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# Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables)

## Selina Textbook Solutions Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables)

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables).

All Selina textbook questions of Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam.

Exercise/Page

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Page/Excercise 21(E)

Solution 1

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(xiv)

(xv)

(xvi)

(xvii)

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

(i) 2 sinA - 1 = 0

(ii)

Solution 7

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

Solution 8

(i)

(ii)

(iii)

(iv)

(v)

Solution 9

Since, A and B are complementary angles, A + B = 90°

(i)

(ii)

(iii)

= cosec2A [sec(90Â° - B)]2

= cosec2A cosec2B

(iv)

Solution 10

Solution 11

4 cos2A - 3 = 0

Solution 12

(i)

(ii) sin 3A - 1 = 0

(iii)

(iv)

(v)

Solution 13

(i)

(ii)

Solution 14

Solution 15

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Page/Excercise 21(B)

Solution 1

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Solution 2

Solution 3

Solution 4

Solution 5

Given:

and

Solution 6

Solution 7

LHS = (m2 + n2) cos2B

Hence, (m2 + n2) cos2B = n2.

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Page/Excercise 21(C)

Solution 1

(i)

(ii)

(iii)

Solution 2

Solution 3

(i)

(ii)

Solution 4

(i) We know that for a triangle ABC

A + B + C = 180°

(ii) We know that for a triangle ABC

A + B + C = 180°

B + C = 180° - A

Solution 5

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Solution 6

Since, ABC is a right angled triangle, right angled at B.

So, A + C = 90

Solution 7

(i)

Hence, x =

(ii)

Hence, x =

(iii)

Hence, x =

(iv)

Hence, x =

(v)

Hence, x =

(vi)

Hence, x =

(vii)

Hence,

Solution 8

(i)

(ii)

Solution 9

(i)

(ii)

Solution 10

Solution 11

Solution 12

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Page/Excercise 21(D)

Solution 1

(i) sin 21o = 0.3584

(ii) sin 34o 42'= 0.5693

(iii) sin 47o 32'= sin (47o 30' + 2') =0.7373 + 0.0004 = 0.7377

(iv) sin 62o 57' = sin (62o 54' + 3') = 0.8902 + 0.0004 = 0.8906

(v) sin (10o 20' + 2045') = sin 30o65' = sin 31o5' = 0.5150 + 0.0012 = 0.5162

Solution 2

(i) cos 2° 4’ = 0.9994 - 0.0001 = 0.9993

(ii) cos 8° 12’ = cos 0.9898

(iii) cos 26° 32’ = cos (26° 30’ + 2’) = 0.8949 - 0.0003 = 0.8946

(iv) cos 65° 41’ = cos (65° 36’ + 5’) = 0.4131 -0.0013 = 0.4118

(v) cos (9° 23’ + 15° 54’) = cos 24° 77’ = cos 25° 17’ = cos (25° 12’ + 5’) = 0.9048 - 0.0006 = 0.9042

Solution 3

(i) tan 37= 0.7536

(ii) tan 4218' = 0.9099

(iii) tan 17o  27' = tan (1724' + 3') = 0.3134 + 0.0010 = 0.3144

Solution 4

(i) From the tables, it is clear that sin 29o = 0.4848

Hence, = 29o

(ii) From the tables, it is clear that sin 2230' = 0.3827

Hence, = 2230'

(iii) From the tables, it is clear that sin 4042' = 0.6521

sin - sin 40o 42' = 0.6525 -; 0.6521 = 0.0004

From the tables, diff of 2' = 0.0004

Hence, = 40o  42' + 2' = 4044'

Solution 5

(i) From the tables, it is clear that cos 10° = 0.9848

Hence, = 10°

(ii) From the tables, it is clear that cos 16° 48’ = 0.9573

cos - cos 16° 48’ = 0.9574 - 0.9573 = 0.0001

From the tables, diff of 1’ = 0.0001

Hence, = 16° 48’ - 1’ = 16° 47’

(iii) From the tables, it is clear that cos 46° 30’ = 0.6884

cos q - cos 46° 30’ = 0.6885 - 0.6884 = 0.0001

From the tables, diff of 1’ = 0.0002

Hence, = 46° 30’ - 1’ = 46° 29’

Solution 6

(i) From the tables, it is clear that tan 13° 36’ = 0.2419

Hence, = 13° 36’

(ii) From the tables, it is clear that tan 25° 18’ = 0.4727

tan - tan 25° 18’ = 0.4741 - 0.4727 = 0.0014

From the tables, diff of 4’ = 0.0014

Hence, = 25° 18’ + 4’ = 25° 22’

(iii) From the tables, it is clear that tan 36° 24’ = 0.7373

tan - tan 36° 24’ = 0.7391 - 0.7373 = 0.0018

From the tables, diff of 4’ = 0.0018

Hence, = 36° 24’ + 4’ = 36° 28’

## Browse Study Material

TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables)  for ICSE Class 10 Mathematics free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Mathematics textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well.

# Text Book Solutions

ICSE X - Mathematics

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