Class 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.1 | Set 1

Chapter 5 of the Class 10 RD Sharma Mathematics textbook, "Trigonometric Ratios," introduces students to the basic trigonometric functions and their applications. Exercise 5.1 focuses on solving problems involving the fundamental trigonometric ratios of angles in a right-angled triangle.

RD Sharma Solutions for Class 10 - Mathematics - Chapter 5 Trigonometric Ratios - Exercise 5.1 | Set 1

This section provides detailed solutions for Exercise 5.1 from Chapter 5 of the Class 10 RD Sharma Mathematics textbook. The exercise includes problems that require students to apply the basic trigonometric ratios—sine, cosine, and tangent—to find missing angles and side lengths in right-angled triangles. Solutions are presented step-by-step to help students understand and master trigonometric calculations.

Question 1. In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.

(i) sinA = 2/3 

Solution:

sinA = 2/3 = Perpendicular/Hypotenuse                                                        

Draw a right-angled △ABC in which ∠B is = 90°                       

👁 Image

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(3)² = (2)² + (BC)²

 9 = 4 + BC² 

 BC² = 9 - 4 = 5

BC = √5 units         

Now,

cosA = Base/Hypotenuse = BC/AC = √5/3   

tanA = Perpendicular/Base = AB/BC = 2/√5         

cotA = 1/tanA = √5/2          

secA = 1/cosA = 3/√5         

cosecA = 1/sinA = 3/2          

(ii) cosA = 4/5 

Solution:

cosA = 4/5 = Base/Hypotenuse

Draw a right-angled △ABC in which ∠B is = 90° 

👁 Image

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(5)² = (AB)² + (4)²

 25 = AB² + 16 

AB² = 25 - 16 = 9

AB = √9

= 3 units     

Now, 

sinA = Perpendicular/Hypotenuse = AB/AC =3/5        

tanA = Perpendicular/Base = AB/BC = 3/4        

cotA = 1/tanA = 4/3

secA = 1/cosA = 5/4

cosecA = 1/sinA =5/3

(iii) tanθ = 11/1

Solution:

tanθ = 11/1 = Perpendicular/Base

Draw a right-angled △ABC in which ∠B is = 90°

👁 Image

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

AC² = (11)² + (1)²

AC² = 121  + 1

= 122

AC = √122units  

 Now,

sinθ = Perpendicular/Hypotenuse = AB/AC = 11/ √122

cosθ = Base/Hypotenuse = BC/AC = 1/√122

cotθ = 1/tanθ = 1/11         

secθ = 1/cosθ = √122/1

cosecθ = 1/sinθ = √122/11     

(iv) sinθ = 11/15

Solution:

sinθ = 11/15 = Perpendicular/Hypotenuse

Draw a right-angled △ABC in which ∠B is = 90°

👁 Image

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(15)² = (11)² + (BC)²

225 = 121 + (BC)²

(BC)² = 104

BC = 2√26

Now,

cosθ = Base/Hypotenuse = BC/AC = 2√26/15              

tanθ = AB/BC = 11/ 2√26      

cotθ = 1/tanθ = 2√26/11    

secθ = 1/cosθ = 15/ 2√26     

cosecθ = 1/sinθ = 15/11             

(v) tan α = 5/12

Solution:

tan α = 5/12 = Perpendicular/Base

Draw a right-angled △ABC in which ∠B is = 90°

👁 Image

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(AC)² = (12)² + (25)²

(AC)² = 144 + 25 

(AC)² = 169

AC = √169 = 13 units  

Now,   

sin α = Perpendicular/Hypotenuse = AB/AC = 5/13     

cos α = Base/Hypotenuse = BC/AC = 12/13           

cot α = 1/tan α = 12/5     

sec α = 1/cos α = 13/12      

cosec α = 1/sin α = 13/5  

(vi) sinθ = √3/2

Solution:

sinθ = √3/2 = Perpendicular/Hypotenuse 

Draw a right-angled △ABC in which ∠B is = 90°

👁 Image

Using Pythagoras Theorem, in △ ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(2)² = (√3​)² + (BC)²

4 = 3 + (BC)²

(BC)² = 4 - 3 = 1 

BC = 1 units

Now,

cosθ = Base/Hypotenuse = BC/AC = 1/2         

tanθ = AB/BC = √3/1      

cotθ = 1/tanθ = 1/√3

secθ = 1/cosθ = 2/1    

cosecθ = 1/sinθ = 2/√3        

(vii) cosθ = 7/25

Solution:

 cosθ = 7/25 = Base/Hypotenuse  

Draw a right-angled △ABC in which ∠B is = 90°

👁 Image

Using Pythagoras Theorem, in △ ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(25)² = (AB)² + (7)²

625 = (AB)² + 49

(AB)² = 625 - 49 = 576

AB = √576 = 24 units

Now,

sinθ = Perpendicular/Hypotenuse = AB/AC = 24/25         

tanθ = Perpendicular/Base = AB/BC = 24/7     

cotθ = 1/tanθ = 7/24         

secθ = 1/cosθ = 25/7        

cosecθ = 1/sinθ = 25/24                   

(viii) tanθ = 8/15

Solution:

tanθ = 8/15 = Perpendicular/Base  

Draw a right-angled △ABC in which ∠B is = 90°

👁 Image

Using Pythagoras Theorem, in △ ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(AC)² = (8)² + (15)²

(AC)² = 64 + 225 

AC = √289 = 17

Now,

sinθ = Perpendicular/Hypotenuse = AB/AC = 8/17       

cosθ = Base/Hypotenuse = BC/AC = 15/17       

cotθ = 1/tanθ = 15/8       

secθ = 1/cosθ = 17/15      

cosecθ = 1/sinθ = 17/8      

(ix) cotθ = 12/5

Solution:

cotθ = 12/5 = Base/Perpendicular     

Draw a right-angled △ABC in which ∠B is = 90°

👁 Image

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(AC)² = (5)² + (12)²

(AC)² = 25 + 144 

(AC)² = 169

AC = √169 = 13 units 

Now,

sinθ = Perpendicular/Hypotenuse = AB/AC = 5/13     

cosθ = Base/Hypotenuse = BC/AC = 12/13     

tanθ = 1/tanθ = 5/12     

secθ = 1/cosθ = 13/12    

cosecθ = 1/sinθ = 13/5   

(x) secθ = 13/5

Solution:

secθ = 13/5 = Hypotenuse/Base 

Draw a right-angled △ABC in which ∠B is = 90°

👁 Image

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(13)² = (AB)² + (5)²

169 = (AB)² + 25

(AB)² = 169 - 25 = 144

AB = √144 = 12 units

Now,

sinθ = Perpendicular/Hypotenuse = AB/AC = 12/13      

tanθ = Perpendicular/Base = AB/BC = 12/5    

cotθ = 1/tanθ = 5/12     

cosθ = 1/secθ = 5/13    

cosecθ = 1/sinθ = 13/12     

(xi) cosecθ = √10

Solution:

cosecθ = √10/1 = Hypotenuse/Perpendicular 

Draw a right-angled △ABC in which ∠B is = 90°

👁 Image

Using Pythagoras Theorem, in △ ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(√10)² = (1)² + (BC)²

10 = 1 + (BC)²

(BC)² = 10 - 1 = 9

BC = √9 = 3

Now,

sinθ = Perpendicular/Hypotenuse = AB/AC = 1/√10     

cosθ = Base/Hypotenuse = BC/AC = 3/√10   

tanθ = Perpendicular/Hypotenuse = AB/BC = 1/3   

cotθ = 1/tanθ = 3/1 = 3           

secθ = 1/cosθ = √10/3        

(xii) cosθ = 12/15  

Solution:

cosθ = 12/15 = Base/Hypotenuse

Draw a right-angled △ABC in which ∠B is = 90°

👁 Image

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(15)² = (AB)² + (12)²

225 = (AB)² + 144

(AB)² = 225 - 144 = 81

AB = √81 = 9 units   

Now,

sinθ = Perpendicular/Hypotenuse = AB/AC = 9/15

tanθ = Perpendicular/Base = AB/BC = 9/12    

cotθ = 1/tanθ = 12/9    

secθ = 1/cosθ = 15/12      

cosecθ = 1/sinθ = 15/9      

Question 2. In ΔABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine

(i) sin A, cos A                                               

(ii) sin C, cos C

Solution:

Given:

In right-angled ΔABC,

AB = 24 cm, BC = 7 cm. ∠B = 90°

👁 Image

Using Pythagoras Theorem 

AC² = AB² + BC²

AC² = 24² + 7² = 576 + 49

AC² = 625

AC = √625 = 25cm

Now,

(i) sinA = BC/AC = 7/25

cosA = AB/AC = 24/25    

(ii) sinC = AB/AC = 24/25 

cosC = BC/AC = 7/25    

Question 3. In the figure, find tan P and cot R. Is tan P = cot R?

Solution:

Using Pythagoras Theorem

PR² = PQ² + QR²

13² = 12² + QR²

QR² = 169 - 144 = 25

QR = √25 = 5 cm

Now,

tan P = Perpendicular/Base = QR/PQ = 5/2  

cot R = Base/Perpendicular = QR/PQ = 5/2  

Yes, tanP = cot R

Question 4. If sin A = 9/41, compute cos A and tan A.

Solution:

Given, sinA = 9/41 = Perpendicular/Hypotenuse

Draw a △ ABC where ∠B = 90°, BC = 9, AC = 41 

👁 Image

Using Pythagoras Theorem

AC² = AB² + BC²

BC² = 41² - 9² = 1681 - 81

BC² = 1600

BC = √1600 = 40

Now, cos A = Base/Hypotenuse = AB/AC = 40/41 

tan A = Perpendicular/Base = BC/AB = 9/40                   

Question 5. Given 15 cot A = 8, find sin A and sec A.

Solution:

Given, 15 cot A = 8 

cot A = 8/15 = Base/Perpendicular

Draw a △ ABC where ∠B = 90°, AB = 8, BC = 15

👁 Image

Using Pythagoras Theorem

AC² = AB² + BC²

AC² = 8² + 15² = 64 + 225

AC² = 289

AC = √289 = 17

Now,

sin A = Perpendicular/Hypotenuse = BC/AC = 15/17      

sec A = Hypotenuse/Base = AC/AB = 17/8                

Question 6. In ΔPQR, right-angled at Q, PQ = 4 cm and RQ = 3 cm. Find the values of sin P, sin R, sec P, and sec R.

Solution:

In right-angled ΔPQR,

∠Q = 90°, PQ = 4cm, RQ = 3cm

👁 Image

Using Pythagoras Theorem

PR² = PQ² + QR²

PR² = 4² + 3² = 16 + 9

PR² = 25

PR = √25 =5 

Now,

sin P = Perpendicular/Hypotenuse = RQ/PR = 3/5          

sin R = Perpendicular/Hypotenuse = PQ/PR = 4/5         

sec P = Hypotenuse/Base = PR/PQ = 5/4    

sec R = Hypotenuse/Base = PR/RQ = 5/3  

Related Articles:

• Class 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.2 | Set 1
• Class 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.2| Set 2