Class 10 Maths Ex 8.3 Assamese Medium focuses on applying trigonometric identities in different types of problems. By solving অনুশীলনী 8.3 Class 10, students can build a strong conceptual base in trigonometry and prepare effectively for the SEBA Class 10 final exams. These solutions are explained in simple Assamese for better understanding.
Class 10 Maths Ex 8.3 Assamese Medium Solution
Here we provide Class 10 Maths Exercise 8.3 Assamese Medium solutions with detailed steps. Each question is solved in Assamese so that SEBA students can learn easily and improve their problem-solving skills.
| SEBA Class 10 Maths Chapter 8: Exercises | Total Number of Questions |
|---|---|
| Class 10 Maths Ex 8.1 Assamese Medium | 11 |
| Class 10 Maths Ex 8.2 Assamese Medium | 4 |
| Class 10 Maths Ex 8.3 Assamese Medium | 7 |
| Class 10 Maths Ex 8.4 Assamese Medium | 5 |
অনুশীলনী 8.3 Class 10 Solution
1. মান নিৰ্ণয় কৰা
(i) \frac{\sin 18^\circ}{\cos 72^\circ}
সমাধান :
= \frac{\sin 18^\circ}{\cos (90^\circ - 18^\circ)}
= \frac{\sin 18^\circ}{\sin 18^\circ}
= 1
(ii) \frac{\tan 26^\circ}{\cot 64^\circ}
সমাধান :
= \frac{\tan (90^\circ - 64^\circ)}{\cot 64^\circ}
= \frac{\cot 64^\circ}{\cot 64^\circ}
= 1
(iii) \cos 48^\circ - \sin 42^\circ
সমাধান :
= \cos (90^\circ - 42^\circ) - \sin 42^\circ
= \sin 42^\circ - \sin 42^\circ
= 0
(iv) \csc 31^\circ - \sec 59^\circ
সমাধান :
= \csc (90^\circ - 59^\circ) - \sec 59^\circ
= \sec 59^\circ - \sec 59^\circ
= 0
2. দেখুওৱা যে
(i) \tan 48^\circ \, \tan 23^\circ \, \tan 42^\circ \, \tan 67^\circ = 1
(ii) \cos 38^\circ \cos 52^\circ - \sin 38^\circ \sin 52^\circ = 0
(i) সমাধান :
LHS = \tan 48^\circ \, \tan 23^\circ \, \tan 42^\circ \, \tan 67^\
= \tan 48^\circ \times \tan 23^\circ \times \tan (90^\circ - 48^\circ) \times \tan (90^\circ - 23^\circ)
= \tan 48^\circ \times \tan 23^\circ \times \cot 48^\circ \times \cot 23^\circ
= \tan 48^\circ \times \tan 23^\circ \times \frac{1}{\tan 48^\circ} \times \frac{1}{\tan 23^\circ}
= 1
∴ RHS = LHS
(ii) সমাধান :
LHS = \cos 38^\circ \cos 52^\circ - \sin 38^\circ \sin 52^\circ
= \cos (90^\circ - 52^\circ) \cos 52^\circ - \sin (90^\circ - 52^\circ) \sin 52^\circ
= \sin 52^\circ \cos 52^\circ - \cos 52^\circ \sin 52^\circ
= 0
∴ RHS = LHS
3. যদি \tan 2A = \cot (A-18^\circ) , য’ত 2A সূক্ষ্মকোণ, তেন্তে A ৰ মান উলিওৱা।
সমাধান :
\tan 2A = \cot (A - 18^\circ)⇒ \cot (90^\circ - 2A) = \cot (A - 18^\circ)
⇒ 90^\circ - 2A = A - 18^\circ
⇒ 108^\circ = 3A
⇒ A = \frac {108^\circ}{3}
⇒ A = 36^\circ
4. যদি \tan A = \cot B , প্ৰমাণ কৰা যে A + B = 90^\circ .
সমাধান : \tan A = \cot B
⇒ \tan A = \tan (90^\circ - B)
\therefore A = 90^\circ - B⇒ A + B = 90^\circ
5. যদি \sec 4A = \cosec (A - 20^\circ) , য’ত 4A সূক্ষ্মকোণ, তেন্তে A ৰ মান উলিওৱা।
সমাধান : \sec 4A = \csc (A - 20^\circ)
⇒ \cosec (90^\circ - 4A) = \csc (A - 20^\circ)
⇒ { \cosec (90^\circ - \theta) = \sec \theta }
⇒ 90^\circ - 4A = A - 20^\circ
⇒ 5A = 110^\circ
⇒ A = 22^\circ
6. যদি A, B আৰু C কোণকেইটা ABC ত্ৰিভূজৰ অন্তঃকোণ হয়, তেন্তে দেখুওৱা যে
\sin \left(\frac{B + C}{2}\right) = \cos \left(\frac{A}{2}\right)সমাধান :
আমি জানো যে, A + B + C = 180^\circ
⇒ B + C = 180^\circ - A
⇒ \frac{B + C}{2} = \frac{180^\circ - A}{2}
⇒ \frac{B + C}{2} = 90^\circ - \frac{A}{2}
⇒ \sin \left(\frac{B + C}{2}\right) = \sin \left(90^\circ - \frac{A}{2}\right)
⇒ \sin \left(\frac{B + C}{2}\right) = \cos \left(\frac{A}{2}\right)
7. \sin 67^\circ + \cos 75^\circ ক 0^\circ আৰু 45^\circ ৰ মাজৰ কোণৰ ত্ৰিকোণমিতিক অনুপাত হিচাপে প্ৰকাশ কৰা।
সমাধান : \sin 67^\circ + \cos 75^\circ
= \sin (90^\circ - 23^\circ) + \cos (90^\circ - 15^\circ)
= \cos 23^\circ + \sin 15^\circ
Q1. What does Class 10 Maths Exercise 8.3 Assamese Medium cover?
Assamese Medium Class 10 Maths Ex 8.3 mainly focuses on solving problems using trigonometric identities.
Q2. Where can I get Class 10 Maths Ex 8.3 Assamese Medium solution?
You can find complete Class 10 Maths Exercise 8.3 Assamese Medium answers here in step-by-step format.
Q3. How is অনুশীলনী 8.3 Class 10 useful?
It strengthens the understanding of trigonometric relations, which is important for both exams and higher studies.
Q4. Is solving Exercise 8.3 enough for exams?
Along with Ex 8.3 solutions in Assamese, students should also revise all formulas and practice other exercises for the best results.
Benefits of Class 10 Maths Ex 8.3 Assamese Medium Answer
Builds Strong Conceptual Clarity
Practicing Class 10 Maths Exercise 8.3 Assamese Medium helps students clearly understand the application of trigonometric identities and their real-world uses.
Improves Step-by-Step Problem-Solving
With detailed Class 10 Maths Ex 8.3 Assamese Medium solutions, students learn the proper step-by-step approach, which boosts accuracy in exams.
Enhances Exam Preparation
Solving অনুশীলনী 8.3 Class 10 in Assamese gives students confidence and makes them well-prepared for SEBA Class 10 board exams.
Conclusion
The Class 10 Maths Ex 8.3 Assamese Medium solution provides students with clear, step-by-step answers to all problems from অনুশীলনী 8.3. By practicing these solutions, learners can easily understand the use of trigonometric identities and improve their problem-solving skills. Regular practice not only builds a strong foundation in trigonometry but also boosts confidence for the SEBA Class 10 final exams.