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

ML Aggarwal Trigonometrical Ratios of Standards Angles Chapter Test Class 9 ICSE Maths Solutions Ch-18. Step by Step Solutions of Ch-Test 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 Chapter Test Class 9 ICSE Maths Solutions

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

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

ML Aggarwal Class 9 ICSE Maths Solutions Ch-18

#### Question 1 Find the values of:

(i) sin2 60° – cos2 45° + 3tan2 30°
(ii)……………..
(iii) sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60°

##### (i) sin2 60° – cos2 45° + 3tan2 30°

Therefore, sin2 60° – cos2 45° + 3tan2 30° = 1¼

##### (iii) sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60°

= 2 + 1 + ½ = 3 + ½ = (6 + 1)/2

= 7/2 = 3½

Thus,

sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60° = 3½

#### Question 2. Taking A = 30°, verify that

(i) cos4 A – sin4 A = cos 2A
(ii) 4cos A cos (60° – A) cos (60° + A) = cos 3 A.

##### (i) cos4 A – sin4 A = cos 2A

Let’s take A = 30°

so,

we have

L.H.S.= cos4 A – sin4 A = cos4 30° – sin4 30°

R.H.S. = cos 2A = cos 2(30o) = ½
Therefore, L.H.S. = R.H.S. hence verified.

##### (ii) 4 cos A cos (60°- A) cos (60° + A) = cos 3 A

Let’s take A = 30°

L.H.S. = 4 cos A cos (60° – A) cos (60° + A)

= 4 cos 30° cos (60° – 30°) cos (60° + 30°)

= 4 cos 30° cos 30° cos 90°

= 4 × (√3/2) × (√3/2) × 0

= 0

R.H.S. = cos 3A

= cos (3 × 30°) = cos 90° = 0

Therefore, L.H.S. = R.H.S.

Hence proofed

Hence proofed

#### Question 4. Taking A = 60° and B = 30°, verify that

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

(ii) sin (A – B)/sin A sin B = cot B – cot A

#### Question 5. If √2 tan 2θ = √6 and θ° < 2θ < 90°, find the value of sin θ + √3 cos θ – 2 tan2 θ.

√2 tan 2θ = √6

⇒ tan 2θ = √6/ √2

= √3

= tan 60o

⇒ 2θ = 60o

⇒ θ = 30o

sin θ + √3 cos θ – 2 tan2 θ

= sin 30o + √3 cos 30o – 2 tan2 30o

= ½ + √3 x √3/2 – 2 (1/√3)2

= ½ + 3/2 – 2/3

= 4/2 – 2/3

= (12 – 4)/6

= 8/6

= 4/3

#### Question 6. If 3θ is an acute angle, solve the following equations for θ:

(i) (cosec 3θ – 2) (cot 2θ – 1) = 0

(ii) (tan θ – 1) (cosec 3θ – 1) = 0

##### (i) (cosec 3θ – 2) (cot 2θ – 1) = 0

cosec 3θ – 2 or cot 2θ – 1 = 0
⇒ cosec 3θ = 2 or cot 2θ = 1

So,

cosec 3θ = cosec 3θ° or cot 2θ =cot 45°

⇒ 3θ = 30° or 2θ = 45°

Thus,

θ = 30° or 45°.

##### (ii) (tan θ – 1) (cosec 3θ – 1) = 0

tan θ – 1 = 0 or cosec 3θ – 1 = 0

⇒ tan θ = 1 or cosec 3θ = 1

So,

tan θ = tan45° or cosec 3 θ = cosec 90°
⇒ θ = 45° or 3θ = 90° i.e. θ = 30°
Thus,

θ = 45° or 30°.

#### Question 7. If tan (A + B) = √3and tan (A – B) = 1 and A, B (B < A) are acute angles, find the values of A and B.

tan (A + B) = √3

So, tan (A + B) = tan 60° [Since, tan 60° = √3]

⇒ A + B = 60° …(i)

tan (A – B) = 1

tan (A – B) = tan 45° [tan 45° = 1]

⇒ A – B = 45° …(ii)

From equation (1) and (2), we get

A + B + A – B = 60° + 45°

⇒ 2A = 105o

⇒ A = 52½o

on substituting the value of A in equation (i), we get

52½o + B = 60°

⇒ B = 60° – 52½o = 7½o

Therefore, the value of A = 52½and B = 7½o

#### Question 8. Without using trigonometrical tables, evaluate the following:

(i) sin2 28° + sin2 62° – tan2 45°
(ii)…………………………

(iii) cos 18° sin 72° + sin 18° cos 72°

(iv) 5 sin 50° sec 40° – 3 cos 59° cosec 31°

##### (i) sin2 28° + sin2 62° – tan2 45°

= sin2 28° + sin2 (90° – 28°) – tan2 45°

= sin2 28° + cos2 28° – tan2 45°

= 1 – (1)2 (∵ sin2 θ + cos2 θ = 1 and tan 45° = 1)

= 1 – 1

= 0

= (2×1) + 1 + 1

= 2 + 1 + 1

= 4

##### (iii) cos 18° sin 12° + sin 18° cos 12°

= cos (90° – 12°) sin 72° + sin (90° – 12°) cos 12°

= sin 72°.sin 12° + cos 12° cos 12°

= sin2 12° + cos2 12°

= 1 (∵ sin2 θ + cos2 θ = 1)

= 5 – 3

= 2

……………………..

L.H.S. = R.H.S.

Hence proved.

#### Question 10. When 0° < A < 90°, solve the following equations:

(i) sin 3A = cos 2A
(ii) tan 5A = cot A

##### (i) sin 3A = cos 2A

⇒ sin 3A = sin (90° – 2A)

3A = 90° – 2A

⇒ 3 A + 2A = 90°

⇒ 5A = 90°

∴ A = 90°/5 = 18°

⇒  tan 5A = cot A

⇒ tan 5A = tan (90° – A)

5A = 90°- A

⇒ 5A + A = 90°

⇒ 6A = 90°

∴ A = 90°/6 = 15°

#### Question 11. Find the value of θ if

(i) sin (θ + 36°) = cos θ, where θ and θ + 36° are acute angles.
(ii) sec 4θ = cosec (θ – 20°), where 4θ and θ – 20° are acute angles.

##### (i) Given, θ and (θ + 36°) are acute angles

sin (θ + 36°) = cos θ = sin (90° – θ) [As, sin (90° – θ) = cos θ]

θ + 36° = 90° – θ

⇒ θ + θ = 90° – 36°

⇒ 2θ = 54°

⇒ θ = 54°/2

∴ θ = 27°

##### (ii) Given, θ and (θ – 20°) are acute angles

And,

sec 4θ = cosec (θ – 20°)

⇒ cosec (90° – 4θ) = cosec (θ – 20°) [Since, cosec (90° – θ) = sec θ]

On comparing, we get

90° – 4θ = θ – 20°

⇒ 90° + 20° = θ + 4θ

⇒ 5θ = 110°

⇒ θ = 110°/5

∴ θ = 22°

#### Question 12. In the adjoining figure, ABC is right-angled triangle at B and ABD is right angled triangle at A. If BD ⊥ AC and BC = 2√3cm, find the length of AD.

∆ABC and ∆ABD are right angled triangles in which ∠A = 90° and ∠B = 90°

BC = 2√3 cm. AC and BD intersect each other at E at right angle and ∠CAB = 30°.

Now in right ∆ABC, we have

tan θ = BC/AB

⇒ tan 30° = 2√3/ AB

⇒ 1/√3 = 2√3/ AB

⇒ AB = 2√3 × √3 = 2 × 3 = 6 cm.

In ∆ABE, ∠EAB = 30° and ∠EAB = 90°

∠ABE or ∠ABD = 180° – 90° – 30°

= 60°
Now in right ∆ABD, we have