Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days
8104911739
For Business Enquiry

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

RD Sharma Solution for Class 10 Mathematics Chapter 10 - Trigonometric Ratios

Share this:

Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 10 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 10 - Trigonometric Ratios.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 10 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 10 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

Read more

RD Sharma Solution for Class 10 Mathematics Chapter 10 - Trigonometric Ratios Page/Excercise 10.1

Solution 1 (i)

Solution 1 (ii)



Solution 1 (iii)



Solution 1 (iv)

Solution 1 (v)

Solution 1 (vi)

Solution 1 (vii)

Solution 1 (viii)

Solution 1 (ix)

Solution 1 (x)

Solution 1 (xi)

Solution 1 (xii)

Solution 2

In ABC by applying Pythagoras theorem
AC2 = AB2 + BC2
= (24)2 + (7)2
= 576 + 49
= 625
AC =  = 25 cm






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

RD Sharma Solution for Class 10 Mathematics Chapter 10 - Trigonometric Ratios Page/Excercise 10.2

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 (i)

Solution 26 (ii)

Solution 26 (iii)

Solution 26 (iv)

Solution 27 (i)

Solution 27 (ii)

Solution 27 (iii)

Solution 28 (i)

Solution 28 (ii)

Solution 29

Solution 30

Solution 31

Solution 32



Solution 33

Solution 34




Solution 35

Solution 36



Solution 37



Solution 38



Solution 39

Solution 40

RD Sharma Solution for Class 10 Mathematics Chapter 10 - Trigonometric Ratios Page/Excercise 10.3

Solution 1

Solution 2 (i)

Solution 2 (ii)

Solution 2 (iii)

Solution 2 (iv)

Solution 2 (v)

Solution 2 (vi)

Solution 2 (vii)

Solution 2 (viii)

Solution 2 (ix)

Solution 2 (x)

Solution 2 (xi)

Solution 3


Solution 4

Solution 5

Solution 6(i)

Solution 6(ii)

Solution 7 (i)

Solution 7 (ii)

Solution 7 (iii)

Solution 7 (iv)

Solution 8 (i)

Solution 8 (ii)

Solution 8 (iii)

Solution 8 (iv)

Solution 8 (v)

Solution 9 (i)

Solution 9 (ii)

Solution 9 (iii)

Solution 9 (iv)

Solution 9 (v)

Solution 9 (vi)

Solution 9 (vii)

Solution 9 (viii)

Solution 9 (ix)

Solution 9 (x)

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

RD Sharma Solution for Class 10 Mathematics Chapter 10 - Trigonometric Ratios Page/Excercise 10.56

Solution 1

begin mathsize 11px style It space is space given space that space straight theta space is space an space acute space angle space so space straight theta space lies space in space 1 to the power of st space quadrant.
In space the space first space quadrant space all space trigonometric space functions space are space positive.
sin space straight theta space equals space square root of 1 minus cos squared end root straight theta
space space space space space space space space space space equals square root of 1 minus open parentheses 3 over 5 close parentheses squared end root
space space space space space space space space space space equals square root of 1 minus 9 over 25 end root equals 4 over 5
tan space straight theta space equals space fraction numerator sin space straight theta over denominator cos space straight theta end fraction
space space space space space space space space space space space equals fraction numerator begin display style 4 over 5 end style over denominator begin display style 3 over 5 end style end fraction equals 4 over 3
Now space fraction numerator sin space straight theta space tan space straight theta space minus 1 over denominator 2 space tan space squared straight theta end fraction
equals fraction numerator begin display style 4 over 5 cross times 4 over 3 minus 1 end style over denominator 2 open parentheses begin display style 4 over 3 end style close parentheses squared end fraction
equals fraction numerator begin display style 1 over 15 end style over denominator begin display style 32 over 9 end style end fraction equals fraction numerator 9 over denominator 15 cross times 32 end fraction equals 3 over 160
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Solution 2

begin mathsize 11px style fraction numerator straight a space sin space straight theta space plus space straight b space cos space straight theta over denominator straight a space sin space straight theta space minus space straight b space cos space straight theta end fraction
equals space fraction numerator begin display style fraction numerator straight a space sin space straight theta over denominator cos space straight theta end fraction plus straight b end style over denominator begin display style fraction numerator straight a space sin space straight theta over denominator cos space straight theta end fraction minus space straight b end style end fraction
equals fraction numerator straight a space tan space straight theta plus straight b over denominator straight a space tan space straight theta space minus space straight b end fraction space space space space space space space space space space space space space space space space space space space space space space space space open curly brackets fraction numerator sin space straight theta over denominator cos space straight theta end fraction equals space tan space straight theta close curly brackets
equals fraction numerator straight a open parentheses begin display style straight a over straight b end style close parentheses plus straight b over denominator straight a open parentheses begin display style straight a over straight b end style close parentheses minus straight b end fraction equals fraction numerator straight a squared plus straight b squared over denominator straight a squared minus straight b squared end fraction
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Solution 3

begin mathsize 11px style 5 space tan space straight theta space minus 4 space equals space 0
box enclose tan space straight theta space equals space 4 over 5 end enclose space minus space circle enclose 1
To space find space colon space fraction numerator 5 space sin space straight theta space minus space 4 space cos space straight theta over denominator 5 space sin space straight theta space plus space 4 space cos space straight theta end fraction
space space space space space space space space space space space equals fraction numerator 5 space tan space straight theta minus 4 over denominator 5 space tan space straight theta plus 4 end fraction space space space...... space left parenthesis after space taking space cosθ space common right parenthesis
from space circle enclose 1 comma space
fraction numerator 5 space tan space straight theta minus 4 over denominator 5 space tan space straight theta plus 4 end fraction equals space fraction numerator 5 open parentheses begin display style 4 over 5 end style close parentheses minus 4 over denominator 5 open parentheses begin display style 4 over 5 end style close parentheses plus 4 end fraction equals space 0
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Solution 4

begin mathsize 12px style 16 space cot space straight x space equals space 12
rightwards double arrow space cot space straight x space equals space 12 over 16
rightwards double arrow box enclose cot space straight x space equals space 3 over 4 end enclose space minus space circle enclose 1
To space find space colon space fraction numerator space sin space straight x space minus space cos space straight x over denominator sin space straight x space plus space cos space straight x end fraction
equals space fraction numerator open parentheses 1 minus begin display style cosx over sinx end style close parentheses over denominator open parentheses 1 plus fraction numerator cos space straight x over denominator sin space straight x end fraction close parentheses end fraction
equals space fraction numerator 1 space minus space cot space straight x over denominator 1 space plus space cot space straight x end fraction
from space circle enclose 1
equals space fraction numerator 1 minus begin display style 3 over 4 end style over denominator 1 plus begin display style 3 over 4 end style end fraction
equals fraction numerator begin display style 1 fourth end style over denominator begin display style 7 over 4 end style end fraction equals 1 over 7 end style

So, the correct option is (a).

Solution 5

begin mathsize 11px style 8 space tan space straight x space equals space 15
space tan space straight x space equals space 15 over 8
square root of 8 squared plus 15 squared end root space equals space square root of 64 space plus 225 end root
space space space space space space space space space space space space space space space space space space space equals space square root of 289
space space space space space space space space space space space space space space space space space space space equals space 17 space space space space space space
Sin space straight x space equals space 15 over 17 space space space space space space space space space space space space space space cos space straight x space equals space 8 over 17
To space find space colon space sin space straight x space minus space cos space straight x
space space space space space space space space space space space space space space space equals 15 over 17 minus 8 over 17
space space space space space space space space space space space space space space space equals 7 over 17
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Solution 6

begin mathsize 11px style We space know comma space tan space straight theta space equals space fraction numerator 1 over denominator cot space straight theta end fraction space minus space circle enclose 1
space space space space space space space space space space space space space space space space space space 1 plus tan squared straight theta space equals space sec squared straight theta minus circle enclose 2
space space space space space space space space space space space space space space space space space space 1 space plus space cot squared straight theta space equals space cosec squared straight theta space minus space circle enclose 3
from space circle enclose 1
space space space space space space space space space space box enclose cot space straight theta space equals space square root of 7 end enclose
from space circle enclose 2
space space space space space space space space space 1 space plus space open parentheses fraction numerator 1 over denominator square root of 7 end fraction close parentheses squared space equals space sec squared straight theta
space rightwards double arrow space space sec squared straight theta space equals space 1 plus 1 over 7 equals 8 over 7
rightwards double arrow space box enclose sec squared straight theta equals 8 over 7 end enclose minus space circle enclose 4
from space circle enclose 3
space space space space 1 space plus space open parentheses square root of 7 close parentheses squared equals space cosec squared straight theta
rightwards double arrow space space box enclose cosec squared straight theta equals 8 end enclose space minus space circle enclose 5
To space find colon space fraction numerator cosec squared straight theta minus sec squared straight theta over denominator cosec squared straight theta plus sec squared straight theta end fraction
space space space space space space space equals fraction numerator 8 minus begin display style 8 over 7 end style over denominator 8 plus begin display style 8 over 7 end style end fraction equals fraction numerator begin display style 48 over 7 end style over denominator begin display style 64 over 7 end style end fraction
space space space space space space space equals space 48 over 64 equals space 3 over 4
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

RD Sharma Solution for Class 10 Mathematics Chapter 10 - Trigonometric Ratios Page/Excercise 10.57

Solution 7

Error converting from MathML to accessible text.

begin mathsize 11px style from space circle enclose 1 comma circle enclose 2 comma circle enclose 3
box enclose sin space straight theta space equals space 3 over 5 end enclose minus space circle enclose 4
box enclose cos space straight theta space equals space 4 over 5 space end enclose minus space circle enclose 5
To space find space colon space space cos squared straight theta space minus space space sin squared straight theta
space space space space space space space space space space space space space space space space equals open parentheses 4 over 5 close parentheses squared space minus space open parentheses 3 over 5 close parentheses squared... left parenthesis from space circle enclose 4 space and space circle enclose 5 right parenthesis
space space space space space space space space space space space space space space space space equals space 7 over 25
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Solution 8

Error converting from MathML to accessible text.

begin mathsize 11px style sin space straight theta space equals space square root of 8 over 15 end root
cos space straight theta space equals space square root of 7 over 15 end root
To space find space colon space fraction numerator open parentheses 1 plus sin space straight theta close parentheses open parentheses 1 minus sin space straight theta close parentheses over denominator open parentheses 1 plus cos space straight theta close parentheses open parentheses 1 minus cosθ close parentheses end fraction
space space space space space space space space space space space space space equals fraction numerator 1 minus sin squared straight theta over denominator 1 minus cos squared straight theta end fraction
space space space space space space space space space space space space space space equals fraction numerator 1 minus open parentheses square root of begin display style 8 over 15 end style end root close parentheses squared over denominator 1 minus open parentheses begin display style 7 over 15 end style close parentheses squared end fraction
space space space space space space space space space space space space space space equals fraction numerator 1 minus begin display style 8 over 15 end style over denominator 1 minus begin display style 7 over 15 end style end fraction equals fraction numerator begin display style 7 over 15 end style over denominator begin display style 8 over 15 end style end fraction
space space space space space space space space space space space space space space equals 7 over 8
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Solution 9

Error converting from MathML to accessible text.

begin mathsize 11px style sin space straight theta space equals space fraction numerator 3 over denominator square root of 34 end fraction
cos space straight theta space equals space fraction numerator 5 over denominator square root of 34 end fraction
We space know space sec space straight theta space equals space fraction numerator 1 over denominator cos space straight theta end fraction space equals space fraction numerator square root of 34 over denominator 5 end fraction
To space find space colon space fraction numerator 5 space sin space straight theta space minus space 2 space sec cubed straight theta space plus space 2 space cosθ over denominator 5 space sin space straight theta space plus space 2 space sec cubed straight theta space minus space 2 space cos space straight theta end fraction

space space space space space space space space space space space rightwards double arrow fraction numerator 5 open parentheses begin display style fraction numerator 3 over denominator square root of 34 end fraction end style close parentheses minus 2 open parentheses square root of begin display style 34 over 5 end style end root close parentheses cubed plus 2 open parentheses begin display style fraction numerator 5 over denominator square root of 34 end fraction end style close parentheses over denominator 5 open parentheses begin display style fraction numerator 3 over denominator square root of 34 end fraction end style close parentheses plus 2 open parentheses begin display style fraction numerator square root of 34 over denominator 5 end fraction end style close parentheses cubed minus 2 open parentheses begin display style fraction numerator 5 over denominator square root of 34 end fraction end style close parentheses end fraction
space space space space space space space space space space rightwards double arrow fraction numerator begin display style fraction numerator 25 over denominator square root of 34 end fraction minus fraction numerator 68 square root of 34 over denominator 125 end fraction end style over denominator begin display style fraction numerator 5 over denominator square root of 34 end fraction plus fraction numerator 68 square root of 34 over denominator 125 end fraction end style end fraction
space space space space space space space space space space space rightwards double arrow fraction numerator space 125 space cross times space 25 space minus space 68 space cross times space 34 over denominator 5 space cross times space 125 space plus space 68 space cross times 34 end fraction
space space space space space space space space space space space rightwards double arrow space 813 over 2937
space space space space space space space space space space space space space rightwards double arrow 271 over 979
space So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Solution 10

begin mathsize 11px style We space know
space space space space space space space space space space space tan space 45 to the power of straight o space equals space 1 space space space space minus space circle enclose 1
space space space space space space space space space space cos space 30 to the power of straight o space space equals space fraction numerator square root of 3 over denominator 2 end fraction space minus space circle enclose 2
space space space space space space space space space sin space 45 to the power of straight o space equals space fraction numerator 1 over denominator square root of 2 end fraction space space space minus space circle enclose 3 space space space space
space space space space space space space cos space 45 to the power of straight o space end exponent equals space fraction numerator 1 over denominator square root of 2 end fraction space minus space circle enclose 4
Given space tan squared 45 to the power of straight o space minus space cos squared 30 to the power of straight o space end exponent space equals space straight x space sin space 45 to the power of straight o space cos space 45 to the power of straight o
from space circle enclose 1 comma circle enclose 2 comma circle enclose 3
rightwards double arrow 1 space minus space open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses squared space equals space straight x open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses
rightwards double arrow 1 space minus space 3 over 4 equals space straight x over 2
rightwards double arrow space straight x over 2 space equals space 1 fourth
rightwards double arrow space straight x space equals space 2 over 4 space equals space 1 half end style

So, the correct option is (d).

Solution 11

begin mathsize 11px style We space know space sin space left parenthesis 90 space minus space straight theta right parenthesis space equals space cos space straight theta
space space space space space space space space space space space so space space sin space left parenthesis 90 minus space 17 right parenthesis space equals space cos 17
space space space space space space space space space space space space space rightwards double arrow space sin space left parenthesis 73 to the power of straight o right parenthesis space equals space cos 17 to the power of straight o space space minus space circle enclose 1
To space find space colon space cos squared 17 to the power of straight o space minus space sin squared 73 to the power of straight o
from space circle enclose 1
space space space space space space space space space space rightwards double arrow cos squared 17 to the power of straight o space minus space cos squared 17 to the power of straight o
space space space space space space space space space space rightwards double arrow space 0 end style

So, the correct option is (c).

Solution 12

begin mathsize 11px style We space know space sin space left parenthesis 90 minus straight theta right parenthesis equals cos space straight theta
so space space space space space space space space space space space space space space sin space left parenthesis 90 degree minus 20 to the power of straight o right parenthesis space equals space cos 20 to the power of straight o
space space space space space space space space space space space space space space space rightwards double arrow box enclose sin space 70 to the power of straight o space equals space cos space 20 to the power of straight o end enclose space minus space circle enclose 1
space space space space space space space space space space space space space space space space space space sin space left parenthesis 90 degree minus 70 to the power of straight o right parenthesis space equals space ws space 70 to the power of straight o
space space space space space space space space space space space space space space space space space box enclose sin space left parenthesis 20 to the power of straight o right parenthesis space equals space cos space left parenthesis 70 to the power of straight o right parenthesis end enclose space minus space circle enclose 2
To space find space fraction numerator cos cubed 20 to the power of straight o space minus space cos cubed 70 to the power of straight o over denominator sin space cubed 70 to the power of straight o space minus space sin space cubed 20 to the power of straight o end fraction
space space space from space circle enclose 1 space end enclose space space and space circle enclose 2
rightwards double arrow space fraction numerator cos cubed 20 to the power of straight o minus cos cubed 70 to the power of straight o over denominator cos cubed 20 to the power of straight o minus cos cubed 70 to the power of straight o end fraction equals 1
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Solution 13

begin mathsize 11px style We space know
rightwards double arrow space tan space 60 to the power of straight o equals space square root of 3
rightwards double arrow tan space 30 to the power of straight o space equals space fraction numerator 1 over denominator square root of 3 end fraction
rightwards double arrow cos space 45 to the power of straight o space equals space fraction numerator 1 over denominator square root of 2 end fraction
rightwards double arrow sec space 45 to the power of straight o space equals space square root of space 2 end root
rightwards double arrow sin space 60 to the power of straight o space equals space fraction numerator square root of 3 over denominator 2 end fraction
rightwards double arrow cosec space 30 to the power of straight o space equals space 2
given space fraction numerator xcosec squared 30 to the power of straight o sec squared 45 to the power of straight o over denominator 8 cos squared 45 to the power of straight o space sin space squared 60 to the power of straight o end fraction equals space tan squared 6 to the power of straight o minus tan squared 30 to the power of straight o
rightwards double arrow fraction numerator straight x space left parenthesis 4 right parenthesis space left parenthesis 2 right parenthesis over denominator 8 space open parentheses begin display style 1 half end style close parentheses open parentheses begin display style 3 over 4 end style close parentheses end fraction equals 3 minus 1 third
rightwards double arrow fraction numerator 8 straight x over denominator 3 end fraction space equals space 8 over 3
rightwards double arrow box enclose straight x space equals space 1 end enclose
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Solution 14

begin mathsize 11px style Complementary space angles space means space that space sum space of space two space angles space is space 90 to the power of straight o
straight A space and space straight B space are space complementary space angles space so
straight A space plus space straight B space equals space 90 to the power of straight o
rightwards double arrow space straight A space equals space 90 to the power of straight o space minus space straight B space space minus space circle enclose 1
left parenthesis straight a right parenthesis space Applying space sin space on space both space sides space in space circle enclose 1
space space space space space space space space sin space straight A space equals space sin space left parenthesis 90 to the power of straight o space minus space straight B right parenthesis
space space space space space space space rightwards double arrow sin space straight A space equals space cos space straight B
left parenthesis straight b right parenthesis space Applying space cos space on space both space sides space in space circle enclose 1
space space space space space space cos space straight A space equals space cos space left parenthesis 90 degree space minus space straight B right parenthesis
space space space space space space rightwards double arrow cos space straight A space equals space sin space straight B
left parenthesis straight c right parenthesis space Applying space tan space on space both space sides space in space circle enclose 1
space space space space space space space space space tan space straight A space equals space tan space left parenthesis 90 degree minus straight B right parenthesis
space space space space space space rightwards double arrow space tan space straight A space equals space cot space straight B
left parenthesis straight d right parenthesis space Applying space sec space on space both space sides space in space circle enclose 1
space space space space space space space space sec space straight A space equals space sec space left parenthesis 90 degree minus straight B right parenthesis
space space space space space space space box enclose sec space straight A space equals space cosec space straight B end enclose
So comma space correct space option space is space left parenthesis straight d right parenthesis end style

Solution 15

begin mathsize 11px style We space know space that
space space space space space space space space space space space space space space space space sin space left parenthesis 90 degree space minus space straight theta right parenthesis space equals space cos space straight theta
space space space space space space space space space space space space space space space space cos space left parenthesis 90 degree space minus straight theta right parenthesis space equals space sin space straight theta
space space space space space space space space space space space space space space space space cot space left parenthesis 90 degree space minus space straight theta right parenthesis space equals space tan space straight theta
rightwards double arrow space straight x space sin space left parenthesis 90 degree space minus space straight theta right parenthesis space cot space left parenthesis 90 degree space minus space straight theta right parenthesis space equals space cos space left parenthesis 90 degree space minus space straight theta right parenthesis
rightwards double arrow straight x space cos space straight theta space tan space straight theta space equals space sin space straight theta
rightwards double arrow space straight x space cos space straight theta space open parentheses fraction numerator sin space straight theta over denominator cos space straight theta end fraction close parentheses space equals space sin space straight theta
rightwards double arrow space box enclose straight x space equals space 1 end enclose end style

So, the correct option is (b).

Solution 16

begin mathsize 11px style We space know
space space space space space space space space space space space space space space rightwards double arrow space tan space 45 to the power of straight o space equals space 1
space space space space space space space space space space space space space space rightwards double arrow cos space 60 to the power of straight o space equals space 1 half
space space space space space space space space space space space space space space rightwards double arrow sin space 60 to the power of straight o space equals space fraction numerator square root of 3 over denominator 2 end fraction
space space space space space space space space space space space space space space rightwards double arrow cot space 60 to the power of straight o space equals space fraction numerator 1 over denominator square root of 3 end fraction
To space find space colon space straight x space tan space 45 to the power of straight o cos space 60 to the power of straight o space equals space sin space 60 to the power of straight o cot 60 to the power of straight o
space space space space space space space space space space space space space space space rightwards double arrow straight x space left parenthesis 1 right parenthesis space open parentheses 1 half close parentheses space equals space open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses
space space space space space space space space space space space space space space space space rightwards double arrow box enclose straight x space equals space 1 end enclose end style

So, the correct option is (a).

Solution 17

begin mathsize 11px style We space know comma space sum space of space all space angles space of space straight a space triangle space equals space 180
space space space space space space space space space space space space space rightwards double arrow space straight A space plus space straight B space plus space straight C space equals space 180
it space is space given space that space straight A comma straight B comma straight C space are space in space AP comma
So comma space let space straight A space equals space straight B minus straight r space space and space straight C space equals space straight B space plus space straight r
Now space angles space are space straight B space minus space straight r comma space straight B comma space straight B space plus space straight r
space space space space space space space space space From space circle enclose 1 space and space circle enclose 2
space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis straight B minus straight r right parenthesis plus space straight B space plus space left parenthesis straight B space plus straight r right parenthesis space equals space 180 to the power of straight o
space space space space space space space space space space space space space space rightwards double arrow space 3 straight B space equals space 180 to the power of straight o space rightwards double arrow space straight B space equals space 60 to the power of straight o
Now space sin space straight B space rightwards double arrow space box enclose sin space 60 to the power of straight o space equals space fraction numerator square root of 3 over denominator 2 end fraction end enclose end style

So, the correct option is (b).

Solution 18

Error converting from MathML to accessible text.

begin mathsize 11px style space space space space space space space space space space space space space space space space space space space space space space space space space space space space sin space straight theta space equals space fraction numerator square root of 2 over denominator square root of 3 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space cosec space straight theta space equals space fraction numerator 1 over denominator sin space straight theta end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow box enclose cosec space straight theta space equals space fraction numerator square root of 3 over denominator square root of 2 end fraction end enclose space minus space circle enclose 2
To space find space colon space space fraction numerator tan squared straight theta space minus space cosec squared straight theta over denominator tan squared straight theta space plus space cosec squared straight theta end fraction
space space space space space space space space space space space space From space circle enclose 2 space end enclose space and space circle enclose 1
space space space space space space space space space space space rightwards double arrow fraction numerator 2 minus open parentheses begin display style 3 over 2 end style close parentheses over denominator 2 plus open parentheses begin display style 3 over 2 end style close parentheses end fraction equals fraction numerator begin display style 1 half end style over denominator begin display style 7 over 2 end style end fraction
space space space space space space space space space equals space 1 over 7 end style

So, the correct option is (d).

RD Sharma Solution for Class 10 Mathematics Chapter 10 - Trigonometric Ratios Page/Excercise 10.58

Solution 19

begin mathsize 11px style We space know space tan space left parenthesis 90 minus straight theta right parenthesis space equals space cot space straight theta space minus space circle enclose 1
space space space space space space space space space space space space and space tan space open parentheses straight theta close parentheses space equals space fraction numerator 1 over denominator cot space straight theta end fraction space minus space circle enclose 2
from space circle enclose 1 space and space circle enclose 2
space space space space space space space space space space space space box enclose tan space left parenthesis 90 space minus space straight theta right parenthesis space equals space fraction numerator 1 over denominator tan space straight theta end fraction end enclose
or space space space space space space space space space box enclose tan space straight theta space tan space left parenthesis 90 space minus space straight theta right parenthesis space equals space 1 end enclose space minus space circle enclose 3
To space find space colon space space tan space 1 to the power of straight o space tan space 2 to the power of straight o space tan space 3 to the power of straight o straight _ space straight _ space straight _ space straight _ space straight _ thin space straight _ space straight _ space tan space 89 to the power of straight o
space space space space space space space space space space space space space it space can space also space be space rewritten space space as
space space space space space space space space space space rightwards double arrow left parenthesis tan space 1 to the power of straight o space tan 89 to the power of straight o right parenthesis space left parenthesis tan 2 to the power of straight o space tan 88 to the power of straight o right parenthesis space straight _ space straight _ thin space straight _ space straight _ thin space straight _ open parentheses tan space 44 to the power of straight o space tan space 46 to the power of straight o close parentheses space tan space 45 to the power of straight o
space space space space space space space space space space rightwards double arrow from space circle enclose 3 space and space tan space 45 to the power of straight o space equals space 1
space space space space space space space space space space rightwards double arrow left parenthesis 1 right parenthesis space left parenthesis 1 right parenthesis space left parenthesis 1 right parenthesis space straight _ space straight _ space straight _ space straight _ space straight _ thin space straight _ space straight _ left parenthesis 1 right parenthesis
space space space space space space space space space space space equals space 1 space
space So comma space the space correct space option space is space left parenthesis straight a right parenthesis. space space space space space space space space space end style

Solution 20

begin mathsize 11px style We space know space box enclose cos space 90 to the power of straight o space equals space 0 end enclose space minus space circle enclose 1
space space space space space space space space So space cos 1 to the power of straight o space space cos 2 to the power of straight o space straight _ space straight _ space straight _ thin space straight _ thin space straight _ space space cos 90 to the power of straight o space straight _ space straight _ space straight _ space straight _ thin space space cos space 180 to the power of straight o space minus space circle enclose 2
space space space space space space space space space from space circle enclose 1 space and space circle enclose 2
space space space space space space space rightwards double arrow space 0 end style

So, the correct option is (b).

Solution 21

begin mathsize 11px style space space space space space space space We space know
space space space space space space space space space space space space space space tan space left parenthesis 90 space minus space straight theta right parenthesis space equals space cot space straight theta space space space minus space circle enclose 1
space space and space space space space space tan space straight theta space equals space fraction numerator 1 over denominator cot space straight theta end fraction space minus space space space space space space space space space space circle enclose 2
From space circle enclose 1 space and space circle enclose 2
space space space space space space space space space box enclose tan space left parenthesis 90 space minus space straight theta right parenthesis space equals space fraction numerator 1 over denominator tan space straight theta end fraction end enclose space minus space circle enclose 3
For space straight theta space equals space 10 to the power of straight o space space rightwards double arrow tan space left parenthesis 80 to the power of straight o right parenthesis space equals space fraction numerator 1 over denominator tan space 10 to the power of straight o end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space tan space left parenthesis 80 to the power of straight o right parenthesis space tan space left parenthesis 10 to the power of straight o right parenthesis space equals space 1 space space space space space space space space minus space circle enclose 4
For space straight theta space equals space 15 to the power of straight o space space rightwards double arrow space tan space left parenthesis 75 to the power of straight o right parenthesis space equals space fraction numerator 1 over denominator tan space left parenthesis 15 to the power of straight o right parenthesis end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space tan space left parenthesis 75 to the power of straight o right parenthesis space tan space left parenthesis 15 to the power of straight o right parenthesis space equals space 1 space space space space space minus space circle enclose 5
To space Find space colon space space space tan space 10 to the power of straight o space tan space 15 to the power of straight o space tan space 75 to the power of straight o space tan space 80 to the power of straight o
space space space space space space space space space space space space space space space space space rightwards double arrow space open parentheses tan space 10 to the power of straight o space tan space 80 to the power of straight o close parentheses space left parenthesis tan space 15 to the power of straight o space tan space 75 to the power of straight o right parenthesis
space space space space space space space space space space space from space circle enclose 4 space and space circle enclose 5
space space space space space space space space space space space space space space space space space space space rightwards double arrow 1
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Solution 22

begin mathsize 11px style We space know comma space space space space space space space cos space left parenthesis 90 space minus space straight theta right parenthesis space equals space sin space straight theta space space minus space circle enclose 1
space space space space space space space space space space space space space space space space space space space space space space space space sec space left parenthesis 90 space minus space straight theta right parenthesis space equals space cosec space straight theta space minus space circle enclose 2
space space space space space space space space space space space space space space space space space space space space space space space space cosec space left parenthesis 90 space minus space straight theta right parenthesis space equals space sec space straight theta space minus space circle enclose 3
space space space space space space space space space space space space space space space space space space space space space space space sin space left parenthesis 90 space minus space straight theta right parenthesis space space equals space cos space straight theta space space minus space circle enclose 4
space space space space space space space space space space space space space space space space space space space space space space cot space left parenthesis 90 space minus space straight theta right parenthesis space equals space tan space straight theta space space minus space circle enclose 5
space space space space space space space space space space space space space space space space space space space space space space tan space left parenthesis 90 space minus space straight theta right parenthesis space equals space cot space straight theta space space minus space circle enclose 6
To space find space colon space space fraction numerator cos space left parenthesis 90 degree space minus space straight theta right parenthesis space sec space left parenthesis 90 degree space minus space straight theta right parenthesis space tan space straight theta over denominator cosec space left parenthesis 90 degree minus space straight theta right parenthesis space sin space left parenthesis 90 degree minus space straight theta right parenthesis cot left parenthesis 90 degree minus space straight theta right parenthesis end fraction space plus space fraction numerator tan space left parenthesis 90 degree space minus space straight theta right parenthesis over denominator cot space left parenthesis straight theta right parenthesis end fraction
space space space space space space space rightwards double arrow space with space the space help space of space eq space circle enclose 1 comma space circle enclose 2 comma circle enclose 3 comma circle enclose 4 comma circle enclose 5 comma circle enclose 6
space space space space space space space rightwards double arrow fraction numerator sin space straight theta space cosec space straight theta space up diagonal strike tan space straight theta end strike over denominator sin space straight theta space cosec space straight theta space up diagonal strike tan space straight theta end strike end fraction space plus space fraction numerator up diagonal strike cot space straight theta end strike over denominator up diagonal strike cot space straight theta end strike end fraction
space space space space space space space rightwards double arrow space fraction numerator sin space straight theta space cosec space straight theta over denominator sec space straight theta space cos space straight theta end fraction plus 1
We space also space know space that
space space space space space space space space space space space sin space straight theta space equals space fraction numerator 1 over denominator cosec space straight theta end fraction rightwards double arrow space sin space straight theta space cosec space equals space 1 space space minus space circle enclose 8
and space space space cos space straight theta space equals space fraction numerator 1 over denominator sec space straight theta end fraction space rightwards double arrow sec space straight theta space cos space straight theta space equals space 1 space space space minus space circle enclose 9
From space circle enclose 7 comma circle enclose 8 comma circle enclose 9
rightwards double arrow space we space get
space space space space space space space space space fraction numerator sin space straight theta space cosec space straight theta over denominator cos space straight theta space sec space straight theta end fraction plus 1
space space rightwards double arrow space 2
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Solution 23

begin mathsize 11px style space space space space space space space space space space space space We space know space cos space left parenthesis 90 space minus space straight theta right parenthesis space equals space sin space straight theta space space minus space circle enclose 1
space space space space space space space space space space space space rightwards double arrow Given space space space space space space space sin space straight theta space equals space cos space left parenthesis 2 straight theta space minus space 45 to the power of straight o right parenthesis space minus space circle enclose 2
from space circle enclose 1 space and space circle enclose 2
space space space space space space space space space space space cos space left parenthesis 90 space minus space straight theta right parenthesis space equals space cos space left parenthesis 2 straight theta space minus space 45 to the power of straight o right parenthesis
space space space space space rightwards double arrow space 90 space minus space straight theta space equals space 2 straight theta space minus 45 to the power of straight o
space space space space space rightwards double arrow 90 to the power of straight o space plus space 45 to the power of straight o space equals space 3 straight theta
space space space space space space space rightwards double arrow 3 straight theta space equals space 135 to the power of straight o
space space space space space space space rightwards double arrow box enclose straight theta space equals space 45 to the power of straight o end enclose
Then space tan space straight theta
space space space space space space rightwards double arrow tan space 45 to the power of straight o
space space space space space space rightwards double arrow space 1
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Solution 24

begin mathsize 11px style space space space space space space We space know
space space space space space space space space space space space space space space space space space space sin space left parenthesis 90 space minus space 4 straight A right parenthesis space equals space cos space 4 straight A space minus space circle enclose 1
space space space space given space space space space sin space 5 straight theta space space equals space cos space 4 straight theta space space space space minus space circle enclose 2
From space circle enclose 1 space end enclose space and space circle enclose 2
space space space space space space space space space space space space space space space space sin space left parenthesis 5 straight theta right parenthesis space equals space sin space left parenthesis 90 space minus space 4 straight theta right parenthesis
space space space space space space rightwards double arrow space space space space space space space 5 straight theta space equals space 90 space minus space 4 straight theta
space space space space space space space rightwards double arrow space space space space space space 9 straight theta equals space 90
space space space space space space space rightwards double arrow space space box enclose straight theta space equals space 10 to the power of straight o end enclose
To space find space 2 space sin space 3 straight theta space minus space square root of 3 space tan space 3 straight theta
space space space space space space space rightwards double arrow space 2 space sin space 30 to the power of straight o space minus space square root of 3 space end root space tan space 30 to the power of straight o
space space space space space space space rightwards double arrow 2 space open parentheses 1 half close parentheses space minus space square root of 3 open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space
space space space space space space space rightwards double arrow space 1 space minus space 1 space equals space 0
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Solution 25

begin mathsize 11px style space space space space space space space Given space straight A space plus space straight B space equals space 90 to the power of straight o
space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space straight A equals space 90 to the power of straight o space minus space straight B space minus space circle enclose 1
space space space space Apply space sin space both space side space in space circle enclose 1
space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space sin space straight A space equals space space sin space left parenthesis 90 to the power of straight o space minus space straight B right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space box enclose sin space straight A space equals space cos space straight B end enclose space space minus space circle enclose 2
space space space Apply space cos space both space side space in space circle enclose 1
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space cos space straight A space equals space cos space left parenthesis 90 to the power of straight o minus straight B right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space box enclose cos space straight A space equals space sin space straight B end enclose space minus space circle enclose 3
space space space Apply space cosec space both space side space in space circle enclose 1
space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space cosec space straight A space equals space cosec space left parenthesis 90 space minus space straight B right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow box enclose cosec space straight A space equals space sec space straight B end enclose space minus space circle enclose 4
To space find space colon thin space space fraction numerator tan space straight A space tan space straight B space plus space tan space straight A space cot space straight B over denominator space space space space sin space straight A space sec space straight B end fraction space minus space fraction numerator sin squared straight B over denominator cos squared straight A end fraction
space space space space space space space space space space space space space space space space space rightwards double arrow fraction numerator open parentheses begin display style fraction numerator sin space straight A over denominator cos space straight A end fraction end style close parentheses open parentheses begin display style fraction numerator sin space straight B over denominator cos space straight B end fraction end style close parentheses plus open parentheses begin display style fraction numerator sin space straight A over denominator cos space straight A end fraction end style close parentheses open parentheses begin display style fraction numerator cos space straight B over denominator sin space straight B end fraction end style close parentheses over denominator sin space straight A. space cosec space straight A end fraction space space minus fraction numerator sin squared straight B over denominator cos squared straight A end fraction
space space space space space space space space space space space space space space space space space With space help space of space circle enclose 2 comma circle enclose 3 comma circle enclose 4
space space space space space space space rightwards double arrow fraction numerator 1 plus begin display style fraction numerator sin squared straight A over denominator cos squared straight A end fraction end style over denominator sin space straight A. open parentheses begin display style fraction numerator 1 over denominator sin space straight A end fraction end style close parentheses end fraction minus 1
space space space space space space space equals 1 space space plus space tan squared straight A space minus space 1
space space space space space space space equals tan squared straight A
space space space space space space space equals open square brackets tan space left parenthesis 90 minus straight B right parenthesis close square brackets squared space space space space space space.... open curly brackets straight A space equals space 90 minus straight B close curly brackets
space space space space space space space equals space open square brackets cot space straight B close square brackets squared
space space space space space space space equals cot squared straight B
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. space space space space space space space space space space space end style

Solution 26

begin mathsize 11px style We space know
space space space space space space space space space space space space space tan space 30 to the power of straight o space equals space fraction numerator 1 over denominator square root of 3 end fraction
space space To space find space colon space space fraction numerator 2 space tan 30 to the power of straight o over denominator 1 plus tan squared 30 to the power of straight o end fraction
space space space space space space space space space space space space rightwards double arrow fraction numerator 2 open parentheses begin display style fraction numerator 1 over denominator square root of 3 end fraction end style close parentheses over denominator 1 plus open parentheses begin display style 1 third end style close parentheses end fraction equals space fraction numerator begin display style fraction numerator 2 over denominator square root of 3 end fraction end style over denominator begin display style 4 over 3 end style end fraction
space space space space space space space space space space space space space rightwards double arrow fraction numerator square root of 3 over denominator 2 end fraction
Value space of space sin space 60 to the power of straight o space equals space fraction numerator square root of 3 over denominator 2 end fraction
space space space space space space space space space space space cos space 60 to the power of straight o space equals space 1 half
space space space space space space space space space space space tan space 60 to the power of straight o space equals space square root of 3
space space space space space space space space space space space sin space 30 to the power of straight o space equals space 1 half
Hence space correct space option space is space left parenthesis straight a right parenthesis end style

Solution 27

begin mathsize 11px style tan space 45 to the power of straight o space equals space 1
To space find space colon space minus space fraction numerator 1 minus tan squared 45 to the power of straight o over denominator 1 plus tan squared 45 to the power of straight o end fraction
space space space space space space space space space space space space space space space space space rightwards double arrow fraction numerator 1 space minus space 1 over denominator 1 space plus space 1 end fraction equals 0
We space know space sin space left parenthesis 0 to the power of straight o right parenthesis space equals space 0 end style

So, the correct option is (d).

Solution 28

begin mathsize 11px style We space know space
space space space space space space space space space space space space space space sin space 2 straight A space equals space 2 sin space straight A space cos space straight A space minus space circle enclose 1
Given space space sin space 2 straight A space equals space 2 space sin space straight A space minus space circle enclose 2
space space space space from space circle enclose 1 space and space circle enclose 2
space space space space space space space rightwards double arrow 2 space sin space straight A space cos space straight A space equals space 2 space sin space straight A
space space space space space space space rightwards double arrow 2 space sin space straight A space left parenthesis cos space straight A space minus space 1 right parenthesis space equals space 0
space space space space space space space rightwards double arrow sin space straight A space equals space 0 space space space space space space or space cos space straight A space equals space 1
space We space knows space sin space 0 to the power of straight o space equals space 0 space and space cos space 0 to the power of straight o space equals space 1
space space space space space space space Hence space box enclose straight A space equals space 0 to the power of straight o end enclose end style

So, the correct option is (a).

Solution 29

begin mathsize 11px style We space know comma space tan space 30 to the power of straight o space equals space fraction numerator 1 over denominator square root of 3 end fraction
space space space space space Given comma space fraction numerator 2 space tan space 30 to the power of straight o over denominator 1 minus space tan squared 30 end fraction
space space space space space space space space space space space rightwards double arrow fraction numerator 2 open parentheses begin display style fraction numerator 1 over denominator square root of 3 end fraction end style close parentheses over denominator 1 minus open parentheses begin display style 1 third end style close parentheses end fraction
space space space space space space space space space space space rightwards double arrow fraction numerator begin display style fraction numerator 2 over denominator square root of 3 end fraction end style over denominator begin display style 2 over 3 end style end fraction
space space space space space space space space space space space rightwards double arrow square root of 3
We space also space know space that space tan space 60 to the power of straight o space equals space square root of 3
So comma space correct space answer space is space left parenthesis straight c right parenthesis end style

Solution 30

begin mathsize 11px style We space know comma space if space straight A comma straight B comma straight C space are space angles space of space straight a space triangle space then
space space space space space straight A space plus space straight B space plus space straight C space equals space 180 to the power of straight o
space space rightwards double arrow space straight B space plus thin space straight C space equals space 180 to the power of straight o space minus space straight A
space space rightwards double arrow fraction numerator straight B plus straight C over denominator 2 end fraction equals 90 to the power of straight o space minus straight A over 2
space space space rightwards double arrow sin space open parentheses fraction numerator straight B plus straight C over denominator 2 end fraction close parentheses equals sin space open parentheses 90 to the power of straight o minus straight A over 2 close parentheses
space space space space rightwards double arrow sin space open parentheses fraction numerator straight B plus straight C over denominator 2 end fraction close parentheses space equals space cos space open parentheses straight A over 2 close parentheses
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Solution 31

Error converting from MathML to accessible text.

begin mathsize 11px style space space space space space space tan space straight theta space equals space fraction numerator square root of 5 over denominator 2 end fraction
space space space space space space space sec space straight theta space equals space 3 over 2
To space find space colon space space 2 space sec squared straight theta space plus space 2 space tan squared straight theta space minus space 7
space space space space space space space space space space space space space space space space rightwards double arrow 2 space open parentheses 3 over 2 close parentheses squared space space plus space 2 open parentheses fraction numerator square root of 5 over denominator 2 end fraction close parentheses squared space minus space 7
space space space space space space space space space space space space space space space rightwards double arrow 9 over 2 space plus space 5 over 2 space minus space 7
space space space space space space space space space space space space space space rightwards double arrow space 7 space minus space 7
space space space space space space space space space space space space space space rightwards double arrow space 0 space
space So comma space the space correct space answer space is space left parenthesis straight b right parenthesis. end style

Solution 32

begin mathsize 11px style We space know space tan space left parenthesis 90 minus straight theta right parenthesis space equals space cot space straight theta space minus space circle enclose 1
and space space space cot space straight theta space equals space fraction numerator 1 over denominator tan space straight theta end fraction space space minus space circle enclose 2
from space circle enclose 1 space and space circle enclose 2
tan space left parenthesis 90 space minus space straight theta right parenthesis space equals space fraction numerator 1 over denominator tan space straight theta end fraction
rightwards double arrow tan space left parenthesis straight theta right parenthesis space tan space left parenthesis 90 space minus space straight theta right parenthesis space equals space 1
for space straight theta space equals space 5 to the power of straight o
space space space space space space rightwards double arrow box enclose tan space left parenthesis 5 to the power of straight o right parenthesis space tan space left parenthesis 85 to the power of straight o right parenthesis space equals space 1 end enclose space space space space minus space circle enclose 3
Also space space tan space 30 to the power of straight o space equals space fraction numerator 1 over denominator square root of 3 end fraction space space minus space circle enclose 4
To space find colon space tan space 5 to the power of straight o space cross times space tan space 30 to the power of straight o space cross times space 4 space tan space 85 to the power of straight o
space space space space space space space space rightwards double arrow space tan space 5 to the power of straight o space tan space 85 to the power of straight o space cross times space 4 space tan space 30 to the power of straight o
space space space space space space space space rightwards double arrow 1 space cross times space 4 space cross times space fraction numerator 1 over denominator square root of 3 end fraction
space space space space space space space space space rightwards double arrow fraction numerator 4 over denominator square root of 3 end fraction
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

RD Sharma Solution for Class 10 Mathematics Chapter 10 - Trigonometric Ratios Page/Excercise 10.59

Solution 33

begin mathsize 11px style We space know space tan space left parenthesis 90 minus straight theta right parenthesis space equals space cot space straight theta
space space space For space straight theta space equals space 35 to the power of straight o space rightwards double arrow box enclose tan space left parenthesis 55 to the power of straight o right parenthesis space equals space cot space left parenthesis 35 to the power of straight o right parenthesis end enclose space minus space circle enclose 1
Also space cot space 90 to the power of straight o space equals space 0 space space minus space circle enclose 2
To space find space colon negative space fraction numerator tan space 55 to the power of straight o over denominator cot space 35 to the power of straight o end fraction plus cot space 1 to the power of straight o space cot space 2 to the power of straight o space straight _ space straight _ thin space straight _ space straight _ thin space straight _ space cot space 90 to the power of straight o
space space space space space space space space space rightwards double arrow space from space circle enclose 1 space and space circle enclose 2
space space space space space space space space space rightwards double arrow space 1 space plus space 0 space equals space 1 end style

So, the correct option is (c).

Solution 34

begin mathsize 11px style ECB space is space straight a space straight space line
so space straight ϕ space plus space 90 to the power of straight o space plus space straight theta space equals space 180 to the power of straight o
space space space space space rightwards double arrow straight ϕ space plus space straight theta space equals space 90 to the power of straight o
space space space space space space rightwards double arrow straight ϕ space equals space 90 minus straight theta
space space space space space space rightwards double arrow cos space left parenthesis straight ϕ right parenthesis space equals space cos space left parenthesis 90 space minus straight theta right parenthesis
space space space space space space space space space space cos space straight ϕ space equals space sin space straight theta space space space space minus space circle enclose 1
In space increment space ABC
space space space space space space space space space sin space straight theta space equals space 4 over 5 space space space space space minus space circle enclose 2
from space circle enclose 1 space and space circle enclose 2
space space space space space space space space space space rightwards double arrow box enclose cos space straight ϕ space equals space 4 over 5 end enclose
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Solution 35

begin mathsize 11px style In space increment space ABD
AB squared space equals space BD squared space plus space AD squared
space space space space space space space space space space space equals space 3 squared space plus space 4 squared
space space space space space space space space space space space space equals space 25
space space space space AB space equals space 5 space cm
In space increment space ABC
AC squared space equals space AB squared space plus space BC squared
rightwards double arrow space AC squared space equals space 5 squared space plus space 12 squared
space space space space space space space space space space space space space space space equals space 169
rightwards double arrow AC space equals space 13 space cm
space cot space straight theta space equals space BC over AB space equals space 12 over 5
box enclose cot space straight theta space equals space 12 over 5 end enclose
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 10. Access the CBSE Class 10 Mathematics Chapter 10 - Trigonometric Ratios for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 10 - Trigonometric Ratios.

Text Book Solutions

CBSE X - Mathematics

This content is available for subscribed users only.

OR

Call us

1800-212-7858 (Toll Free) to speak to our academic expert.
OR

Let us get in touch with you