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

Last Updated : 23 Jul, 2025

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 ratiosā€”sine, cosine, and tangentā€”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)Ā sinA = 2/3Ā 

Solution:

sinA = 2/3 = Perpendicular/Hypotenuse Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā 

Draw a right-angled ā–³ABC in which āˆ B is = 90Ā° Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā Ā 

Using Pythagoras Theorem, in ā–³ABC,

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

AC2 = AB2 + BC2

(3)2 = (2)2 + (BC)2Ā 

Ā 9 = 4 + BC2Ā 

Ā BC2 = 9 - 4 = 5

BC = āˆš5 units Ā  Ā  Ā  Ā Ā 

Now,

cosA = Base/Hypotenuse = BC/AC = āˆš5/3 Ā Ā 

tanA = Perpendicular/Base = AB/BC = 2/āˆš5 Ā  Ā  Ā  Ā Ā 

cotA = 1/tanA = āˆš5/2 Ā  Ā  Ā  Ā  Ā 

secA = 1/cosA = 3/āˆš5 Ā  Ā  Ā  Ā Ā 

cosecA = 1/sinA = 3/2 Ā  Ā  Ā  Ā  Ā 

(ii)Ā cosA = 4/5Ā 

Solution:

cosA = 4/5 = Base/Hypotenuse

Draw a right-angled ā–³ABC in which āˆ B is = 90Ā°Ā 

Using Pythagoras Theorem, in ā–³ABC,

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

AC2 = AB2 + BC2

(5)2 = (AB)2 + (4)2

Ā 25 = AB2 + 16Ā 

AB2 = 25 - 16 = 9

AB = āˆš9

= 3 units Ā  Ā Ā 

Now,Ā 

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

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

cotA = 1/tanA = 4/3

secA = 1/cosA = 5/4

cosecA = 1/sinA =5/3

(iii) tanĪø = 11/1

Solution:

tanĪø = 11/1 = Perpendicular/Base

Draw a right-angled ā–³ABC in which āˆ B is = 90Ā°

Using Pythagoras Theorem, in ā–³ABC,

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

AC2 = AB2 + BC2

AC2 = (11)2 + (1)2

AC2 = 121 Ā + 1

= 122

AC = āˆš122units Ā 

Ā Now,

sinĪø = Perpendicular/Hypotenuse = AB/AC = 11/ āˆš122

cosĪø = Base/Hypotenuse = BC/AC = 1/āˆš122

cotĪø = 1/tanĪø = 1/11 Ā  Ā  Ā  Ā Ā 

secĪø = 1/cosĪø = āˆš122/1

cosecĪø = 1/sinĪø = āˆš122/11 Ā  Ā Ā 

(iv) sinĪø = 11/15

Solution:

sinĪø = 11/15 = Perpendicular/Hypotenuse

Draw a right-angled ā–³ABC in which āˆ B is = 90Ā°

Using Pythagoras Theorem, in ā–³ABC,

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

AC2 = AB2 + BC2

(15)2 = (11)2 + (BC)2

225 = 121 + (BC)2

(BC)2 = 104

BC = 2āˆš26

Now,

cosĪø = Base/Hypotenuse = BC/AC = 2āˆš26/15 Ā  Ā  Ā  Ā  Ā  Ā  Ā 

tanĪø = AB/BC = 11/ 2āˆš26 Ā  Ā  Ā 

cotĪø = 1/tanĪø = 2āˆš26/11 Ā  Ā 

secĪø = 1/cosĪø = 15/ 2āˆš26 Ā  Ā Ā 

cosecĪø = 1/sinĪø = 15/11 Ā  Ā  Ā  Ā  Ā  Ā Ā 

(v) tan Ī± = 5/12

Solution:

tan Ī± = 5/12 = Perpendicular/Base

Draw a right-angled ā–³ABC in which āˆ B is = 90Ā°

Using Pythagoras Theorem, in ā–³ABC,

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

AC2 = AB2 + BC2

(AC)2 = (12)2 + (25)2

(AC)2 = 144 + 25Ā 

(AC)2 = 169

AC = āˆš169 = 13 units Ā 

Now, Ā Ā 

sin Ī± = Perpendicular/Hypotenuse = AB/AC = 5/13 Ā  Ā Ā 

cos Ī± = Base/Hypotenuse = BC/AC = 12/13 Ā  Ā  Ā  Ā  Ā Ā 

cot Ī± = 1/tan Ī± = 12/5 Ā  Ā Ā 

sec Ī± = 1/cos Ī± = 13/12 Ā  Ā  Ā 

cosec Ī± = 1/sin Ī± = 13/5 Ā 

(vi) sinĪø = āˆš3/2

Solution:

sinĪø = āˆš3/2 = Perpendicular/HypotenuseĀ 

Draw a right-angled ā–³ABC in which āˆ B is = 90Ā°

Using Pythagoras Theorem, in ā–³ ABC,

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

AC2 = AB2 + BC2

(2)2 = (āˆš3ā€‹)2 + (BC)2

4 = 3 + (BC)2

(BC)2 = 4 - 3 = 1Ā 

BC = 1 units

Now,

cosĪø = Base/Hypotenuse = BC/AC = 1/2 Ā  Ā  Ā  Ā Ā 

tanĪø = AB/BC = āˆš3/1 Ā  Ā  Ā 

cotĪø = 1/tanĪø = 1/āˆš3

secĪø = 1/cosĪø = 2/1 Ā  Ā 

cosecĪø = 1/sinĪø = 2/āˆš3 Ā  Ā  Ā  Ā 

(vii) cosĪø = 7/25

Solution:

Ā cosĪø = 7/25 = Base/Hypotenuse Ā 

Draw a right-angled ā–³ABC in which āˆ B is = 90Ā°

Using Pythagoras Theorem, in ā–³ 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 = āˆš576 = 24 units

Now,

sinĪø = Perpendicular/Hypotenuse = AB/AC = 24/25 Ā  Ā  Ā  Ā Ā 

tanĪø = Perpendicular/Base = AB/BC = 24/7 Ā  Ā Ā 

cotĪø = 1/tanĪø = 7/24 Ā  Ā  Ā  Ā Ā 

secĪø = 1/cosĪø = 25/7 Ā  Ā  Ā  Ā 

cosecĪø = 1/sinĪø = 25/24Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā 

(viii) tanĪø = 8/15

Solution:

tanĪø = 8/15 = Perpendicular/Base Ā 

Draw a right-angled ā–³ABC in which āˆ B is = 90Ā°

Using Pythagoras Theorem, in ā–³ ABC,

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

AC2 = AB2 + BC2

(AC)2 = (8)2 + (15)2

(AC)2 = 64 + 225Ā 

AC = āˆš289 = 17

Now,

sinĪø = Perpendicular/Hypotenuse = AB/AC = 8/17 Ā  Ā  Ā Ā 

cosĪø = Base/Hypotenuse = BC/AC = 15/17 Ā  Ā  Ā Ā 

cotĪø = 1/tanĪø = 15/8 Ā  Ā  Ā Ā 

secĪø = 1/cosĪø = 17/15 Ā  Ā  Ā 

cosecĪø = 1/sinĪø = 17/8 Ā  Ā  Ā 

(ix) cotĪø = 12/5

Solution:

cotĪø = 12/5 = Base/Perpendicular Ā  Ā Ā 

Draw a right-angled ā–³ABC in which āˆ B is = 90Ā°

Using Pythagoras Theorem, in ā–³ABC,

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

AC2 = AB2 + BC2

(AC)2 = (5)2 + (12)2

(AC)2 = 25 + 144Ā 

(AC)2 = 169

AC = āˆš169 = 13 unitsĀ 

Now,

sinĪø = Perpendicular/Hypotenuse = AB/AC = 5/13 Ā  Ā Ā 

cosĪø = Base/Hypotenuse = BC/AC = 12/13 Ā  Ā Ā 

tanĪø = 1/tanĪø = 5/12 Ā  Ā Ā 

secĪø = 1/cosĪø = 13/12 Ā  Ā 

cosecĪø = 1/sinĪø = 13/5 Ā Ā 

(x) secĪø = 13/5

Solution:

secĪø = 13/5 = Hypotenuse/BaseĀ 

Draw a right-angled ā–³ABC in which āˆ B is = 90Ā°

Using Pythagoras Theorem, in ā–³ABC,

(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 = āˆš144 = 12 units

Now,

sinĪø = Perpendicular/Hypotenuse = AB/AC = 12/13 Ā  Ā  Ā 

tanĪø = Perpendicular/Base = AB/BC = 12/5 Ā  Ā 

cotĪø = 1/tanĪø = 5/12 Ā  Ā Ā 

cosĪø = 1/secĪø = 5/13 Ā  Ā 

cosecĪø = 1/sinĪø = 13/12 Ā  Ā Ā 

(xi) cosecĪø = āˆš10

Solution:

cosecĪø = āˆš10/1 = Hypotenuse/PerpendicularĀ 

Draw a right-angled ā–³ABC in which āˆ B is = 90Ā°

Using Pythagoras Theorem, in ā–³ ABC,

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

AC2 = AB2 + BC2

(āˆš10)2 = (1)2 + (BC)2

10 = 1 + (BC)2

(BC)2 = 10 - 1 = 9

BC = āˆš9 = 3

Now,

sinĪø = Perpendicular/Hypotenuse = AB/AC = 1/āˆš10 Ā  Ā Ā 

cosĪø = Base/Hypotenuse = BC/AC = 3/āˆš10 Ā Ā 

tanĪø = Perpendicular/Hypotenuse = AB/BC = 1/3 Ā Ā 

cotĪø = 1/tanĪø = 3/1 = 3 Ā  Ā Ā  Ā  Ā  Ā 

secĪø = 1/cosĪø = āˆš10/3 Ā  Ā  Ā  Ā 

(xii) cosĪø = 12/15 Ā 

Solution:

cosĪø = 12/15 = Base/Hypotenuse

Draw a right-angled ā–³ABC in which āˆ B is = 90Ā°

Using Pythagoras Theorem, in ā–³ABC,

(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 = āˆš81 = 9 units Ā Ā 

Now,

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

tanĪø = Perpendicular/Base = AB/BC = 9/12 Ā  Ā 

cotĪø = 1/tanĪø = 12/9 Ā  Ā 

secĪø = 1/cosĪø = 15/12 Ā  Ā  Ā 

cosecĪø = 1/sinĪø = 15/9 Ā  Ā  Ā 

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

(i) sin A, cos A Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā Ā 

(ii) sin C, cos C

Solution:

Given:

In right-angled Ī”ABC,

AB = 24 cm, BC = 7 cm. āˆ B = 90Ā°

Using Pythagoras TheoremĀ 

AC2 = AB2 + BC2

AC2 = 242 + 72 = 576 + 49

AC2 = 625

AC = āˆš625 = 25cm

Now,

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

cosA = AB/AC = 24/25 Ā  Ā 

(ii) sinC = AB/AC = 24/25Ā 

cosC = BC/AC = 7/25 Ā  Ā 

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 = āˆš25 = 5 cm

Now,

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

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

Yes, tanP = cot R

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

Solution:

Given,Ā sinA = 9/41 = Perpendicular/Hypotenuse

Draw a ā–³ ABC where āˆ B = 90Ā°, BC = 9, AC = 41Ā 

Using Pythagoras Theorem

AC2 = AB2 + BC2

BC2 = 412 - 92 = 1681 - 81

BC2 = 1600

BC = āˆš1600 = 40

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

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

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

Solution:

Given, 15 cot A = 8Ā 

cot A = 8/15 = Base/Perpendicular

Draw a ā–³ ABC where āˆ B = 90Ā°, AB = 8, BC = 15

Using Pythagoras Theorem

AC2 = AB2 + BC2

AC2 = 82 + 152 = 64 + 225

AC2 = 289

AC = āˆš289 = 17

Now,

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

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

Question 6. In Ī”PQR, 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 Ī”PQR,

āˆ Q = 90Ā°, PQ = 4cm, RQ = 3cm

Using Pythagoras Theorem

PR2 = PQ2 + QR2

PR2 = 42 + 32 = 16 + 9

PR2 = 25

PR = āˆš25 =5Ā 

Now,

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

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

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

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

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