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

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

Question 1

find the value of:

(i) sin 30o cos 30o

(ii) tan 30o tan 60o

(iii) cos2 60o + sin2 30o

(iv) cosec2 60o - tan2 30o

(v) sin2 30o + cos2 30o + cot2 45o

(vi) cos2 60o + sec2 30o + tan2 45o.

Solution 1

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Question 2

find the value of :

(i) tan2 30o + tan2 45o + tan2 60o

(ii)

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

Solution 2

(i)

(ii)

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

 

Question 3

Prove that:

(i) sin 60o cos 30o + cos 60o. sin 30o = 1

(ii) cos 30o. cos 60o - sin 30o. sin 60o = 0

(iii) cosec2 45o - cot2 45o = 1

(iv) cos2 30o - sin2 30o = cos 60o.

(v)

(vi) 3 cosec2 60o - 2 cot2 30o + sec2 45o = 0.

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

Question 4

prove that:

(i) sin (2 30o) =

(ii) cos (2 30o) =

(iii) tan (2 30o) =

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

 

 

Question 5

ABC is an isosceles right-angled triangle. Assuming of AB = BC = x, find the value of each of the following trigonometric ratios:

(i) sin 45o

(ii) cos 45o

(iii) tan 45o

Solution 5

Given that AB = BC = x

(i)

(ii)

(iii)

Question 6

Prove that:

(i) sin 60o = 2 sin 30o cos 30o.

(ii) 4 (sin4 30o + cos4 60o)

-3 (cos2 45o - sin2 90o) = 2

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

Question 7

(i) If sin x = cos x and x is acute, state the value of x.

(ii) If sec A = cosec A and 0o A 90o, state the value of A.

(iii) If tan = cot and 0o 90o, state the value of.

(iv) If sin x = cos y; write the relation between x and y, if both the angles x and y are acute.

Solution 7

(i)

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

 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

Question 8

(i) If sin x = cos y, then x + y = 45o ; write true of false.

(ii) sec. Cot = cosec; write true or false.

(iii) For any angle , state the value of :

Sin2 + cos2.

Solution 8

(i)

if x and y are acute angles,

is false.

 

(ii)

 


Sec. Cot = cosec is true


(iii)


Question 9

State for any acute angle whether:

(i) sin increases or decreases as increases:

(ii) cos increases or decreases as increases.

(iii) tan increases or decreases as decreases.

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.

Question 10

If = 1.732, find (correct to two decimal place) the value of each of the following:

(i) sin 60o (ii)

Solution 10

(i)

(ii)

Question 11

Evaluate:

(i) , when A = 15o.

(ii) ; when B = 20o.

Solution 11

(i) Given that A=

(ii) Given that B=

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

Question 1

Given A = 60o and B = 30o, prove that:

(i) sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos A cos B - sin A sin B

(iii) cos (A - B) = cos A cos B + sin A sin B

(iv) tan (A - B) =

Solution 1

Given A = 60o and B = 30o

(i)

(ii)

(iii)

(iv)

Question 2

If A =30o, then prove that:

(i) sin 2A = 2sin A cos A =

(ii) cos 2A = cos2A - sin2A

=

(iii) 2 cos2 A - 1 = 1 - 2 sin2A

(iv) sin 3A = 3 sin A - 4 sin3A.

Solution 2

Given A=

(i)

(ii)

(iii)

(iv)

Question 3

If A = B = 45o, show that:

(i) sin (A - B) = sin A cos B - cos A sin B

(ii) cos (A + B) = cos A cos B - sin A sin B

Solution 3

Given that A = B = 45o

(i)

(ii)

Question 4

If A = 30o; show that:

(i) sin 3 A

= 4 sin A sin (60o - A) sin (60o + A)

(ii) (sin A - cos A)2 = 1 - sin 2A

(iii) cos 2A = cos4 A - sin4 A

(iv)

(v) = 2 cos A.

(vi) 4 cos A cos (60o - A). cos (60o + A)

= cos 3A

(vii)

Solution 4

Given that A = 30o

(i)

 

(ii)

 

(iii)

(iv)

(v)

(vi)

(vii)

 

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

Question 1

Solve the following equations for A, if :

(i) 2 sin A = 1 (ii) 2 cos 2 A = 1

(iii) sin 3 A = (iv) sec 2 A = 2

(v) tan A = 1 (vi) tan 3 A = 1

(vii) 2 sin 3 A = 1 (viii) cot 2 A = 1

Solution 1

(i)

(ii)

(iii)

(iv)

(V)

(vi)

(vii)

(viii)

Question 2

Calculate the value of A, if :

(i) (sin A - 1) (2 cos A - 1) = 0

(ii) (tan A - 1) (cosec 3A - 1) = 0

(iii) (sec 2A - 1) (cosec 3A - 1) = 0

(iv) cos 3A. (2 sin 2A - 1) = 0

(v) (cosec 2A - 2) (cot 3A - 1) = 0

Solution 2

(i)

(ii)

(iii)

(iv)

(v)

Question 3

If 2 sin xo - 1 = 0 and xo is an acute angle; find :

(i) sin xo (ii) xo (iii) cos xo and tan xo.

Solution 3

(i)

 

(ii)

(iii)

Question 4

If 4 cos2 xo - 1 = 0 and 0 xo 90o, find:

(i) xo (ii) sin2 xo + cos2 xo

(iii)

Solution 4

(i)

(ii)

(iii)

Question 5

If 4 sin2 - 1= 0 and angle is less than 90o, find the value of and hence the value of cos2 + tan2.

Solution 5

Question 6

If sin 3A = 1 and 0 A 90o, find:

(i) sin A (ii) cos 2A

(iii) tan2A -

Solution 6

(i)

(ii)

(iii)

Question 7

If 2 cos 2A = and A is acute, find:

(i) A (ii) sin 3A

(iii) sin2 (75o - A) + cos2 (45o +A)

Solution 7

(i)

 

(ii)

(iii)

Question 8

(i) If sin x + cos y = 1 and x = 30o, find the value of y.

(ii) If 3 tan A - 5 cos B= and B = 90o, find the value of A.

Solution 8

(i)

Given that x = 30o

 

(ii)

Given that B = 90o

Question 9

From the given figure, find:

(i) cos xo(ii) xo

(iii)

(iv) Use tan xo, to find the value of y.

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

Question 10

Use the given figure to find:

(i) tan (ii) (iii) sin2o - cos2o

(iv) Use sin o to find the value of x.

Solution 10

(i)

(ii)

(iii)

(iv)

Question 11

Find the magnitude of angle A, if:

(i) 2 sin A cos A - cos A - 2 sin A + 1 = 0

(ii) tan A - 2 cos A tan A + 2 cos A - 1 = 0

(iii) 2 cos2 A - 3 cos A + 1 = 0

(iv) 2 tan 3A cos 3A - tan 3A + 1 = 2 cos 3A

Solution 11

(i)

 

(ii)

(iii)

(iv)

Question 12

Solve for x:

(i) 2 cos 3x - 1 = 0 (ii) cos = 0

(iii) sin (x + 10o) = (iv) cos (2x - 30o) = 0

(v) 2 cos (3x - 15o) = 1 (vi) tan2 (x - 5o) = 3

(vii) 3 tan2 (2x - 20o) = 1

(viii) cos

(ix) sin2 x + sin2 30o = 1

(x) cos2 30o + cos2 x = 1

(xi) cos2 30o + sin2 2x = 1

(xii) sin2 60o + cos2 (3x- 9o) = 1

Solution 12

(i)

 

(ii)

 

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

Question 13

If 4 cos2 x = 3 and x is an acute angle; find the value of :

(i) x (ii) cos2 x + cot2 x

(iii) cos 3x (iv) sin 2x

Solution 13

(i)

(ii)

(iii)

(iv)

Question 14

In ABC, B = 90o, AB = y units, BC = units, AC = 2 units and angle A = xo, find:

(i) sin xo (ii) xo (iii) tan xo

(iv) use cos xo to find the value of y.

Solution 14

(i)

From

(ii)

(iii)

(iv)

Question 15

If 2 cos (A + B) = 2 sin (A - B) = 1; find the values of A and B.

Solution 15

Adding (1) and (2)

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