Friday, March 24, 2023

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

ML Aggarwal Trigonometrical Ratios of Standards Angles MCQs Class 9 ICSE Maths APC Understanding Solutions. Solutions ofĀ  MCQs . This post is the Solutions ofĀ Ā ML AggarwalĀ Chapter 18 – Trigonometrical Ratios of Standards Angles forĀ ICSEĀ MathsĀ Class-9.Ā  APC UnderstandingĀ ML AggarwalĀ Solutions (APC) Avichal Publication Solutions of Chapter-18 Trigonometrical Ratios of Standards Angles forĀ ICSEĀ BoardĀ Class-9.Ā Visit official websiteĀ CISCEĀ for detail information about ICSE Board Class-9.

## ML Aggarwal Trigonometrical Ratios MCQs Class 9 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 9th Chapter-18 Trigonometrical Ratios of Standards Angles Writer ML Aggarwal Book Name Understanding Topics Solution of MCQs Edition 2021-2022

### MCQs Solutions of ML Aggarwal for ICSE Class-9 Ch-18, Trigonometrical Ratios of Standards Angles

Note:- Before viewing Solutions of Chapter -18 Trigonometrical Ratios of Standards Angles Class-9 of ML AggarwalĀ Solutions . Ā Read the Chapter Carefully. Then solve all example given in Exercise-18.1, Exercise-18.2, MCQs, Chapter Test.

### Trigonometrical Ratios of Standards Angles MCQs

ML Aggarwal Class 9 ICSE Maths Solutions

Page 437

(a) 1/ā2

(b) 1/ā3

(c)Ā ā3

(d) 1

= 1

#### Question 2. the value of (sin 45 + cos 45 ) is

(a)Ā 1/ā2

(b)Ā ā2

(c)Ā ā3/2

(d) 1

sin 45Ā° + cos 45Ā° = 1/ā2 + 1/ā2

ā sin 45Ā° + cos 45Ā° = 2/ā2 = ā2

Thus, sin 45Ā° + cos 45Ā° = ā2.

#### Question 3. the value of tanĀ² 30 ā 4 sinĀ² 45 is

(a) 1

(b) 7/3

(c) -5/3

(d) -11/3

tan Ā²30Ā° ā 4 sinĀ²45Ā°

(1/ā3)Ā² ā 4 Ć (1/ā2)Ā²

1/3 ā 4 Ć 1/2

= -5/3

Trigonometrical Ratios of Standards Angles MCQs

### ML Aggarwal Class 9 ICSE Maths Solutions

Page 438

(a)Ā 1/ā2

(b)Ā ā3/2

(c) 1/2

(d) 1

2Ćsin30Ćcos30

2Ć1/2Ćā3/2

=ā3/2

#### Question 5. The value of (sin 30 + cos 30) ā (sin 60 + cos 60) is

(a) -1

(b) 0

(c) 1

(d) 2

sin 30 + cos 30) ā (sin 60 + cos 60)

#### Question 6. The value ofĀ ā3 cosec 60 ā sec 60 is

(a) 0

(b) 1

(c) 2

(d) -1

ā3 x 2/ā3 ā 2

= 0

#### Question 7. The value of 1/sin30 āĀ ā3/ cos 30 is

(a) 2

(b) 1

(c) 1/2

(d) 0

sin30Ā°=1/2

cos30Ā°= ā3/2

1/sin30Ā° = 1/1/2 = 2 ā¦(eq1)

ā3/cos30Ā° = ā3/ā3/2 = 2 ā¦(eq2)

subtracting eq2 from eq1, we get,

2ā2=0

#### Question 8. If tan A =Ā ā3, then the value of cosec A is

(a) 1/2

(b) 2

(c) 1/ā2

(d)Ā ā3/2

tan A=ā3Ā āā-1

But, tan 60Ā°=ā3Ā āāā-2

FromĀ 1Ā andĀ 2, A=60Ā°

cosec A=cosec 60Ā°

=2/ā3

#### Question 9. If secĀ Īø.sinĀ Īø = 0, then the value of cosĀ Īø isĀ

(a) 0

(b) 1/ā2

(c) 1/2

(d) 1

Sec A Ć sin A = 0

( 1/cos A ) Ć sin A = 0

( Sin A/Cos A ) = 0

tan A = 0

tan A = tan 0Ā°
Therefore ,
A = 0Ā°
Now ,
Cos A = Cos 0Ā° = 1

#### Question 10. If sin alpha = 1/2 then the value of 3 cos alpha ā 4 cosĀ³ alpha is

since sin alpha is = 1/2

sin 30 = 1/2

so,

alpha = 30

now,

3 cos30 ā 4 cosĀ³ 30

=> 3(ā3)/2 ā 4(ā3/2)Ā³

=> 3 xĀ ā3/2 ā 4(3 xĀ ā3)/8

=> 12 xĀ ā3/8 ā 12 xĀ ā3/8 Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  [taking common denominator]

=> 0

#### Question 11. The value of 1-tanĀ² 45/ 1+ tanĀ² 45 is equal to

(a) tan 60

(b) tan 30

(c) sin 45

(d) tan 0

1-tanĀ² 45/ 1+ tanĀ² 45

1-1/1+1

0/2

1-tanĀ² 45/ 1+ tanĀ² 45 = 0

tan 0

#### Question 12. if sin alpha = 1/2 and cos beta = 1/2, then the value of (alpha + beta) is

(a) 0

(b) 30

(c) 60

(d) 90

Sin alpha = 1/2
Sin alpha = sin 30Ā°
Alpha = 30Ā°
Cos alpha = 1/2
Cos beta = cos 60Ā°
Beta = 60Ā°
Now, alpha + beta
= 30Ā°+60Ā°
= 90Ā°

#### Question 13. If triangle ABC is right angled at C, then the value of cos (A + B) is

(a) 0

(b) 1

(c) 1/2

(d) ā3/2

since ABC is right angled and angle C is 90Ā°

therefore,

A+B=180Ā° ā C

A+B=180Ā°-90Ā°

A+B= 90Ā°

Hence, cos (A+B)=cos90Ā°

=0

#### Question 14. In the adjoining figure, ABC is a right triangle rightĀ angled at B. If AB = 10 cm and angle C = 30Ā°, then theĀ length of the side BC is

(a) 10/ā3 cm

(b) 10ā3 cm

(c) 20 cm

(d) 5 cm

In the givenĀ
Sum of all angles isĀ

#### Question 15. In the adjoining figure, PQR is a right triangle rightĀ angled at Q. If PQ = 4 cm and PR = 8 cm then angle P isĀ equal to

(a) 60

(b) 45

(c) 30

(d) 15

cos P = PQ/PR

cos P = 4/8

= 1/2

cos P = 1/2

cos P = cos 60

so, P = 60

āĀ  : End of ML Aggarwal Trigonometrical Ratios of Standards Angles MCQs Class 9 ICSE Maths Solutions :ā

Thanks

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