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

• Last Updated : 07 Dec, 2021

### (i) sinA = 2/3

Solution:

sinA = 2/3 = Perpendicular/Hypotenuse

Draw a right-angled â–³ABC in which âˆ B is = 90Â°

Using Pythagoras Theorem, in â–³ABC,

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

AC2 = AB2 + BC2

(3)2 = (2)2 + (BC)2

9 = 4 + BC

BC2 = 9 – 4 = 5

BC = âˆš5 units

Now,

cosA = Base/Hypotenuse = BC/AC = âˆš5/3

tanA = Perpendicular/Base = AB/BC = 2/âˆš5

cotA = 1/tanA = âˆš5/2

secA = 1/cosA = 3/âˆš5

cosecA = 1/sinA = 3/2

### (ii) cosA = 4/5

Solution:

cosA = 4/5 = Base/Hypotenuse

Draw a right-angled â–³ABC in which âˆ B is = 90Â°

Using Pythagoras Theorem, in â–³ABC,

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

AC2 = AB2 + BC2

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

25 = AB2 + 16

AB2 = 25 – 16 = 9

AB = âˆš9

= 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) tanÎ¸ = 11/1

Solution:

tanÎ¸ = 11/1 = Perpendicular/Base

Draw a right-angled â–³ABC in which âˆ B is = 90Â°

Using Pythagoras Theorem, in â–³ABC,

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

AC2 = AB2 + BC2

AC2 = (11)2 + (1)2

AC2 = 121  + 1

= 122

AC = âˆš122units

Now,

sinÎ¸ = Perpendicular/Hypotenuse = AB/AC = 11/ âˆš122

cosÎ¸ = Base/Hypotenuse = BC/AC = 1/âˆš122

cotÎ¸ = 1/tanÎ¸ = 1/11

secÎ¸ = 1/cosÎ¸ = âˆš122/1

cosecÎ¸ = 1/sinÎ¸ = âˆš122/11

### (iv) sinÎ¸ = 11/15

Solution:

sinÎ¸ = 11/15 = Perpendicular/Hypotenuse

Draw a right-angled â–³ABC in which âˆ B is = 90Â°

Using Pythagoras Theorem, in â–³ABC,

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

AC2 = AB2 + BC2

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

225 = 121 + (BC)2

(BC)2 = 104

BC = 2âˆš26

Now,

cosÎ¸ = Base/Hypotenuse = BC/AC = 2âˆš26/15

tanÎ¸ = AB/BC = 11/ 2âˆš26

cotÎ¸ = 1/tanÎ¸ = 2âˆš26/11

secÎ¸ = 1/cosÎ¸ = 15/ 2âˆš26

cosecÎ¸ = 1/sinÎ¸ = 15/11

### (v) tan Î± = 5/12

Solution:

tan Î± = 5/12 = Perpendicular/Base

Draw a right-angled â–³ABC in which âˆ B is = 90Â°

Using Pythagoras Theorem, in â–³ABC,

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

AC2 = AB2 + BC2

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

(AC)2 = 144 + 25

(AC)2 = 169

AC = âˆš169 = 13 units

Now,

sin Î± = Perpendicular/Hypotenuse = AB/AC = 5/13

cos Î± = Base/Hypotenuse = BC/AC = 12/13

cot Î± = 1/tan Î± = 12/5

sec Î± = 1/cos Î± = 13/12

cosec Î± = 1/sin Î± = 13/5

### (vi) sinÎ¸ = âˆš3/2

Solution:

sinÎ¸ = âˆš3/2 = Perpendicular/Hypotenuse

Draw a right-angled â–³ABC in which âˆ B is = 90Â°

Using Pythagoras Theorem, in â–³ ABC,

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

AC2 = AB2 + BC2

(2)2 = (âˆš3â€‹)2 + (BC)2

4 = 3 + (BC)2

(BC)2 = 4 – 3 = 1

BC = 1 units

Now,

cosÎ¸ = Base/Hypotenuse = BC/AC = 1/2

tanÎ¸ = AB/BC = âˆš3/1

cotÎ¸ = 1/tanÎ¸ = 1/âˆš3

secÎ¸ = 1/cosÎ¸ = 2/1

cosecÎ¸ = 1/sinÎ¸ = 2/âˆš3

### (vii) cosÎ¸ = 7/25

Solution:

cosÎ¸ = 7/25 = Base/Hypotenuse

Draw a right-angled â–³ABC in which âˆ B is = 90Â°

Using Pythagoras Theorem, in â–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 = âˆš576 = 24 units

Now,

sinÎ¸ = Perpendicular/Hypotenuse = AB/AC = 24/25

tanÎ¸ = Perpendicular/Base = AB/BC = 24/7

cotÎ¸ = 1/tanÎ¸ = 7/24

secÎ¸ = 1/cosÎ¸ = 25/7

cosecÎ¸ = 1/sinÎ¸ = 25/24

### (viii) tanÎ¸ = 8/15

Solution:

tanÎ¸ = 8/15 = Perpendicular/Base

Draw a right-angled â–³ABC in which âˆ B is = 90Â°

Using Pythagoras Theorem, in â–³ ABC,

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

AC2 = AB2 + BC2

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

(AC)2 = 64 + 225

AC = âˆš289 = 17

Now,

sinÎ¸ = Perpendicular/Hypotenuse = AB/AC = 8/17

cosÎ¸ = Base/Hypotenuse = BC/AC = 15/17

cotÎ¸ = 1/tanÎ¸ = 15/8

secÎ¸ = 1/cosÎ¸ = 17/15

cosecÎ¸ = 1/sinÎ¸ = 17/8

### (ix) cotÎ¸ = 12/5

Solution:

cotÎ¸ = 12/5 = Base/Perpendicular

Draw a right-angled â–³ABC in which âˆ B is = 90Â°

Using Pythagoras Theorem, in â–³ABC,

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

AC2 = AB2 + BC2

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

(AC)2 = 25 + 144

(AC)2 = 169

AC = âˆš169 = 13 units

Now,

sinÎ¸ = Perpendicular/Hypotenuse = AB/AC = 5/13

cosÎ¸ = Base/Hypotenuse = BC/AC = 12/13

tanÎ¸ = 1/tanÎ¸ = 5/12

secÎ¸ = 1/cosÎ¸ = 13/12

cosecÎ¸ = 1/sinÎ¸ = 13/5

### (x) secÎ¸ = 13/5

Solution:

secÎ¸ = 13/5 = Hypotenuse/Base

Draw a right-angled â–³ABC in which âˆ B is = 90Â°

Using Pythagoras Theorem, in â–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 = âˆš144 = 12 units

Now,

sinÎ¸ = Perpendicular/Hypotenuse = AB/AC = 12/13

tanÎ¸ = Perpendicular/Base = AB/BC = 12/5

cotÎ¸ = 1/tanÎ¸ = 5/12

cosÎ¸ = 1/secÎ¸ = 5/13

cosecÎ¸ = 1/sinÎ¸ = 13/12

### (xi) cosecÎ¸ = âˆš10

Solution:

cosecÎ¸ = âˆš10/1 = Hypotenuse/Perpendicular

Draw a right-angled â–³ABC in which âˆ B is = 90Â°

Using Pythagoras Theorem, in â–³ ABC,

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

AC2 = AB2 + BC2

(âˆš10)2 = (1)2 + (BC)2

10 = 1 + (BC)2

(BC)2 = 10 – 1 = 9

BC = âˆš9 = 3

Now,

sinÎ¸ = Perpendicular/Hypotenuse = AB/AC = 1/âˆš10

cosÎ¸ = Base/Hypotenuse = BC/AC = 3/âˆš10

tanÎ¸ = Perpendicular/Hypotenuse = AB/BC = 1/3

cotÎ¸ = 1/tanÎ¸ = 3/1 = 3

secÎ¸ = 1/cosÎ¸ = âˆš10/3

### (xii) cosÎ¸ = 12/15

Solution:

cosÎ¸ = 12/15 = Base/Hypotenuse

Draw a right-angled â–³ABC in which âˆ B is = 90Â°

Using Pythagoras Theorem, in â–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 = âˆš81 = 9 units

Now,

sinÎ¸ = Perpendicular/Hypotenuse = AB/AC = 9/15

tanÎ¸ = Perpendicular/Base = AB/BC = 9/12

cotÎ¸ = 1/tanÎ¸ = 12/9

secÎ¸ = 1/cosÎ¸ = 15/12

cosecÎ¸ = 1/sinÎ¸ = 15/9

### (ii) sin C, cos C

Solution:

Given:

In right-angled Î”ABC,

AB = 24 cm, BC = 7 cm. âˆ B = 90Â°

Using Pythagoras Theorem

AC2 = AB2 + BC2

AC2 = 242 + 72 = 576 + 49

AC2 = 625

AC = âˆš625 = 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 = âˆš25 = 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 â–³ ABC where âˆ B = 90Â°, BC = 9, AC = 41

Using Pythagoras Theorem

AC2 = AB2 + BC2

BC2 = 412 – 92 = 1681 – 81

BC2 = 1600

BC = âˆš1600 = 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 â–³ ABC where âˆ B = 90Â°, AB = 8, BC = 15

Using Pythagoras Theorem

AC2 = AB2 + BC2

AC2 = 82 + 152 = 64 + 225

AC2 = 289

AC = âˆš289 = 17

Now,

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

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

### Question 6. In Î”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 Î”PQR,

âˆ Q = 90Â°, PQ = 4cm, RQ = 3cm

Using Pythagoras Theorem

PR2 = PQ2 + QR2

PR2 = 42 + 32 = 16 + 9

PR2 = 25

PR = âˆš25 =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

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