ICSE Solutions for Selina Concise Chapter 23 Trigonometrical Ratios of Standard Angles Class 9 Maths

Exercise 23(A)

1. find the value of:

(i) sin 30o cos 30o

(ii) tan 30o tan 60o

(iii) cos260o + sin230o

(iv) cosec260o – tan230o

(v) sin230o + cos230o + cot245o

(vi) cos260o + sec230o + tan245o.

Answer


2. find the value of :

(i) tan2 30o + tan2 45o + tan2 60o

(ii) tan45°/cosec30° + sec60°/cot45° - 5sin90°/2cos0°

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

Answer 


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) [(tan60° + 1)/(tan60° + 1)]2 = (1+cos30°)/(1-cos30°)

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

Answer


4. Without using table prove that:

(i) sin (2 × 30o) = 2tan30°/(1+tan2 30°)

(ii) cos (2 × 30o) = (1-tan2 30°)/(1+tan2 30°)

(iii) tan (2 ×30o) = 2tan30°/(1 - tan2 30°)

Answer


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


Answer

6. Prove that:

(i) sin 60o = 2 sin 30o cos 30o.

(ii) 4 (sin4 30o + cos4 60o-3 (cos2 45o – sin2 90o) = 2

Answer


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 θ = cot θ and 0o ≤ θ ≤ 90o, state the value of θ.

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

Answer


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

(ii) secθ. cotθ = cosecθ; write true or false.

(iii) For any angle, state the value of : sin2θ+ cos2θ

Answer


9. State for any acute angle whether:

(i) sin θ increases or decreases as θ increases:

(ii) cos θ increases or decreases as θ increases.

(iii) tan θ increases or decreases as θ decreases.

Answer

(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.


10. If √3 = 1.732, find (correct to two decimal place) the value of each of the following:

(i) sin 60o

(ii) 2/tan 30°

Answer


11. Evaluate:

(i) (cos 3A - 2cos 4A)/(sin 3A + 2sin 4A), when A = 15o.

(ii) [3 sin 3B + 2cos(2b + 5°)]/[2 cos3B - sin(2B - 10°); when B = 20o.

Answer



Exercise 23(B)

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) = (tan A - tan B)/(1+tan A.tan B)

Answer

2. If A =30o, then prove that:

(i) sin 2A = 2sin A cos A = 2tanA/(1+tan2A)

(ii) cos 2A = cos2A – sin2A = (1-tan2A)/(1+tan2A)

(iii) 2 cos2 A – 1 = 1 – 2 sin2A

(iv) sin 3A = 3 sin A – 4 sin3A.

Answer


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

Answer


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) (1- cos2A)/sin2A = tan A

(v) (1+sin2A + cos2A)/(sin A+ cos A) = 2 cos A.

(vi) 4 cos A cos (60o – A). cos (60o + A) = cos 3A

(vii) (cos3 A - cos 3A)/cosA  + (sin3 A - sin3A)/sinA  = 3

Answer


Exercise 23(C) 

1. Solve the following equations for A, if :

(i) 2 sin A = 1

(ii) 2 cos 2 A = 1

(iii) sin 3 A = √3/2

(iv) sec 2 A = 2

(v) √3 tan A = 1

(vi) tan 3 A = 1

(vii) 2 sin 3 A = 1

(viii) √3 cot 2 A = 1

Answer

(iv)



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

Answer


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.

Answer


4. If 4 cos2xo – 1 = 0 and 0 ≤ xo ≤ 90o, find:

(i) xo 

(ii) sin2 xo + cos2 xo

(iii) 1/cos2 x° - tan2 x°

Answer


5. If 4 sin2θ- 1= 0 and angle is less than 90o, find the value of and hence the value of cos2θ + tan2θ.

Answer


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

(i) sin A 

(ii) cos 2A

(iii) tan2A – 1/cos² A

Answer

sin 3A = 1 
⇒sin 3A = sin90°
⇒ 3A = 90°
⇒ A = 30°



7. If 2 cos 2A = √3 and A is acute, find:

(i) A 

(ii) sin 3A

(iii) sin2 (75o – A) + cos2 (45o +A)

Answer


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

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

Answer


9. From the given figure, find:

(i) cos xo

(ii) xo

(iii) 1/tan2x° - 1/sin2

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

Answer


10. Use the given figure to find:

(i) tan θo

(ii) θ

(iii) sin2θo – cos2θo

(iv) Use sinθo to find the value of x.

Answer


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

Answer


12. Solve for x:

(i) 2 cos 3x – 1 = 0

(ii) cos = 0

(iii) sin (x + 10o) = 1/2 

(iv) cos (2x – 30o) = 0

(v) 2 cos (3x – 15o) = 1 

(vi) tan2 (x – 5o) = 3

(vii) 3 tan2 (2x – 20o) = 1

(viii) cos [(x/2) + 10] =√3/2

(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

Answer



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

Answer


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.

Answer 


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

Answer 15:

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