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 Solutions Ch-18. Step by Step Solutions of MCQs questions on Trigonometrical Ratios 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 MCQs Class 9 ICSE Maths Solutions

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

MCQs Questions on Trigonometrical Ratios of Standards Angles

ML Aggarwal Class 9 ICSE Maths Solutions Ch-18

Question 1. the value of tan 30/cot60 is

(a) 1/√2

(b) 1/√3

(c) √3

(d) 1

Answer :

ML aggarwal class-10 Trigonimetrical Ratio of standerd angle chapter 18 img 38

= 1

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

(a) 1/√2

(b) √2

(c) √3/2

(d) 1

Answer :

sin 45° + cos 45° = 1/√2 + 1/√2

⇒ sin 45° + cos 45° = 2/√2 = √2

Thus, sin 45° + cos 45° = √2.

Question 3. the value of tan² 30 – 4 sin² 45 is

(a) 1

(b) 7/3

(c) -5/3

(d) -11/3

Answer:

tan ²30° – 4 sin²45°

(1/√3)² – 4 × (1/√2)²

1/3 – 4 × 1/2

= -5/3

Question 4. If A = 30, then the value of 2 sin A Cos A is

(a) 1/√2

(b) √3/2

(c) 1/2

(d) 1

Answer :

2×sin30×cos30

2×1/2×√3/2

=√3/2

Question 5. The value of (sin 30 + cos 30) – (sin 60 + cos 60) is

(a) -1

(b) 0

(c) 1

(d) 2

Answer :

sin 30 + cos 30) – (sin 60 + cos 60)

ML aggarwal class-10 Trigonimetrical Ratio of standerd angle chapter 18 img 39

Question 6. The value of √3 cosec 60 – sec 60 is

(a) 0

(b) 1

(c) 2

(d) -1

Answer :

√3 x 2/√3 – 2

= 0

Question 7. The value of 1/sin30 – √3/ cos 30 is

(a) 2

(b) 1

(c) 1/2

(d) 0

Answer :

sin30°=1/2

cos30°= √3/2

1/sin30° = 1/1/2 = 2 …(eq1)

√3/cos30° = √3/√3/2 = 2 …(eq2)

subtracting eq2 from eq1, we get,

2–2=0

Question 8. If tan A = √3, then the value of cosec A is

(a) 1/2

(b) 2

(c) 1/√2

(d) √3/2

Answer :

tan A=√3 ——-1

But, tan 60°=√3 ———-2

From 1 and 2, A=60°

cosec A=cosec 60°

=2/√3

Question 9. If sec θ.sin θ = 0, then the value of cos θ is 

(a) 0

(b) 1/√2

(c) 1/2

(d) 1

Answer :

Sec A × sin A = 0

( 1/cos A ) × sin A = 0

( Sin A/Cos A ) = 0

tan A = 0

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

Question 10. If sin alpha = 1/2 then the value of 3 cos alpha – 4 cos³ alpha is

Answer :

since sin alpha is = 1/2

sin 30 = 1/2

so,

alpha = 30

now,

3 cos30 – 4 cos³ 30

=> 3(√3)/2 – 4(√3/2)³

=> 3 x √3/2 – 4(3 x √3)/8

=> 12 x √3/8 – 12 x √3/8                       [taking common denominator]

=> 0

Question 11. The value of 1-tan² 45/ 1+ tan² 45 is equal to

(a) tan 60

(b) tan 30

(c) sin 45

(d) tan 0

Answer :

1-tan² 45/ 1+ tan² 45

1-1/1+1

0/2

1-tan² 45/ 1+ tan² 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

Answer :

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

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) √3/2

Answer :

since ABC is right angled and angle C is 90°

therefore,

A+B=180° – C

A+B=180°-90°

A+B= 90°

Hence, cos (A+B)=cos90°

=0

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

(a) 10/√3 cm

(b) 10√3 cm

(c) 20 cm

(d) 5 cm

Answer :

In the given 
Sum of all angles is 

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

(a) 60

(b) 45

(c) 30

(d) 15

Answer :

cos P = PQ/PR

cos P = 4/8

= 1/2

cos P = 1/2

cos P = cos 60

so, P = 60

—  : End of ML Aggarwal Trigonometrical Ratios of Standards Angles MCQs Class 9 ICSE Maths Solutions :–

Return to :-  ML Aggarawal Maths Solutions for ICSE  Class-9

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