ML Aggarwal Trigonometrical Ratios of Standards Angles MCQs Class 9 ICSE Maths APC Understanding Solutions. Solutions of MCQs . This post is the Solutions of ML Aggarwal Chapter 18 – Trigonometrical Ratios of Standards Angles for ICSE Maths Class-9. APC Understanding ML Aggarwal Solutions (APC) Avichal Publication Solutions of Chapter-18 Trigonometrical Ratios of Standards Angles for ICSE Board Class-9. Visit official website CISCE 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 Aggarwal Solutions . 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
Question 1. the value of tan 30/cot60 is
(a) 1/√2
(b) 1/√3
(c) √3
(d) 1
Answer :
= 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
Trigonometrical Ratios of Standards Angles MCQs
ML Aggarwal Class 9 ICSE Maths Solutions
Page 438
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)
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 :
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
Thanks
Please Share with Your Friends
