Selina Chapter 22 Trigonometrical Ratios ICSE Solutions Class 9 Maths

ICSE Solutions for Selina Concise Chapter 22 Trigonometrical Ratios Class 9 Maths

Exercise 22(A)

1. From the following figure, find the values of :

(i) sin A

(ii) cos A

(iii) cot A

(iv) sec C

(v) cosec C

(vi) tan C.

Answer

Given angle ABC = 90°

⇒ AC² = AB²  + BC²  (AC is hypotenuse)
⇒ AC² = 3²  + 4²  
∴ AC² = 9+16 = 25 and AC = 5
(i) sin A = perpendicular/hypotenuse = BC/AC = 4/5
(ii) cos A = base/hypotenuse = AB/AC = 3/5 
(iii) cot A = base/perpendicular = AB/BC = 3/4
(iv) sec C = hypotenuse/base = AC/BC = 5/4
(v) cosec C = hypotenuse/perpendicular = AC/AB = 5/3
(vi) tan C = perpendicular/base = AB/BC = 3/4

2. Form the following figure, find the values of :

(i) cos B

(ii) tan C

(iii) sin²B + cos²B

(iv) sin B. cos C + cos B. sin C

Answer

Given angle BAC = 90°

⇒ BC²  = AB²  + AC²  (BC is hypotenuse)
⇒ 17² = 8² + AC²  
∴ AC²  = 289 - 64 = 225 and AC = 15 
(i) cos B = base/hypotenuse = AB/BC = 8/17
(ii) tan C = perpendicular/base = AB/AC = 8/15
(iii) sin B = perpendicular/hypotenuse = AC/BC = 15/17
cos B = base/hypotenuse = AB/BC = 8/17
sin² B + cos² B = (15/17)²  + (8/17)²  

= (225 + 64)/289
= 289/289
= 1
(iv) sin B = perpendicular/hypotenuse = AC/BC = 15/17
cos B = base/hypotenuse = AB/BC = 8/17
sin C = perpendicular/hypotenuse = AB/BC = 8/17
cos C = base/hypotenuse = AC/BC = 15/17
sinB. cosC + cosB. sinC = (15/17)×(15/17) + (8/17)×(8/17)
= (225+64)/289
= 289/289
= 1 

3. From the following figure, find the values of :

(i) cos A (ii) cosec A

(iii) tan²A – sec²A (iv) sin C

(v) sec C (vi) cot² C – 1/sin² C

Answer

Consider the diagram as 

Given,

angle ADB = 90° and BDC = 90°
⇒ AB²  = AD²  + BD²  (AB is hypotenuse is △ABD)
⇒ AB²  = 3²  + 4²  
∴ AB²  = 9+ 16 = 25 and AB = 5 
⇒ BC²  = BD²  + DC²  (BC is hypotenuse in △BDC)
⇒ DC²  = 12²  - 4²  
∴ DC²  = 144 - 16 = 128 and DC = 8√2
(i) cos A = base/hypotenuse = AD/AB = 3/5 
(ii) cosec A = hypotenuse/perpendicular  = AB/BD = 5/4 
(iii) tan A = perpendicular/base = BD/AD = 4/3
sec A = hypotenuse/base = AB/AD = 5/3 
tan²  A - se²  A = (4/3)²  - (5/3)²  
= (16/9) - (25/9)
= -9/9
= -1
(iv) sin C = perpendicular/hypotenuse = BD/BC = 4/12 = 1/3 
(v) sec C = hypotenuse/base = BC/DC = 12/8√2 = 3/2√2 = 3√2/4
(vi) cot C = base/perpendicular = DC/BD = 8√2/4 = 2√2
sin C = perpendicular/hypotenuse = BD/BC = 4/12 = 1/3
cot² C - 1/sin² C = (2√2)²  - 1/(1/3)²  
= 8- 9
= -1

4. From the following figure, find the values of :

(i) sin B 

(ii) tan C

(iii) sec² B – tan²B 

(iv) sin²C + cos²C

Answer

Given angle ADB = 90° and ADC = 90°

5. Given: sin A = 3/5 , find :

(i) tan A

(ii) cos A

Answer

Consider the diagram below : 

6. From the following figure, find the values of :

(i) sin A

(ii) sec A

(iii) cos² A + sin²A

Answer

Given angle ABC = 90° in the figure 

7. Given: cos A = 5/13

Evaluate: (i)( sin A – Cot A )/2 Tan A

(ii) cot A+ (1 / cos A)

Answer

Consider the diagram below : 

8. Given: sec A =29/11, evaluate: sin A – 1/ Tan A

Answer

Consider the diagram below : 

9. Given: tan A =4/3, find : cosecA/(cot A-sec A)

Answer

Consider the diagram below : 

10. Given: 4 cot A = 3 find;

(i) sin A

(ii) sec A

(iii) cosec² A – cot²A.

Answer

Consider the diagram below : 

11. Given: cos A = 0.6; find all other trigonometrical ratios for angle A.

Answer

Consider the diagram below : 

12. In a right-angled triangle, it is given that A is an acute angle and tan A =. 5/12

find the value of :

(i) cos A

(ii) sin A

(iii) (cos A + sin A)/(cos A – Sin A)

Answer

Consider the diagram below : 

tan A = 5/12 

13. Given: sin θ = p/q

Find cos θ + sin θ in terms of p and q.

Answer

Consider the diagram below : 

14. If cos A =1/2 and sin B = 1/√2  , find the value of : .

tan A–tan B/(1+tan A tan B)

Are angles A and B from the same triangle? Explain.

Answer

Consider the diagram below : 

15. If 5 cot θ = 12, find the value of : Cosec θ + sec θ

Answer

Consider the diagram below : 

16. If tan x = 1 1/3, find the value of : 4 sin²x – 3 cos²x + 2

Answer

Consider the diagram below : 

17. If cosec θ= √5, find the value of:

(i) 2 – sin²θ- cos²θ

(ii) 2 + (1+sin²θ) - (cos²θ/sin²θ)

Answer

Consider the diagram below : 

18. If sec A =√2 , find the value of : (3cos² A + 5 tan² A)/(4tan⁴ A - sin² A)

Answer

Consider the diagram below : 

19. If cot = 1; find the value of: 5 tan²θ + 2 sin²θ – 3

Answer

Consider the diagram below : 

20. In the following figure:

AD⟂BC, AC = 26 CD = 10, BC = 42,

∠DAC = x and B = y.

Find the value of :

(i) cot x

(ii) 1/sin² y - 1/tan² y

(iii) 6/cos x - 5/cos y + 8 tan y.

Answer

Given,

∠DAC = 90° and ∠ADB = 90° in the figure 

= 9/9
= 1 

Exercise 22(B)

1. From the following figure, find:

(i) y

(ii) sin x^o

(iii) (sec x^o – tan x^o) (sec x^o + tan x^o)

Answer

Consider the given figure 

(i) Since the triangle is a right angled triangle, so using Pythagorean Theorem

2. Use the given figure to find:

(i) sin x^o

(ii) cos y^o

(iii) 3 tan x^o – 2 sin y^o + 4 cos y^o.

Answer

Since, the triangle is a right angled triangle, so using Pythagorean Theorem 

= 12/5

3. In the diagram, given below, triangle ABC is right-angled at B and BD is perpendicular to AC.

Find:

(i) cos ∠DBC

(ii) cot ∠DBA

Answer

Since the triangle is a right angled triangle.

So using Pythagorean Theorem,

AC² = 5² + 12²

⇒ AC² = 25 + 144

⇒ AC² = 169

⇒ AC = 13

In △CBD and △CBA, the C is common to both the triangles, ∠CDB = ∠CBA = 90°, so therefore ∠CBD = ∠CAB.

Therefore, △CBD and △CBA are similar triangles according to AAA Rule.

4. In the given figure, triangle ABC is right-angled at B. D is the foot of the perpendicular from B to AC. Given that BC = 3 cm and AB = 4 cm. find:

(i) tan DBC

(ii) sin DBA

Answer

Consider the given figure 

Since the triangle is a right angled triangle, so using Pythagorean Theorem. 

5. In triangle ABC, AB = AC = 15 cm and BC = 18 cm, find cos ABC.

Answer

Consider the figure below 

In the isosceles △ABC, AB = AC = 15 cm and BC = 18cm the perpendicular drawn from angle A to the side BC divides the side BC into two equal parts BD = DC = 9 cm 

6. In the figure given below, ABC is an isosceles triangle with BC = 8 cm and AB = AC = 5 cm. Find:

(i) sin B

(ii) tan C

(iii) sin² B + cos²B 

(iv) tan C – cot B

Answer

Consider the figure below 

In the isosceles △ABC, AB = AC = 5cm and BC = 8cm the perpendicular drawn from angle A to the side BC divides the side BC into two equal parts  BD = DC = 4cm 
Since ∠ADB = 90°

Therefore,
tan C - cot B
= 3/4 - 4/3
= -7/12

7. In triangle ABC; ABC = 90^o, CAB = x^o, tan x^o = and BC = 15 cm. Find the measures of AB and AC.

Answer

Consider the figure 

= 25 cm

8. Using the measurements given in the following figure:

(i) Find the value of sin and tan.

(ii) Write an expression for AD in terms of

Answer

Consider the figure 

A perpendicular is drawn from D to the side AB at point E which makes BCDE is a rectangle. 
Now in right angled triangle BCD using Pythagorean Theorem 

9. In the given figure;

BC = 15 cm and sin B =.

(i) Calculate the measure of AB and AC.

(ii) Now, if tan ADC = 1; calculate the measures of CD and AD.

Also, show that: tan²B – 1/cos² B = -1

Answer

Given  
Sin B = 4/5 
i.e.   perpendicular /hypotenuse  = AC/AB = 4/5
Therefore if length of perpendicular  = 4x, length of hypotenuse = 5x 
Since, 

Therefore if length of perpendicular = x, length of hypotenuse  = x 
Since,

10. If sin A + cosec A = 2;

Find the value of sin² A + cosec² A.

Answer

sin A + cosec A = 2 
Squaring both sides 

11. If tan A + cot A = 5;

Find the value of tan² A + cot² A.

Answer

tan A + cot A = 5 
Squaring both sides

12. Given: 4 sin = 3 cos ; find the value of:

(i) sin 

(ii) cos

(iii) cot² – cosec².

(iv) 4 cos²– 3 sin²+ 2

Answer

13. Given : 17 cos = 15;

Find the value of : tan + 2 sec.

Answer

Consider the diagram below : 

= 2(4/5)

14. Given : 5 cos A – 12 sin A = 0; evaluate :
(sin A + cos A)/(2cos A - sin A)

Answer

5 cos A - 12 sin A = 0 
⇒ 5 cos A = 12 sin A 

15. In the given figure; C = 90^o and D is mid-point of AC. Find

(i) tan CAB/tan CDB

(ii) tan ABC/tan CDB

Answer

Since D is mid - point of AC so AC = 2DC 

16. If 3 cos A = 4 sin A, find the value of :

(i) cos A

(ii) 3 – cot² A + cosec²A.

Answer

Consider the diagram below : 

= 36/9 
= 4 

17. In triangle ABC, B = 90^o and tan A = 0.75. If AC = 30 cm, find the lengths of AB and BC.

Answer

Consider the figure 

18. In rhombus ABCD, diagonals AC and BD intersect each other at point O.

If cosine of angle CAB is 0.6 and OB = 8 cm, find the lengths of the the side and the diagonals of the rhombus.

Answer

Consider the figure 

Since the sides of a rhombus are equal so the length of the side of the rhombus = 10cm 
The diagonals are 
BD = 8×2 = 16 cm 
AC = 6×2 = 12 cm 

19. In triangle ABC, AB = AC = 15 cm and BC = 18 cm. Find:

(i) cos B

(ii) sin C

(iii) tan² B – sec² B + 2

Answer

Consider the figure below 

20. In triangle ABC, AD is perpendicular to BC. sin B = 0.8, BD = 9 cm and tan C = 1. Find the length of AB, AD, AC and DC.

Answer

Consider the figure below 

21. Given q tan A = p, find the value of :
(p sinA - q cosA)/(p sinA + q cosA)

Answer

22. If sin A = cos A, find the value of 2 tan²A – 2 sec² A + 5.

Answer

Consider the figure  

23. In rectangle ABCD, diagonal BD = 26 cm and cotangent of angle ABD = 1.5. Find the area and the perimeter of the rectangle ABCD.

Answer

Consider the diagram 

24. If 2 sin x = , evaluate.

(i) 4 sin³ x – 3 sin x.

(ii) 3 cos x – 4 cos³ x.

Answer

Consider the figure 

25. If sin A = √3/2 and cos B = √3/2, find the value of : 
(tan A - tan B)/(1+tanA . tan B) . 

Answer

Consider the diagram below : 

26. Use the information given in the following figure to evaluate:

10/sin x + 6/sin y -  6 cot y

Answer

Consider the given diagram as 

27. If sec A = √2 , find: (3 cot² A + 2 sin² A)/(tan² A - cos² A) . 

Answer

Consider the figure 

= 4/(1/2)
= 8 

28. If 5 cosθ = 3, evaluate : (cosecθ - cotθ)/(cosecθ + cotθ) . 

Answer

29. If cosec A + sin A = 5,1/5 find the value of cosec²A + sin²A.

Answer

30. If 5 cosθ = 6 sinθ ; evaluate:

(i) tan θ 

(ii) (12sinθ - 3cosθ)/(12sinθ + 3cosθ)

Answer