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

### 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 â–³ABC in which âˆ B is = 90Â°

Using Pythagoras Theorem, in â–³ABC,

(Hypotenuse)

^{2}= (Perpendicular)^{2}+ (Base)^{2}AC

^{2}= AB^{2 }+ BC^{2}(3)

^{2}= (2)^{2}+ (BC)^{2}9 = 4 + BC

^{2 }BC

^{2 }= 9 – 4 = 5BC = 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}AC

^{2}= AB^{2 }+ BC^{2}(5)

^{2}= (AB)^{2 }+ (4)^{2}25 = AB

^{2 }+ 16AB

^{2 }= 25 – 16 = 9AB = 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}AC

^{2}= AB^{2 }+ BC^{2}AC

^{2 }= (11)^{2}+ (1)^{2}AC

^{2 }= 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}AC

^{2}= AB^{2 }+ BC^{2}(15)

^{2}= (11)^{2}+ (BC)^{2}225 = 121 + (BC)

^{2}(BC)

^{2}= 104BC = 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}AC

^{2}= AB^{2 }+ BC^{2}(AC)

^{2}= (12)^{2}+ (25)^{2}(AC)

^{2}= 144 + 25(AC)

^{2}= 169AC = 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}AC

^{2}= AB^{2 }+ BC^{2}(2)

^{2 }= (âˆš3â€‹)^{2 }+ (BC)^{2}4 = 3 + (

BC)^{2}

(BC)^{2 }= 4 – 3 = 1BC = 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}AC

^{2}= AB^{2 }+ BC^{2}(25)

^{2}= (AB)^{2}+ (7)^{2}625 = (AB)

^{2}+ 49(AB)

^{2}= 625 – 49 = 576AB = 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}AC

^{2}= AB^{2 }+ BC^{2}(AC)

^{2}= (8)^{2}+ (15)^{2}(AC)

^{2}= 64 + 225AC = 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}AC

^{2}= AB^{2 }+ BC^{2}(AC)

^{2}= (5)^{2}+ (12)^{2}(AC)

^{2}= 25 + 144(AC)

^{2}= 169AC = 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}AC

^{2}= AB^{2 }+ BC^{2}(13)

^{2}= (AB)^{2}+ (5)^{2}169 = (AB)

^{2}+ 25(AB)

^{2}= 169 – 25 = 144AB = 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}AC

^{2}= AB^{2 }+ BC^{2}(âˆš10)

^{2}= (1)^{2}+ (BC)^{2}10 = 1 + (BC)

^{2}(BC)

^{2}= 10 – 1 = 9BC = 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}AC

^{2}= AB^{2 }+ BC^{2}(15)

^{2}= (AB)^{2}+ (12)^{2}225 = (AB)

^{2 }+ 144(AB)

^{2}= 225 – 144 = 81AB = 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

### Question 2. In Î”ABC, 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 Î”ABC,

AB = 24 cm, BC = 7 cm. âˆ B = 90Â°

Using Pythagoras Theorem

AC

^{2 }= AB^{2 }+ BC^{2}AC

^{2 }= 24^{2 }+ 7^{2 }= 576 + 49AC

^{2 }= 625AC = âˆš625 = 25cm

Now,

(i)sinA = BC/AC = 7/25cosA = AB/AC = 24/25

(ii)sinC = AB/AC = 24/25cosC = BC/AC = 7/25

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

**Solution:**

Using Pythagoras Theorem

PR

^{2 }= PQ^{2 }+ QR^{2}13

^{2 }= 12^{2 }+ QR^{2}QR

^{2 }= 169 – 144 = 25QR = 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

AC

^{2 }= AB^{2 }+ BC^{2}BC

^{2 }= 41^{2 }– 9^{2 }= 1681 – 81BC

^{2 }= 1600BC = 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

AC

^{2 }= AB^{2 }+ BC^{2}AC

^{2 }= 8^{2 }+ 15^{2 }= 64 + 225AC

^{2 }= 289AC = 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

PR

^{2 }= PQ^{2 }+ QR^{2}PR

^{2 }= 4^{2 }+ 3^{2 }= 16 + 9PR

^{2 }= 25PR = 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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