Class 10 Maths Ex 8.1 Assamese Medium is the first exercise of Chapter 8 – Introduction to Trigonometry. This section is particularly important for SEBA students, as it lays the foundation for understanding trigonometric ratios. In this post, we have provided the Class 10 Maths Exercise 8.1 Assamese Medium solutions with step-by-step answers so that students can easily understand each problem.

Whether you are searching for অনুশীলনী 8.1 Class 10, Class 10 Maths Ex 8.1 Assamese Medium answer, or Class 10 Maths Ex 8.1 in Assamese, here you will find accurate and well-explained solutions that will strengthen your exam preparation.

Class 10 Maths Ex 8.1 Assamese Medium Solution

Here we provide Class 10 Maths Exercise 8.1 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 8.1 Class 10 Solution

1. \triangle ABC ত্রিভুজৰ B কোণ সমকোণ আৰু AB = 24cm, BC = 7cm হ’লে তলত দিয়াবিলাক উলিওৱাঃ

(i) sinA, cosA
(ii) sinC, cosC

সমাধান :

দিয়া আছে : AB = 24cm, BC = 7cm, \angle B = 90^\circ

∵ AC^2 = AB^2 + BC^2 (পাইথাগোৰাছ সূত্ৰ)
⇒ AC^2 = (24)^2 + (7)^2
⇒ AC^2 = 576 + 49 = 625
⇒ AC = \sqrt{625} = 25cm

(i)
\sin A = \frac{BC}{AC} = \frac{7}{25}
\cos A = \frac{AB}{AC} = \frac{24}{25}

(ii)
\sin C = \frac{AB}{AC} = \frac{24}{25}
\cos C = \frac{BC}{AC} = \frac{7}{25}

2. চিত্ৰ 8.13 ৰ পৰা tanP – cotR নিৰ্ণয় কৰা।

সমাধান :

∵ PR = 13cm, PQ = 12cm

PR^2 = PQ^2 + QR^2
⇒ (13)^2 = (12)^2 + (QR)^2
⇒ 169 = 144 + QR^2
⇒ QR^2 = 169 - 144 = 25
⇒ QR = \sqrt{25} = 5cm

\tan P = \frac{QR}{PQ} = \frac{5}{12}
\cot R = \frac{PQ}{QR} = \frac{12}{5}

∴ \tan P - \cot R = \frac{5}{12} - \frac{12}{5} = 0

3. যদি sinA = \frac{3}{4} , তেন্তে cosA আৰু tanA উলিওৱা।

সমাধান :

∵ ABC এটা সমকোণী ত্ৰিভুজ \angle B = 90^\circ

\sin A = \frac{3}{4} = \frac{BC}{AC}=k , ধৰা হল (সমানুপাতৰ ভিত্তিত)

∴ BC = 3k, AC = 4k

∵ AC^2 = AB^2 + BC^2 (পাইথাগোৰাছ সূত্ৰ)
⇒ (4k)^2 = AB^2 + (3k)^2
⇒ 16k^2 = AB^2 + 9k^2
⇒ AB^2 = 16k^2 - 9k^2 = 7k^2
⇒ AB = \sqrt{7}k

∴ \cos A = \frac{AB}{AC} = \frac{\sqrt{7}k}{4k} = \frac{\sqrt{7}}{4}
∴ \tan A = \frac{BC}{AB} = \frac{3k}{\sqrt{7}k} = \frac{3}{\sqrt{7}}

4. দিয়া আছে যে, 15 cotA = 8, সেই অনুসৰি sinA আৰু secA উলিওৱা।

সমাধান :

∵ ABC এটা সমকোণী ত্ৰিভুজ \angle B = 90^\circ

15 \cot A = 8
⇒ \cot A = \frac{8}{15} = \frac{AB}{BC}=k , ধৰা হল (সমানুপাতৰ ভিত্তিত)

AB = 8k, BC = 15k

∵ AC^2 = AB^2 + BC^2 (পাইথাগোৰাছ সূত্ৰ)
⇒ AC^2 = (8k)^2 + (15k)^2
⇒ AC^2 = 64k^2 + 225k^2
⇒ AC^2 = 289k^2
⇒ AC = \sqrt{289k^2} = 17k

∴ \cos A = \frac{AB}{AC} = \frac{8k}{17k} = \frac{8}{17}
\sec A = \frac{AC}{AB} = \frac{17k}{8k} = \frac{17}{8}

5. দিয়া আছে যে, secθ = \frac{13}{12}, আন ত্ৰিকোণমিতিক অনুপাতসমূহৰ গণনা কৰা।

সমাধান :

∵ ABC এটা সমকোণী ত্ৰিভুজ, \angle B = 90^\circ, \angle ABC = \theta (ধরা হৈছে)

⇒ \sec \theta = \frac{13}{12} = \frac{AC}{AB} = k, ধৰা হল (সমানুপাতৰ ভিত্তিত)
AC = 13k, \quad AB = 12k

∵ AC^2 = AB^2 + BC^2 (পাইথাগোৰাছ সূত্ৰ)
⇒ (13k)^2 = (12k)^2 + BC^2
⇒ 169k^2 = 144k^2 + BC^2
⇒ BC^2 = 169k^2 - 144k^2
⇒ BC^2 = 25k^2
⇒ BC = \sqrt{25k^2} = 5k

∴ \sin \theta = \frac{BC}{AC} = \frac{5k}{13k} = \frac{5}{13}
\cos \theta = \frac{AB}{AC} = \frac{12k}{13k} = \frac{12}{13}
\tan \theta = \frac{BC}{AB} = \frac{5k}{12k} = \frac{5}{12}
\csc \theta = \frac{AC}{BC} = \frac{13k}{5k} = \frac{13}{5}
\cot \theta = \frac{AB}{BC} = \frac{12k}{5k} = \frac{12}{5}

6. যদি \angle A আৰু \angle B সুক্ষ্মকোণ হয় যাতে \cos A = \cos B , তেতিয়া দেখুৱাব যে \angle A = \angle B

সমাধান :

ধৰা হল, \triangle ABC ত \angle C = 90^\circ

এতিয়া, \cos A = \dfrac{AB}{AC}

আৰু \cos B = \dfrac{BC}{AC}

প্ৰশ্নমতে \cos A = \cos B

⇒ \dfrac{AB}{AC} = \dfrac{BC}{AC}

⇒ AB = AC

∴ ABC এটা সমদ্বিবাহু ত্ৰিভুজ।
∵ আমি জানো যে, ত্ৰিভুজৰ সমান বাহু দুটাৰ বিপৰীত কোণ দুটা সমান।

∴ প্ৰমাণিত হ’ল যে \angle A = \angle B

7. যদি cot\theta=\frac{7}{8} তলৰ মান উলিওৱা –

(i) \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}
(ii) \cot^2\theta

সমাধান :

(i) ধৰা হল, \triangle ABC এটা সমকোণী ত্ৰিভুজ, \angle C = 90^\circ.

\cot \theta = \frac{7}{8}=\frac{BC}{AC}=k ধৰা হল (সমানুপাতৰ ভিত্তিত)
⇒ BC = 7k,\ AC = 8k
⇒ AB^2 = (8k)^2 + (7k)^2
⇒ AB^2= 64k^2 + 49k^2
⇒ AB^2= 113k^2
⇒ AB = \sqrt{113}k

∴ \sin \theta = \frac{AC}{AB} = \frac{8k}{\sqrt{113}k} = \frac{8}{\sqrt{113}}
\cos \theta = \frac{BC}{AB} = \frac{7k}{\sqrt{113}k} = \frac{7}{\sqrt{113}}

∵ \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)} = \frac{1-\sin^2\theta}{1-\cos^2\theta}

= \frac{1-(\frac{8}{\sqrt{113}})^2}{1-(\frac{7}{\sqrt{113}})^2}

= \frac{1-\frac{64}{113}}{1-\frac{49}{113}}

= \frac{\frac{49}{113}}{\frac{64}{113}}

= \frac{49}{64}

(ii) দিয়া আছে, \cot \theta = \frac{7}{8}
⇒ \cot^2\theta = (\frac{7}{8})^2
⇒ \cot^2\theta= \frac{49}{64}

8. যদি 3 \cot A = 4 , তেন্তে \frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A হবনে নহয় পৰীক্ষা কৰা।

সমাধান :
3 \cot A = 4
⇒ \cot A = \frac{4}{3}
⇒ \frac{AB}{BC} = \frac{4}{3}=k , ধৰা হল (সমানুপাতৰ ভিত্তিত)
∴ AB = 4k \quad আৰু \quad BC = 3k

এতিয়া,
AC = \sqrt{(AB)^2 + (BC)^2}
⇒ AC = \sqrt{(4k)^2 + (3k)^2}
⇒ AC = 5k

∵ \sin A = \frac{BC}{AC}
= \frac{3k}{5k}
= \frac{3}{5}
\cos A = \frac{AB}{AC}
= \frac{4k}{5k}
= \frac{4}{5}
\tan A = \frac{BC}{AB}
= \frac{3k}{4k}
= \frac{3}{4}

LHS = \frac{1 - \tan^2 A}{1 + \tan^2 A}
= \frac{1 - \left(\frac{3}{4}\right)^2}{1 + \left(\frac{3}{4}\right)^2}
= \frac{1 - \tfrac{9}{16}}{1 + \tfrac{9}{16}}
= \frac{16 - 9}{16 + 9}
= \frac{7}{25}

RHS = \cos^2 A - \sin^2 A
= \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2
= \frac{16}{25} - \frac{9}{25}
= \frac{7}{25}

∵ LHS=RHS
অৰ্থাৎ, \frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A

9. \triangle ABC ত কোনো সমকোণ। যদি, \tan A = \frac{1}{\sqrt{3}} তেন্তে তলৰ মান উলিওৱা –

(i) \sin A \cos C + \cos A \sin C
(ii) \cos A \cos C - \sin A \sin C

সমাধান :

∵ ABC এটা সমকোণী ত্ৰিভুজ, \angle B = 90^\circ, \angle A = \theta.

\tan A = \frac{1}{\sqrt{3}} = \frac{BC}{AB}=k, ধৰা হল (সমানুপাতৰ ভিত্তিত)
∴ BC = k,\ AB = \sqrt{3}k

∵ AC^2 = BC^2 + AB^2 (পাইথাগোৰাছ সূত্ৰ)
⇒ AC^2= k^2 + 3k^2
⇒ AC^2= 4k^2
⇒ AC = 2k

∴ \sin A = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}
\cos A = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}
\sin C = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}
\cos C = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}

(i) \sin A \cos C + \cos A \sin C
= \frac{1}{2}\times\frac{1}{2} + \frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}
= \frac{1}{4} + \frac{3}{4}
= 1

(ii) \cos A \cos C - \sin A \sin C
= \frac{\sqrt{3}}{2}\times\frac{1}{2} - \frac{1}{2}\times\frac{\sqrt{3}}{2}
= \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}
= 0

10. \triangle PQR ৰ Q কোণ সমকোণ আৰু PR + QR = 25cm আৰু PQ = 5cm; \cos P আৰু \tan P ৰ মান উলিওৱা।

সমাধান :

দিয়া আছে, PR + QR = 25cm আৰু PQ = 5cm
ধৰা হল, PR = x
\therefore QR = 25-x

PR^2 = QR^2 + PQ^2 (পাইথাগোৰাছ সূত্ৰ)
⇒ x^2 = (25-x)^2 + 25
⇒ x^2 = 625 - 50x + x^2 + 25
⇒ 0 = 650 - 50x
⇒ 50x=650
⇒ x=\frac{650}{50}
⇒ x = 13,
∴ QR = 25-13 = 12

\sin P = \frac{QR}{PR} = \frac{12}{13}
\cos P = \frac{PQ}{PR} = \frac{5}{13}
\tan P = \frac{QR}{PQ} = \frac{12}{5}

Frequently Asked Questions (FAQ)

Benefits of Class 10 Maths Ex 8.1 Assamese Medium Answer

Builds a strong foundation of trigonometric ratios
The Class 10 Maths Ex 8.1 Assamese Medium solution helps students build a clear understanding of the basic trigonometric ratios – sine, cosine, and tangent. These concepts form the backbone of the entire trigonometry chapter and make it easier to solve upcoming exercises.

Helps in solving advanced-level trigonometry problems
Practicing অনুশীলনী 8.1 Class 10 allows students to strengthen their basics so they can easily attempt advanced-level trigonometry questions. This foundation will also be helpful in higher studies, especially in mathematics and physics.

Improves step-by-step problem-solving ability
Each Class 10 Maths Exercise 8.1 Assamese Medium solution is explained in a detailed, step-by-step manner. This improves students’ logical thinking and teaches them how to approach trigonometry problems systematically.

Boosts confidence for SEBA Class 10 final exams
Studying the Class 10 Maths Ex 8.1 Assamese Medium answer regularly helps students gain speed and accuracy. This not only makes revision easier but also boosts confidence to perform well in the SEBA Class 10 final exams.

Conclusion

The Class 10 Maths Ex 8.1 Assamese Medium solutions provided here will guide students to understand trigonometry concepts clearly and prepare better for exams. By practicing these Class 10 Maths Assamese Medium Exercise 8.1 answers, students can score good marks in SEBA Class 10 Mathematics.