Request a call back

Join NOW to get access to exclusive study material for best results

Class 9 SELINA Solutions Maths Chapter 23 - Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios]

Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios] Exercise Ex. 23(A)

Solution 1

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Solution 2

(i)

(ii)

(iii) 3 sin2 30o + 2 tan2 60o - 5 cos2 45o

 

Solution 3

(i) LHS=sin 60o cos 30o + cos 60o. sin 30o

= 

(ii) LHS=cos 30o. cos 60o - sin 30o. sin 60o

==RHS

(iii) LHS= cosec2 45o - cot2 45o

==RHS

(iv) LHS= cos2 30o - sin2 30o

==RHS

(v) LHS=

==RHS

(vi) LHS=

==RHS

Solution 4

(i)

R H S equals
fraction numerator 2 space tan space 30 degree over denominator 1 plus tan squared 30 degree end fraction equals fraction numerator 2 cross times begin display style fraction numerator 1 over denominator square root of 3 end fraction end style over denominator 1 plus open parentheses begin display style fraction numerator 1 over denominator square root of 3 end fraction end style close parentheses squared end fraction equals fraction numerator begin display style fraction numerator 2 over denominator square root of 3 end fraction end style over denominator 1 plus begin display style 1 third end style end fraction equals fraction numerator fraction numerator 2 over denominator square root of 3 end fraction over denominator begin display style 4 over 3 end style end fraction equals fraction numerator square root of 3 over denominator 2 end fraction
L H S equals sin space left parenthesis 2 cross times 30 degree right parenthesis equals sin space 60 degree equals fraction numerator square root of 3 over denominator 2 end fraction
therefore L H S thin space equals R H S

 



(ii)

R H S comma
fraction numerator 1 minus tan squared 30 degree over denominator 1 plus tan squared 30 degree end fraction equals fraction numerator begin display style 1 minus 1 third end style over denominator 1 plus begin display style 1 third end style end fraction equals 1 half
L H S comma
cos space left parenthesis 2 cross times 30 degree right parenthesis equals cos space 60 degree equals 1 half
L H S thin space equals R H S




(iii)

R H S comma
fraction numerator 2 space tan space 30 degree over denominator 1 minus tan squared 30 degree end fraction equals fraction numerator 2 begin display style fraction numerator 1 over denominator square root of 3 end fraction end style over denominator 1 minus begin display style 1 third end style end fraction equals 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 equals square root of 3
L H S comma
tan space left parenthesis 2 cross times 30 degree right parenthesis equals tan space 60 degree equals square root of 3
L H S equals R H S

 

 

Solution 5

Given that AB = BC = x

(i)

(ii)

(iii)

Solution 6

left parenthesis i right parenthesis space L H S equals sin space 60 degree equals fraction numerator square root of 3 over denominator 2 end fraction
R H S equals 2 space sin space 60 degree cos space 60 degree equals 2 cross times fraction numerator square root of 3 over denominator 2 end fraction cross times 1 half equals fraction numerator square root of 3 over denominator 2 end fraction
L H S equals R H S

left parenthesis i i right parenthesis space L H S equals 4 left parenthesis sin to the power of 4 30 degree plus cos to the power of 4 60 degree right parenthesis minus 3 open parentheses cos squared 45 degree minus sin squared 90 degree close parentheses
equals 4 open square brackets open parentheses 1 half close parentheses to the power of 4 plus open parentheses 1 half close parentheses to the power of 4 close square brackets minus 3 open square brackets open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses squared plus open parentheses 1 close parentheses to the power of 4 close square brackets
equals 4 open square brackets 1 over 16 plus 1 over 16 close square brackets minus 3 open square brackets 1 half minus 1 close square brackets equals fraction numerator 4 cross times 2 over denominator 16 end fraction plus 3 cross times 1 half equals 2
R H S equals 2
L H S equals R H S

Solution 7

(i)

The angle, x is acute and hence we have, 0 < x

 W e space k n o w space t h a t
cos squared x plus sin squared x equals 1
rightwards double arrow 2 sin squared x equals 1 space space space space space space open square brackets sin c e space cos x equals sin x close square brackets
rightwards double arrow sin x equals fraction numerator 1 over denominator square root of 2 end fraction
rightwards double arrow x equals 45 degree

 

(ii)

 

(iii)

 

(iv)

sin space straight x equals cos space straight y equals sin space open parentheses 90 degree minus straight y close parentheses
If space straight x space and space straight y space are space acute space angles comma
straight x equals 90 degree minus straight y
rightwards double arrow straight x plus straight y equals 90 degree
Hence space straight x space and space straight y space are space complementary space angles

Solution 8

(i)

if x and y are acute angles,

is false.

 

(ii)

 


Sec. Cot = cosec is true


(iii)


Solution 9

(i)

For acute angles, remember what sine means: opposite over hypotenuse. If we increase the angle, then the opposite side gets larger. That means "opposite/hypotenuse" gets larger or increases.

(ii)

For acute angles, remember what cosine means: base over hypotenuse. If we increase the angle, then the hypotenuse side gets larger. That means "base/hypotenuse" gets smaller or decreases.

(iii)

For acute angles, remember what tangent means: opposite over base. If we decrease the angle, then the opposite side gets smaller. That means "opposite /base" gets decreases.

Solution 10

(i)

(ii)

Solution 11

(i) Given that A=

(ii) Given that B=

Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios] Exercise Ex. 23(B)

Solution 1

Given A = 60o and B = 30o

(i)

(ii)

(iii)

(iv)

Solution 2

Given A=

(i)

(ii)

(iii)

(iv)

Solution 3

Given that A = B = 45o

(i)

(ii)

Solution 4

Given that A = 30o

(i)

 

(ii)

 

(iii)

(iv)

(v)

(vi)

(vii)

 

Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios] Exercise Ex. 23(C)

Solution 1

(i)

(ii)

(iii)

(iv)

(V)

(vi)

(vii)

(viii)

Solution 2

(i)

(ii)

(iii)

(iv)

(v)

Solution 3

(i)

 

(ii)

(iii)

Solution 4

(i)

(ii)

(iii)

Solution 5

Solution 6

(i)

(ii)

(iii)

Solution 7

(i)

 

(ii)

(iii)

Solution 8

(i)

Given that x = 30o

 

(ii)

Given that B = 90o

Solution 9

(i)

(ii)

(iii)

(iv)

W e space k n o w space t h a t space tan x degree equals fraction numerator A B over denominator B C end fraction
rightwards double arrow tan x degree equals y over 10
rightwards double arrow y equals 10 tan x degree
rightwards double arrow y equals 10 tan 60 degree
rightwards double arrow y equals 10 square root of 3

Solution 10

(i)

(ii)

(iii)

(iv)

Solution 11

(i)

 

(ii)

(iii)

(iv)

Solution 12

(i)

 

(ii)

 

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

Solution 13

(i)

(ii)

(iii)

(iv)

Solution 14

(i)

From

(ii)

(iii)

(iv)

Solution 15

Adding (1) and (2)

Get FREE Sample Mock Papers for Final Exams
Download Now
×