 m

—  : End of ML Aggarwal Trigonometrical Ratios Chapter Test Class 9 ICSE Maths Solutions :–