# ML Aggarwal Trigonometrical Ratios of Standards Angles Exe-18.1 Class 9 ICSE Maths Solutions

ML Aggarwal Trigonometrical Ratios of Standards Angles Exe-18.1 Class 9 ICSE Maths Solutions Ch-18. Step by Step Solutions of Exe-18.1 questions on Trigonometrical Ratios of of Standards Angles for ML Aggarwal ICSE Class 9th Mathematics. Visit official website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Trigonometrical Ratios Exe-18.1 Class 9 ICSE Maths Solutions

 Board ICSE Subject Maths Class 9th Chapter-18 Trigonometrical Ratios of Standards Angles Topics Solution of Exe-18.1 Questions Academic Session 2024-2025

### Trigonometrical Ratios of Standards Angles

ML Aggarwal Solution of Exe-18.1 Questions Class 9 ICSE Maths Solutions Ch-18

#### Question 1. Find the values of

(i) 7 sin 30° cos 60°

(ii) 3 sin2 45° + 2 cos2 60°

(iii) cos2 45° + sin2 60° + sin2 30°

(iv) cos 90° + cos2 45° sin 30° tan 45°.

= 7 ×½ ×½

= (7×1×1)/(2×2)

= 7/4

##### (ii) 3 sin2 45° + 2 cos2 60°

= 3×(1/2)2 + [2×(1/2)2]

= (3×½) + (2 × ¼)

= 3/2 + ½

= (3 + 1)/2

= 4/2

= 2

##### (iii) cos2 45° + sin2 60° + sin2 30°

= (1/2)2 + (√3/2)2 + (1/2)2

= ½ + ¾ + ¼

= (2 + 3 + 1)/4

= 6/4

= 3/2

##### (iv) cos 90° + cos2 45° sin 30° tan 45°

= 0 + (1/2)2 × ½ × 1

= ½ × ½ × 1

= ¼

#### Question 2. Find the values of

(i) (sin245° + cos245°)/tan260°

(ii) (sin30° – sin90° + 2cos0°)/tan2 60°

(iii) 4/3 tan230° + sin260° – 3cos260° + 3/4 tan260° – 2tan245°

##### (iii) 4/3 tan230° + sin260° – 3cos260° + ¾tan260° – 2tan245°

= 4/3(1/√3)2 + (√3/2)2 – 3(1/2)2 + ¾×(√3)2 – 2×12

= (4/3× 1/3) + ¾ – (3 × ¼) + (¾ ×3) – (2×1)

= 4/9 + ¾ – 3/4 + 9/4 – 2

= 4/9 + 9/4 – 2

Taking LCM

= (16 + 81 – 72)/36

= (97 – 72)/36

= 25/36

#### Question 3. Find the values of

(i) (sin30°/cos245°) – 3tan30° + 5cos90°

(ii) 2√2 cos45° cos60° + 2√3 sin30° tan60° – cos30°

(iii) 4/5 tan260° – (2/sin230°) – 3/4 tan230°

= 0

##### (ii) 2√2 cos 45° cos 60° + 2√3 sin 30° tan 60° – cos 0°

= 2√2 × 1/√2 × ½ + 2√3 × ½ × √3 – 1

= 2 × 1/1 × 1/2 + 2 × 3 × ½ – 1

= 1 + 3 – 1

= 3

### Trigonometrical Ratios of Standards Angles Exercise-18.1

ML Aggarwal Class 9 ICSE Maths Solutions

Page 429

#### Question 4. Prove That.

(i) cos2 30° + sin 30° + tan2 45° = 2 ¼

(ii) 4 (sin4 30° + cos4 60°) – 3 (cos2 45° – sin2 90°) = 2

(iii) cos 60° = cos2 30° – sin2 30°.

##### (i) cos2 30° + sin 30° + tan2 45° = 2 ¼

LHS = cos2 30° + sin 30° + tan2 45°

= (√3/2)2 + ½ + 12

= ¾ + ½ + 1

= (3 + 2 + 4)/4

= 9/4

= 2 ¼

= RHS

Hence, LHS = RHS.

##### (ii) 4 (sin4 30° + cos4 60°) – 3 (cos2 45° – sin2 90°) = 2

LHS = 4 (sin4 30° + cos4 60°) – 3 (cos2 45° – sin2 90°)

= 4[(½)4 + (½)4] – 3 [(1/√2)2 – 12]

= 4[½ × ½ × ½ × ½ + ½ × ½ × ½ × ½] – 3 [½ – 1]

= 4 [1/16 + 1/16] – 3 (- ½)

= 4[(1 + 1)/16] + 3/2

= (4 × 3)/16 + 3/2

= 8/16 + 3/2

= ½ + 3/2

= (1 + 3)/2

= 4/2

= 2

= RHS

Hence, LHS = RHS.

##### (iii) cos 60° = cos2 30° – sin2 30°

LHS = cos 60° = ½

RHS = cos2 30° – sin2 30°

= (√3/2)2 + (1/2)2

= ¾ – ¼

= (3 – 1)/4

= 2/4

= ½

= RHS

Hence, LHS = RHS.

(ML Aggarwal Trigonometrical Ratios of Standards Angles Exe-18.1 Class 9)

#### Question 5.

(i) If x = 30°, verify that tan 2x = 2tanx/ (1- tan2 x).

(ii) If x = 15°, verify that 4 sin 2x cos 4x sin 6x = 1.

##### (i) x = 30°

Consider LHS = tan 2x

Substituting the value of x

= tan 60°

= √3

Hence, LHS = RHS.

##### (ii) x = 15°

2x = 15 × 2 = 30°

4x = 15 × 4 = 60°

6x = 15 × 6 = 90°

LHS = 4 sin 2x cos 4x sin 6x

= 4 sin 30° cos 60° sin 90°

= 4 × ½ × ½ × 1

= 1

= RHS

Hence, LHS = RHS.

……………….

#### Question 7.  If θ = 30°, verify that

(i) sin 2θ = 2 sin θ cos θ

(ii) cos 2θ = 2 cos2 θ – 1

(iii) sin 3θ = 3 sin θ – 4 sin3 θ

(iv) cos 3θ = 4 cos3 θ – 3 cos θ.

θ = 30°

##### (i) sin 2θ = 2 sin θ cos θ

LHS = sin 2θ

= sin 2 × 30°

= sin 60°

= √3/2

RHS = 2 sin θ cos θ

= 2 sin 30° cos 30°

= 2 × ½ × √3/2

= 1 × √3/2

= √3/2

Hence, LHS = RHS.

##### (ii) cos 2θ = 2 cos2θ – 1

LHS = cos 2θ

= cos 2 × 30°

= cos 60°

= ½

RHS = 2 cos2 θ – 1

= 2 cos2 30° – 1

= 2 (√3/2)2 – 1

= 2 × ¾ – 1

= 3/2 – 1

= (3 – 2)/2

= ½

Hence, LHS = RHS.

##### (iii) sin 3θ = 3 sin θ – 4 sin3 θ

Consider

LHS = sin 3θ

= sin 3 × 30°

= sin 90°

= 1

RHS = 3 sin θ – 4 sin3 θ

= 3 sin 30° – 4 sin3 30°

= (3×½) – 4×(1/2)3

= 3/2 – 4× 1/8

= 3/2 – ½

Taking LCM

= (3 – 1)/2

= 2/2

= 1

Hence, LHS = RHS.

##### (iv) cos 3θ = 4 cos3 θ – 3 cos θ

LHS = cos 3θ

= cos 3 × 30°

= cos 90°

= 0

RHS = 4 cos3 θ – 3 cos θ

= 4 cos3 30° – 3 cos 30°

= 4×(√3/2)3 – 3×(√3/2)

= 4× 3√3/8 – 3√3/2

= 3√3/2 – 3√3/2

= 0

Hence, LHS = RHS.

(ML Aggarwal Trigonometrical Ratios of Standards Angles Exe-18.1 Class 9)

#### Question 8. If 0 = 30°, find the ratio 2 sin 0: sin 2 0.

θ = 30°

2 sin θ: sin 2θ = 2sin 30°: sin(2×30°)

= 2 sin 30°: sin 60°

= 2 sin 30°/sin 60°

Hence,

2 sin θ: sin 2θ = 2: √3.

#### Question 9. By means of an example, show that sin (A + B) ≠ sin A + sin B.

A = 30° and B = 60°

LHS = sin (A + B)

= sin (30° + 60°)

= sin 90°

= 1

RHS = sin A + sin B

= sin 30° + sin 60°

= ½ + √3/2

= (1 + √3)/2

Hence, LHS ≠ RHS i.e. sin (A + B) ≠ sin A + sin B.

#### Question 10.  If A = 60°and B = 30°, verify that

(i) sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos A cos B – sin A sin B

(iii) sin (A – B) = sin A cos B – cos A sin B

(iv) tan (A – B) = (tan A – tan B)/ (1 + tan A tan B).

A = 60° and B = 30°

##### (i) sin (A + B) = sin A cos B + cos A sin B

Here

LHS = sin (A + B)

= sin (60° + 30°)

= sin 90°

= 1

RHS = sin A cos B + cos A sin B

= sin 60° cos 30° + cos 60° sin 30°

= (√3/2 × √3/2) + (½ × ½)

= ¾ + ¼

= 4/4

= 1

Hence, LHS = RHS.

##### (ii) cos (A + B) = cos A cos B – sin A sin B

Here

LHS = cos (A + B)

= cos (60° + 30°)

= cos 90°

= 0

RHS = cos A cos B – sin A sin B

= cos 60° cos 30° – sin 60° sin 30°

= (½× √3/2) – √3/2 + ½

= √3/4 – √3/4

= 0

Hence, LHS = RHS.

##### (iii) sin (A – B) = sin A cos B – cos A sin B

Here

LHS = sin (A – B)

= sin (60° – 30°)

= sin 30°

= ½

RHS = sin A cos B – cos A sin B

= sin 60° cos 30° – cos 60° sin 30°

= (√3/2 × √3/2) – (½ × ½)

= ¾ – ¼

= (3 – 1)/ 4

= 2/4

= ½

Hence, LHS = RHS.

##### (iv) tan (A – B) = (tan A – tan B)/ (1 + tan A tan B)

Here

LHS = tan (A – B)

= tan (60° – 30°)

= tan 30°

= 1/√3

RHS = (tan A – tan B)/ (1 + tan A tan B)

= (tan 60° – tan 30°)/(1 + tan 60° tan 30°)

= 2/√3 × ½

= 1/√3

Hence, LHS = RHS.

(ML Aggarwal Trigonometrical Ratios of Standards Angles Exe-18.1 Class 9)

#### Question 11.

(i) If 2θ is an acute angle and 2 sin 2θ = √3, find the value of θ.

(ii) If 20° + x is an acute angle and cos (20° + x) = sin 60°, then find the value of x.

(iii) If 3 sin2 θ = 2 ¼ and θ is less than 90°, find the value of θ.

##### (i) 2θ is an acute angle

2 sin 2θ = √3

sin 2θ = √3/2 = sin 60°

2θ = 60°

θ = 60°/2 = 30°

Hence, θ = 30°

##### (ii) 20° + x is an acute angle

cos (20° + x) = sin 60°

cos (20° + x) = sin 60° = cos (90° – 60°)

= cos 30°

20° + x = 30°

⇒ x = 30° – 20° = 10°

Hence, x = 100.

##### (iii) 3 sin2 θ = 2 ¼

θ is less than 90°

sin2 θ = 9/(4 × 3) = ¾

sin θ = √3/2 = sin 60°

θ = 60°

Hence, θ = 60°.

#### Question 12.  If θ is an acute angle and sin θ = cos θ, find the value of θ and hence, find the value of 2 tan2 θ + sin2 θ – 1.

sin θ = cos θ

sin θ/cos θ = 1

tan θ = 1

tan 45° = 1

tan θ = tan 45°

θ = 45°

2 tan2 θ + sin2 θ – 1 = 2 tan2 45° + sin2 45° – 1

= 2 (1)2 + (1/√2)2 – 1

= (2×1×1) + ½ – 1

= 2 + ½ – 1

= 5/2 – 1

= (5 – 2)/2

= 3/2

Hence, 2 tan2 θ + sin2 θ – 1 = 3/2.

#### Question 13. From the adjoining figure, find

(i) tan x°

(ii) x

(iii) cos x°

(iv) use sin x° to find y.

= AB/BC

= √3/1

= √3

tan 60° = √3

tan x° = tan 60°

x = 60

cos x° = ½

##### (iv) sin x° = perpendicular/hypotenuse = AB/AC

Substitute x = 60 from (ii)

sin 60° = √3/y

sin 60° = √3/2

√3/2 = √3/y

y = (√3×2)/√3

y = (2×1)/1 = 2

Hence, y = 2.

#### Question 14.  If 3θ is an acute angle, solve the following equations for θ:

(i) 2 sin 3θ = √3

(ii) tan 3θ = 1.

sin 3θ = √3/2

sin 60° = √3/2

sin 3θ = sin 60°

⇒ 3θ = 60°

θ = 60/3 = 20°

tan 45° = 1

tan 3θ = tan 45°

3θ = 45°

⇒ θ = 15°

#### Question 15. If tan 3x = sin 45° cos 45° + sin 30°, find the value of x.

tan 3x = sin 45° cos 45° + sin 30°

= (1/√2 × 1/√2) + ½

= ½ + ½

= 1

tan 3x = tan 45°

3x = 45°

x = 45/3 = 15°

Hence, the value of x is 15°.

#### Question 16. If 4 cos2 x° – 1 = 0 and 0 ≤ x ≤ 90, find

(i) x

(ii) sin2 x° + cos2 x°

(iii) cos2 x° – sin2 x°

4 cos2 x° – 1 = 0

⇒ 4cos2 x° = 1

cos2 x° = ¼

⇒ cos x° = ± √1/4

⇒ cos x° = + √1/4 [0 ≤ x ≤ 90°, then cos x° is positive]

⇒ cos x° = ½

cos 60° = ½

cos x° = cos 60°

x = 60

##### (ii) sin2 x° + cos2 x° = sin2 60° + cos2 60°

= (√3/2)2 + (1/2)2

= ¾ + ¼

= (3 + 1)/4

= 4/4

= 1

Hence, sin2 x° + cos2 x° = 1.

##### (iii) cos2 x° – sin2 x° = cos2 60° – sin2 60°

= (1/2)2 – (√3/2)2

= ¼ – (√3/2 × √3/2)

= ¼ – ¾

= (1 – 3)/4

= -2/4

= – ½

Hence, cos2 x° – sin2 x° = – ½.

### Trigonometrical Ratios of Standards Angles Exercise-18.1

ML Aggarwal Class 9 ICSE Maths Solutions

Page 430

#### Question 17.

(i) If sec θ = cosec θ and 0° ≤ θ ≤ 90°, find the value of θ.

(ii) If tan θ = cot θ and 0° ≤ θ ≤ 90°, find the value of θ

##### (i) sec θ = cosec θ

sec θ = 1/cos θ

cosec θ = 1/sin θ

1/cos θ = 1/sin θ

⇒ sin θ/cos θ = 1

⇒ tan θ = 1

Here tan 45° = 1

tan θ = tan 45°

θ = 45°

##### (ii) tan θ = cot θ

cot θ = 1/tan θ

tan θ = 1/tan θ

tan2 θ = 1

⇒ tan θ = ± √1

⇒ tan θ = + 1 [0 ≤ θ ≤ 90°, tan θ is positive]

⇒ tan θ = tan 45°

θ = 45°

#### Question 18. If sin 3x = 1 and 0° ≤ 3x ≤ 90°, find the values of

(i) sin x

(ii) cos 2x

(iii) tan2 x – sec2 x.

sin 3x = 1

sin 90° = 1

sin 3x = sin 90°

3x = 90

⇒ x = 90/3

⇒ x = 30°

(i) sin x = sin 30° = 1/2

(ii) cos 2x = cos 2 × 30 = cos 60° = 1/2

(iii) tan2 x – sec2 x = tan2 30° – sec2 30°

= (1/√3)2 – (2/√3)2

= 1/3 – 4/3

= (1 – 4)/3

= – 3/3

= -1

Hence, tan2 x – sec2 x = – 1.

#### Question 19. If 3 tan2 θ – 1 = 0, find cos 2θ, given that θ is acute.

3 tan2 θ – 1 = 0

3 tan2 θ = 1

⇒ tan2 θ = 1/3

⇒ tan θ = 1/√3 [θ is acute so tan θ is positive]

θ = 30°

cos 2θ = cos 2 × 30°

= cos 60° = ½

#### Question 21.  If sin x + cos y = 1, x = 30° and y is acute angle, find the value of y.

sin x + cos y = 1

x = 30°

sin 30° + cos y = 1

⇒ 1/2 + cos y = 1

cos y = 1 – ½

cos y = (2 – 1)/2 = ½

cos 60° = ½

cos y = cos 60°

y = 60°

(ML Aggarwal Trigonometrical Ratios of Standards Angles Exe-18.1 Class 9)

#### Question 21. If sin (A + B) = √3/2 = cos (A – B), 0° < A + B ≤ 90° (A > B), find the values of A and B.

sin (A + B) = √3/2 = cos (A – B)

sin (A + B) = √3/2

sin 60 = √3/2

sin (A + B) = sin 60°

A + B = 60° …(1)

cos (A – B) = √3/2

cos 30° = √3/2

cos (A – B) = cos 30°

A – B = 30° …(2)

A + B + A – B = 60° + 30°

2A = 90°

⇒ A = 90°/2 = 45°

Now substitute the value of A in equation (1)

45° + B = 60°

B = 60° – 45° = 15°

Hence, A = 45° and B = 15°.

#### Question 22. If the length of each side of a rhombus is 8 cm and its one angle is 60°, then find the lengths of the diagonals of the rhombus.

Each side of a rhombus = 8 cm

One angle = 60°

∠OAB = 60°/2 = 30°

In right ∠AOB

sin 30° = OB/AB

½ = OB/8

OB = 8/2 = 4 cm

BD = 2OB = 2 × 4 = 8 cm

cos 30° = AO/AB

√3/2 = AO/8

AO = 8√3/2 = 4√3

AC = 4√3 × 2 = 8 √3 cm

Hence,

The length of the diagonals of the rhombus are 8 cm and 8√3 cm.

#### Question 23. In the right-angled triangle ABC, ∠C = 90° and ∠B = 60°. If AC = 6 cm, find the lengths of the sides BC and AB.

In the right-angled triangle ABC, C = 90° and B = 60°

AC = 6 cm

tan B = AC/BC

tan 60° = 6/BC

So we get

√3 = 6/BC

⇒ BC = 6/√3

= 6√3/(√3+√3)

= 6√3/3

= 2√3 cm

sin 60° = AC/AB

√3/2 = 6/AB

AB = (6×2)/√3

AB = (12×√3)/(√3×√3)

= 12√3/3

= 4√3 cm

Hence,

the lengths of the sides BC = 2√3 cm and AB = 4√3 cm.

#### Question 24. In the adjoining figure, AP is a man of height 1.8 m and BQ is a building 13.8 m high. If the man sees the top of the building by focusing his binoculars at an angle of 30° to the horizontal, find the distance of the man from the building.

Height of man AP = 1.8 m

Height of building BQ = 13.8 m

Angle of elevation from the top of the building to the man = 30°

AB = x then PC = x

QC = 13.8 – 1.8 = 12 m

In right △ PQC

tan θ = QC/PC

tan 30° = 12/x

1/√3 = 12/x

x = 12 √3 m

Hence,

the distance of the man from the building is 12 √3 m.

(ML Aggarwal Trigonometrical Ratios of Standards Angles Exe-18.1 Class 9)

#### Question 25. In the adjoining figure, ABC is a triangle in which ∠B = 45° and ∠C = 60°. If AD ⟂ BC and BC = 8 m, find the length of the altitude AD.

ABC

B = 45° and C = 60°

AD ⟂ BC and BC = 8 m

In right △ ABD

In right △ ACD

⇒ 8 = [AD (√3 + 1)]/ √3