ML Aggarwal Solutions For Class 9 Maths Chapter 17 Trigonometric Ratios consists of accurate solutions, which help the students to quickly complete their homework and prepare well for the exams. It ensures that you get all the necessary information of all concepts included in the chapter. Class 9 is a critical level as it forms the base for students for the forthcoming academic years. So, students who aspire to attain good marks in Maths practice ML Aggarwal Solutions for Class 9 Maths. Chapter 17 of ML Aggarwal Solutions deals with trigonometric ratios. In trigonometry, trigonometric ratios are derived from the sides of a right-angled triangle. There are six 6 ratios such as sine, cosine, tangent, cotangent, cosecant, and secant. This chapter of ML Aggarwal Solutions for Class 9 Maths contains one exercise with chapter test. These solutions provided by BYJUâ€™S cover all these concepts, with detailed explanations.

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**1. (a) From the figure (1) given below, find the values of:**

**(i) sin Î¸**

**(ii) cos Î¸**

**(iii) tan Î¸**

**(iv) cot Î¸**

**(v) sec Î¸**

**(vi) cosec Î¸**

**(b) From the figure (2) given below, find the values of:**

**(i) sin Î¸**

**(ii) cos Î¸**

**(iii) tan Î¸**

**(iv) cot Î¸**

**(v) sec Î¸**

**(vi) cosec Î¸**

**Solution:**

(a) From right angled triangle OMP,

By Pythagoras theorem, we get

MP^{2} = OP^{2} + OM^{2}

MP^{2} = (15)^{2} â€“ (12)^{2}

MP^{2} = 225 â€“ 144

MP^{2} = 81

MP^{2} = 9^{2}

MP = 9

(i) sin Î¸ = MP/OP

= 9/15

= 3/5

(ii) cos Î¸ = OM/OP

= 12/15

= 4/5

(iii) tan Î¸ = MP/OP

= 9/12

= Â¾

(iv) cot Î¸ = OM/MP

= 12/9

= 4/3

(v) sec Î¸ = OP/OM

= 15/12

= 5/4

(vi) cosec Î¸ = OP/MP

= 15/9

= /3

(b) From right angled triangle ABC,

By Pythagoras theorem, we get

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

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

AB^{2} = 144 + 25

AB^{2} = 169

AB^{2} = 13^{2}

AB = 13

(i) sin A = BC/AB

= 5/13

(ii) cos A = AC/AB

= 12/13

(iii)

Sin^{2} A + cos^{2} A = (BC/AB)^{2} + (AC/AB)^{2}

= (5/13)^{2} + (12/13)^{2}

= (25/169) + (144/169)

= (25 + 144)/ 169

= 169/169

= 1

Sin^{2} A + cos^{2} A = 1

(iv) Sec^{2} A – tan^{2} A = (AB/AC)^{2} – (BC/AC)^{2}

= (13/12)^{2} – (5/12)^{2}

= (169/144) – (25/144)

= (169 – 25)/ 144

= 144/144

= 1

Sec^{2} A – tan^{2} A = 1

**2. (a) From the figure (1) given below, find the values of:
(i) sin B **

**(i) cos C **

**(iii) sin B + sin C **

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

**(b) From the figure (2) given below, find the values ofâ€¦â€¦..**

**(1) sin B**

**(ii) cos C**

**Solution:**

From right angled triangle ABC,

By Pythagoras theorem, we get

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

AC^{2} = BC^{2} – AB^{2}

AC^{2} = (10)^{2} – (6)^{2}

AC^{2} = 100 – 36

AC^{2} = 64

AC^{2} = 8^{2}

AC = 8

(i) sin B = perpendicular/ hypotenuse

= AC/BC

= 8/10

= 4/5

(ii) cos C = Base/hypotenuse

= AC/BC

= 8/10

= 4/5

(iii) sin B = Perpendicular/hypotenuse

= AC/BC

= 8/10

= 4/5

Sin C = perpendicular/hypotenuse

= AB/BC

= 6/10

= 3/5

Now,

Sin B + sin C = (4/5) + (3/5)

= (4 + 3)/5

= 7/5

(iv) sin B = 4/5

Cos C = 4/5

Sin C = perpendicular/ hypotenuse

= AB/BC

= 6/10

= 3/5

Cos B = Base/Hypotenuse

= AB/BC

= 6/10

= 3/5

sin B cos C + sin C cos B

= (4/5) Ã— (4/5) Ã— (3/5) Ã— (3/5)

= (26/25) Ã— (9/25)

= (16+9)/25

= 25/25

= 1

From Figure

AC = 13, CD = 5, BC =21,

BD = BC â€“ CD

= 21 â€“ 5

= 16

From right angled âˆ†ACD,

By Pythagoras theorem we get

AC = AD^{2} + CD^{2}

AD^{2 }= AC^{2} â€“ CD^{2}

AD^{2 } = (13)^{2} â€“ (5)^{2}

AD^{2 }= 169 â€“ 25

AD^{2 }= 144)

AD^{2 }= (12)^{2 }

AD = 12

From right angled âˆ†ABD,

By Pythagoras angled âˆ†ABD

By Pythagoras theorem we get

AB^{2 }= AD^{2} + BD^{2 }

AB^{2 }= 400

AB^{2 }= (20)^{2 }

AB = 20

(i) tan x = perpendicular/Base (in right angled âˆ†ACD)

=CD/AD

= 5/12

(ii) cos y = Base/Hypotenuse (in right angled âˆ†ABD)

= BD/AB

= (20)/12 â€“ (5/3)

Cot y = Base/Perpendicular (in right âˆ†ABD)

=BD/AB

= 16/20 = 4/5

(iii) ) cos y = Hypotenuse/ perpendicular (in right angled âˆ†ABD)

BD/AB

= 20/12

= 5/3

Cot y = Base/Perpendicular (in right âˆ†ABD)

AB/AD

= 16/12

= 4/3

Cosec^{2} y – cot^{2 }y = (5/3)^{2} â€“ (4/3)^{2 }

= (25/9) â€“ (16/9)

= (25-16)/9

= 9/9

= 1

Hence, cosec^{2} y â€“ cot^{2 }y = 1

(iv) sin x = Perpendicular/Hypotenuse (in right angled âˆ†ACD)

= AD/AB

= 12/20

= 3/5

Cot y = Base/Perpendicular (in right angled âˆ†ABD)

= BD/AD

= 16/12

= 4/3

(5/sin x) + (3/sin y) â€“ 3cot y

= 5/(5/13) + 3/(3/5) â€“ 3 Ã— 4/3

= 5 Ã— 13/5 + 3 Ã— 5/3 â€“ 3 Ã— 4/3

= 1 Ã— 13/1 + 1 Ã— 5/1 â€“ 1 Ã— 4/1

= 13 + 5 â€“ 4 = 18 â€“ 4

= 14

Hence 5/sin x + 3/sin y â€“ 3cot y = 14

**3. From the figure (1) given below, find the value of sec â€¦
(b) From the figure (2) given below, find the values of
(i) sin x **

**(ii) cot x **

**(iii) cot ^{2} x- cosec^{2} x **

**(iv) sec y **

**(v) tan â€¦â€¦**

**Solution:**

(a) From the figure, Sec Î¸ = AB / BD

But in âˆ†ADC, âˆ D = 90^{o}

AC^{2} =AD^{2 }+ DC^{2 }(Pythagoras Theorem)

(13)^{2} =AD^{2 }+ 25

AD^{2 } = 169 -25

= 144

= (12)^{2}

AD = 12

(in right âˆ†ABD)

AB^{2 }= AD^{2 }+ BD^{2}

= (12)^{2 }+ (16)^{2}

= 144 + 256

= 400

= (20)^{2}

AB = 20

Now, Sec Î¸ = AB / BD

= 20/16

= 5/4

(b) let given âˆ†ABC

BD = 3, AC = 12, AD = 4

In right angled âˆ†ABD

By Pythagoras theorem

AB^{2} =AD^{2} + BD^{2 }

AB^{2 }= (4)^{2} + (3)^{2}

AB^{2 }= 16 + 9

AB^{2} = 25

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

AB = 5

In right angled triangle ACD

By Pythagoras theorem,

AC^{2} = AD^{2 }+ CD^{2}

CD^{2} = AC^{2} â€“ AD^{2}

CD^{2} = (12)^{2 }â€“ (4)^{2 }

CD^{2 }= 144 – 16

CD^{2} = 128

CD = âˆš128

CD = âˆš64 Ã— 2 CD

= 8âˆš2

(i) sin x = perpendicular/Hypotenuse

= AD/AB

= 4/5

(ii) cot x = Base/Perpendicular

= BD/AD

= Â¾

(iii) cot x = Base/ Perpendicular

BD/AD

= 3/4

cosec x = Hypotenuse / Perpendicular

AB/BD

= 5/4

Cot^{2} x â€“ cosec^{2} x

= (3/4)^{2} â€“ (5/4)^{2}

= 9/16 â€“ 25/16

(9 -25)/16

= -16/16

= -1

Perpendicular = Hypotenuse/Base (in right angled âˆ†ACD)

= AD/CD

= 12/(8 âˆš2)

= 3/(2 âˆš2)

Cot y = Base/ Hypotenuse

= AD/CD

= 4/8 âˆš 2

= 1/2 âˆš2

Cot y = Base / Hypotenuse ((in right angled âˆ†ACD)

= CD/AC

= 8âˆš2/12

= 2âˆš/3

Now tan^{2} y – 1/cot^{2 }y

= (1/2âˆš2)^{2} â€“ 1/(2âˆš2/3)^{2}

= Â¼ Ã— – Â¼ Ã— 2

= (1/8) â€“ (9/8)

= (1-9)/8

= -8/8

= -1

tan^{2 }y â€“ 1/cot^{2 }y = – 1.

**4. In a right-angled triangle, it is given that angle A is an acute angle and that
Tan A=5/12 Find the values of:
(i) cos A
(ii) cosec A- cot A.**

**Solution:**

Here, ABC is right angled triangle

âˆ A is an acute angle and âˆ C = 90^{o}

tan A = 5/12

BC/AC =5/12

Let BC = 5x and AC = 12x

From right angled âˆ†ABC

By Pythagoras theorem, we get

AB^{2} = (5x)^{2 }+ (12x)^{2}

AB^{2 } = 25x^{2} + 144x^{2 }

AB^{2 }= 169x^{2}

(i) cos A = Base/ Hypotenuse

= AC / AB

= 12x/13x

=12/13

(ii) cosec A = Hypotenuse/perpendicular

= AC / BC

= 13x /5x

= 13/5

cot A = 13/5 â€“ 12/5

= (13-12)/5

= 1/5

**6. In triangle ABC, AB = 15 cm, AC = 15 cm and BC = 18 cm.**

**Solution :**

Here ABC is a triangle in which

AB = 15 cm, AC = 15 cm and BC = 18 cm

Draw AD perpendicular to BC , D is mid-point of BC.

Then, BD â€“ DC = 9 cm

in right angled triangle ABD,

By Pythagoras theorem, we get

AB^{2 }= AD^{2 }+ BD^{2 }

AD^{2 }= AB^{2} â€“ BD^{2}

AD^{2} = (15)^{2} â€“ (9)^{2 }

AD^{2} = 225 â€“ 81

AD^{2 }= 144

AD â€“ 12 cm

(i) cos âˆ ABC = Base/ / Hypotenuse

(In right angled âˆ†ABD, âˆ ABC = âˆ ABD)

= BD / AB

= 9/15

= 3/5

(ii) sin âˆ ACB = sin âˆ ACD

= perpendicular/ Hypotenuse

= AD/AC

= 12/15

= 4/5

**7. If sin â€¦â€¦.=3/5 and â€¦â€¦.**

**Solution:**

Let âˆ† ABC be a right angled at B

Let âˆ ACB = Î¸

Given that, sin Î¸ = 3/5

AB/AC = 3/5

Let AB = 3x

then AC = 5x

In right angled âˆ† ABC,

By Pythagoras theorem,

We get

(5x)^{2} = (3x)^{2 }+ BC^{2 }

BC^{2} = (5x)^{2 }â€“ (3x)^{2}

BC^{2 }= (2x)^{2}

BC = 4x

(i) cos Î¸ = Base/ Hypotenuse

= BC / AC

= 4x /5x

= 4/5

(ii) tan Î¸ = perpendicular/Base

= AB/BC

= 3x/4x

= Â¾

**8. If tan = 4/3**

**find the value of sin .. cos â€¦ (both sin and cos are Positive)**

**Solution :**

Let âˆ†ABC be a right angled

âˆ ACB = Î¸

Given that, tan Î¸ = 4/3

(AB/BC = 4/3)

Give that, tan Î¸ = 4/3

(AB/BC = 4/3)

Let AB = 4x,

then BC = 3x

In right angled âˆ†ABC

By Pythagoras theorem, we get

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

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

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

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

(AC^{2 }= (4x)^{2} + (3x)^{2}

AC^{2 }= 16x^{2 }+ 9x^{2}

AC^{2} = 25x^{2}

AC^{2} = (5x)^{2 }

AC = 5x

Sin Î¸ = perpendicular/Hypotenuse

= AB/AC

= 4x/5x

= 4/5

Cos Î¸ = Base/Hypotenuse

= BC/AC

= 3x/5x

= 3/5

Sin Î¸ + cos Î¸

= 4/5 + 3/5

= (4 + 3)/5

= 7/5

Hence, Sin Î¸ + cos Î¸ = 7/5 = 1 (2/5)

**9. 1f cosec = âˆš5 and is less than 90, find the value of cot – Cos**

**Solution:**

Given cosec Î¸ = âˆš5/1 = OP/PM

OP = âˆš5 and PM = 1

Now OP^{2} = OM^{2} + PM^{2} using Pythagoras theorem

(âˆš5)^{2} = OM^{2} + 1^{2}

5 = OM^{2} + 1

OM^{2} = 5 â€“ 1

OM^{2} = 4

OM = 2

Now cot Î¸ = OM/PM

= 2/1

= 2

Cos Î¸ = OM/OP

= 2/âˆš5

Now cot Î¸ – Cos Î¸ = 2 â€“ (2/âˆš5)

= 2 (âˆš5 â€“ 1)/ 2/âˆš5

**10. Given sin = p/q find sin + cos**

**Solution:**

Given that sin Î¸ = p/q

Which implies,

AB/AC = p/q

Let AB = px

And then AC = qx

In right angled triangle ABC

By Pythagoras theorem,

We get

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

BC^{2} = AC^{2} â€“ AB^{2}

BC^{2} = q^{2}x^{2} â€“ p^{2}x^{2}

BC^{2} = (q^{2} â€“ p^{2})x^{2}

BC = âˆš( q^{2} â€“ p^{2})x

In right angled triangle ABC,

Cos Î¸ = base/ hypotenuse

= BC/AC

= âˆš( q^{2} â€“ p^{2})x/qx

= âˆš( q^{2} â€“ p^{2})/ q

Now,

Sin Î¸ + cos Î¸ = p/q + âˆš( q^{2} â€“ p^{2})/ q

= [p + âˆš( q^{2} â€“ p^{2})]/ q

**11. If O is an acute angle and tan = 8/15 then find sec + cosec**

**Solution:**

Given tan Î¸ = 8/15

Î¸ is an acute angle

in the figure triangle OMP is a right angled triangle,

âˆ M = 90^{o} and âˆ Q = Î¸

Therefore, PM = 8, OM = 15

But OP^{2} = OM^{2} + PM^{2} using Pythagoras theorem,

= 15^{2} + 8^{2}

= 225 + 64

= 289

= 17^{2}

Therefore, OP = 17

Sec Î¸ = OP/OM

= 17/15

Cosec Î¸ = OP/PM

= 17/8

Now,

Sec Î¸ + cosec Î¸ = (17/15) + (17/8)

= (136 + 255)/ 120

= 391/120

**12. Given A is an acute angle and 13 sin A 5, evaluate:** **(5 sin A â€“ 2 cos A)/ tan A**

**Solution:**

Let triangle ABC be a right angled triangle at B and A is an acute angle

Given that 13 sin A = 5

Sin A = 5/13

AB/Ac = 5/13

Let AB = 5x

AC = 13 x

In right angled triangle ABC,

Using Pythagoras theorem,

We get

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

BC^{2} = AC^{2} â€“ BC^{2}

BC^{2} = (13x)^{2} â€“ (5x)^{2}

BC^{2} = 169x^{2} â€“ 25x^{2}

BC^{2} = 144x^{2}

BC = 12x

Sin A = 5/13

Cos A = base/ hypotenuse

= BC/AC

= 12x/ 13x

= 12/13

Tan A = perpendicular/ base

= AB/BC

= 5x/ 12x

= 5/ 12

Now,

(5 sin A â€“ 2 cos A)/ tan A = [(5) (5/13) â€“ (2) (12/13)]/ (5/12)

= (1/13)/ (5/12)

= 12/65

Hence (5 sin A â€“ 2 cos A)/ tan A = 12/65

**13. Given A is an acute angle and cosec A = âˆš2, find the value of 2 sin ^{2 } A + 3 cot^{2} A/ (tan^{2} A â€“ cos^{2} A)**

**Solution:**

Let triangle ABC be a right angled at B and A is a acute angle.

Given that cosec A = âˆš2

Which implies,

AC/BC = âˆš2/1

Let AC = âˆš2x

Then BC = x

In right angled triangle ABC

By using Pythagoras theorem,

We get

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

(âˆš2x)^{2} = AB^{2} + x^{2}

AB^{2} = 2x^{2} â€“ x^{2}

AB = x

Sin A = perpendicular/ hypotenuse

= BC/AC

= 1/ âˆš2

Cot A = base/ perpendicular

= x/x

= 1

Tan A = perpendicular/ base

= BC/AB

= x/x

= 1

Cos A = base/ hypotenuse

= AB/AC

= x/ âˆš2x

= 1/âˆš2

Substituting these values we get

2 sin^{2 } A + 3 cot^{2} A/ (tan^{2} A â€“ cos^{2} A) = 8

Chapter test

**1. (a)From the figure (i) given below, calculate all the six t-ratios for both acuteâ€¦â€¦â€¦**

**(b)From the figure (ii) given below, find the values of x and y in terms of t-ratios**

**Solution:**

(a) From right angled triangle ABC,

By Pythagoras theorem, we get

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

AB^{2} = AC^{2} – BC^{2}

AB^{2} = (3)^{2} â€“ (2)^{2}

AB^{2} = 9 â€“ 4

AB^{2} = 5

AB = âˆš5

(i) sin A = perpendicular/ hypotenuse

= BC/AC

= 2/3

(ii) cos A = base/ hypotenuse

= AB/AC

= âˆš5/3

(iii) tan A = perpendicular/ base

= BC/AB

= 2/ âˆš5

(iv) cot A = base/perpendicular

= AB/ BC

= âˆš5/2

(v) sec A = hypotenuse/ base

= AC/AB

= 3/ âˆš5

(vi) cosec A = hypotenuse/perpendicular

= AC/BC

= 3/2

(b) From right angled triangle ABC,

âˆ BAC = Î¸

Then we know that,

Cot Î¸ = base/ perpendicular

= AB/ BC

= x/ 10

x = 10 cot Î¸

also cosec Î¸ = hypotenuse/ perpendicular

= AC/ BC

= y/ 10

y = 10 cosec Î¸

hence x = 10 cot Î¸ and y = 10 cosec Î¸