# ML Aggarwal Solutions for Class 9 Maths Chapter 17 Trigonometric Ratios

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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Exercise 17

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 A

(ii) cos A

(iii) sin2 A + cos2Â A

(iv) sec2 A â€“ tan2 A.

Solution:

(a) From right angled triangle OMP,

By Pythagoras theorem, we get

OP2 = OM2 +MP2

MP2 = OP2 + OM2

MP2 = (15)2 â€“ (12)2

MP2 = 225 â€“ 144

MP2 = 81

MP2 = 92

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

AB2 = AC2 + BC2

AB2 = (12)2 + (5)2

AB2 = 144 + 25

AB2 = 169

AB2 = 132

AB = 13

(i) sin A = BC/AB

= 5/13

(ii) cos A = AC/AB

= 12/13

(iii)Sin2 A + cos2 A = (BC/AB)2 + (AC/AB)2

= (5/13)2 + (12/13)2

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

= (25 + 144)/ 169

= 169/169

= 1

Sin2 A + cos2 A = 1

(iv) Sec2 A – tan2 A = (AB/AC)2 – (BC/AC)2

= (13/12)2 – (5/12)2

= (169/144) – (25/144)

= (169 – 25)/ 144

= 144/144

= 1

Sec2 A – tan2 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 of:

(i) tan x

(ii) cos y

(iii) cosec2 y â€“ cot2 y

(iv) 5/sin x + 3/sin y â€“ 3 cot y.

Solution:

From right angled triangle ABC,

By Pythagoras theorem, we get

BC2 = AC2 + AB2

AC2 = BC2 – AB2

AC2 = (10)2 – (6)2

AC2 = 100 – 36

AC2 = 64

AC2 = 82

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

From right angled âˆ†ABD,

By Pythagoras angled âˆ†ABD

By Pythagoras theorem we get

AB2 = 400

AB2 = (20)2

AB = 20

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

= 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)

= 16/12

= 4/3

Cosec2 y – cot2 y = (5/3)2 â€“ (4/3)2

= (25/9) â€“ (16/9)

= (25-16)/9

= 9/9

= 1

Hence, cosec2 y â€“ cot2 y = 1

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

= 12/20

= 3/5

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

= 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. (a) From the figure (1) given below, find the value of sec Î¸.

(b) From the figure (2) given below, find the values of:
(i) sin x

(ii) cot x

(iii) cot2 x- cosec2 x

(iv) sec y

(v) tan2 y â€“ 1/cos2 y.

Solution:

(a) From the figure, Sec Î¸ = AB / BD

But in âˆ†ADC, âˆ D = 90o

AC2 =AD2 + DC2 (Pythagoras Theorem)

= 144

= (12)2

(in right âˆ†ABD)

= (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

AB2 = (4)2 + (3)2

AB2 = 16 + 9

AB2 = 25

AB2 = (5)2

AB = 5

In right angled triangle ACD

By Pythagoras theorem,

CD2 = (12)2 â€“ (4)2

CD2 = 144 – 16

CD2 = 128

CD = âˆš128

CD = âˆš64 Ã— 2 CD

= 8âˆš2

(i) sin x = perpendicular/Hypotenuse

= 4/5

(ii) cot x = Base/Perpendicular

= Â¾

(iii) cot x = Base/ Perpendicular

= 3/4

(iv) cosec x = Hypotenuse / Perpendicular

AB/BD

= 5/4

Cot2 x â€“ cosec2 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)

= 12/(8 âˆš2)

= 3/(2 âˆš2)

Cot y = Base/ Hypotenuse

= 4/8 âˆš 2

= 1/2 âˆš2

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

= CD/AC

= 8âˆš2/12

= 2âˆš/3

Now tan2 y = 1/cos2 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

tan2 y â€“ 1/cos2 y = – 1.

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

(i) 2 sin y â€“ cos y

(ii) 2 sin x â€“ cos x

(iii) 1 â€“ sin x + cos y

(iv) 2 cos x â€“ 3 sin y + 4 tan x

(b) In the figure (2) given below, âˆ†ABC is right-angled at B. If AB = y units, BC = 3 units and CA = 5 units, find

(i) sin x0

(ii) y.

Solution:

(a) In a right angled âˆ†BCD,

Using Pythagoras theorem

BC2 = BD2 + CD2

Substituting the values

BC2 = 92 + 122

By further calculation

BC2 = 81 + 144 = 225

BC2 = 152

BC = 15

In a right angled âˆ†ABC,

Using Pythagoras theorem

AC2 = AB2 + BC2

We can write it as

AB2 = AC2Â â€“ BC2

Substituting the values

AB2 = 252 â€“ 152

By further calculation

AB2 = 625 â€“ 225 = 400

So we get

AB2 = 202

AB = 20

(i) We know that

In right angled âˆ†BCD

sin y = perpendicular/ hypotenuse

sin y = BD/ BC

Substituting the values

sin y = 9/15 = 3/5

In right angled âˆ†BCD

cos y = base/hypotenuse

cos y = CD/BC

Substituting the values

cos y = 12/15 = 4/5

Here

2sin y â€“ cos y = 2 Ã— 3/5 â€“ 4/5

We can write it as

= 6/5 â€“ 4/5

= 2/5

Therefore, 2 sin y â€“ cos y = 2/5

(ii) In right angled âˆ†ABC

sin x = perpendicular/ hypotenuse

sin x = BC/AC

Substituting the values

sin x = 15/25 = 3/5

In right angled âˆ†ABC

cos x = base/hypotenuse

cos x = AB/AC

Substituting the values

cos x = 20/25 = 4/5

Here

2 sin x â€“ cos x = 2 Ã— 3/5 â€“ 4/5

We can write it as

= 6/5 â€“ 4/5

= 2/5

Therefore, 2 sin x â€“ cos x = 2/5.

(iii) In right angled âˆ†ABC

sin x = perpendicular/hypotenuse

sin x = BC/AC

Substituting the values

sin x = 12/25 = 3/5

In right angled âˆ†BCD

cos y = base/hypotenuse

cos y = CD/BC

Substituting the values

cos y = 12/15 = 4/5

Here

1 â€“ sin x + cos y = 1 â€“ 3/5 + 4/5

By further calculation

= (5 â€“ 3 + 4)/ 5

So we get

= (9 â€“ 3)/ 5

= 6/5

Therefore, 1 â€“ sin x + cos y = 6/5.

(iv) In right angled âˆ†BCD

cos x = base/hypotenuse

cos x = AB/AC

Substituting the values

cos x = 20/25 = 4/5

In right angled âˆ†BCD

sin y = perpendicular/hypotenuse

sin y = BD/BC

Substituting the values

sin y = 9/15 = 3/5

In right angled âˆ†ABC

tan x = perpendicular/base

tan x = BC/AB

Substituting the values

tan x = 15/20 = Â¾

Here

2 cos x â€“ 3 sin y + 4 tan x = 2 Ã— 4/5 â€“ 3 Ã— 3/5 + 4 Ã— Â¾

By further calculation

= 8/5 â€“ 9/5 3/1

Taking LCM

= (8 â€“ 9 + 15)/5

So we get

= (23 â€“ 9)/ 5

= 14/5

(b) It is given that

AB = y units, BC = 3 units, CA = 5 units

(i) In right angled âˆ†ABC

sin x = perpendicular/hypotenuse

sin x = BC/AC

Substituting the values

sin x = 3/5

(ii) In right angled âˆ†ABC

Using Pythagoras theorem

AC2 = BC2 + AB2

We can write it as

AB2 = AC2 â€“ BC2

Substituting the values

AB2 = 52 â€“ 32

By further calculation

AB2 = 25 â€“ 9 = 16

So we get

AB2 = 42

AB = 4

y = 4 units

Therefore, y = 4 units.

5. 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 = 90o

tan A = 5/12

BC/AC =5/12

Let BC = 5x and AC = 12x

From right angled âˆ†ABC

By Pythagoras theorem, we get

AB2 = (5x)2 + (12x)2

AB2 = 25x2 + 144x2

AB2 = 169x2

(i) cos A = Base/ Hypotenuse

= AC / AB

= 12x/13x

=12/13

(ii) cosec A = Hypotenuse/perpendicular

= AC / BC

= 13x /5x

= 13/5

cosec A – cot A = 13/5 â€“ 12/5

= (13-12)/5

= 1/5

6. (a) In âˆ†ABC, âˆ A = 900. If AB = 7 cm and BC â€“ AC = 1 cm, find:

(i) sin C

(ii) tan B

(b) In âˆ†PQR, âˆ Q = 900. If PQ = 40 cm and PR + QR = 50 cm, find:

(i) sin P

(ii) cos P

(iii) tan R.

Solution:

(a) In right âˆ†ABC

âˆ A = 900

AB = 7 cm

BC â€“ AC = 1 cm

BC = 1 + AC

We know that

BC2 = AB2 + AC2

Substituting the value of BC

(1 + AC)2 = AB2 + AC2

1 + AC2 + 2AC = 72 + AC2

By further calculation

1 + AC2 + 2AC = 49 AC2

2AC = 49 â€“ 1 â€“ 48

So we get

AC = 48/2 = 24 cm

Here

BC = 1 + AC

Substituting the value

BC = 1 + 24 = 25 cm

(i) sin C = AB/BC = 7/25

(ii) tan B = AC/AB = 24/7

(b) In right âˆ†PQR

âˆ Q = 900

PQ = 40 cm

PQ + QR = 50 cm

We can write it as

PQ = 50 â€“ QR

Using Pythagoras theorem

PR2 = PQ2 + QR2

(50 â€“ QR)2 = (40)2 + QR2

By further calculation

2500 + QR2 â€“ 100QR = 1600 + QR2

So we get

2500 â€“ 1600 = 100QR

100QR = 900

By division

QR = 900/100 = 9

We get

PR = 50 â€“ 9 = 41

(i) sin P = QR/PR = 9/41

(ii) cos P = PQ/PR = 40/41

(iii) tan R = PQ/QR = 40/9

7. In triangle ABC, AB = 15 cm, AC = 15 cm and BC = 18 cm. Find

(i) cos âˆ ABC

(ii) sin âˆ ACB.

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

(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

= 12/15

= 4/5

8. (a) In the figure (1) given below, âˆ†ABC is isosceles with AB = AC = 5 cm and BC = 6 cm. Find

(i) sin C

(ii) tan B

(iii) tan C â€“ cot B.

(b) In the figure (2) given below, âˆ†ABC is right-angled at B. Given that âˆ ACB = Î¸, side AB = 2 units and side BC = 1 unit, find the value of sin2 Î¸ + tan2 Î¸.

(c) In the figure (3) given below, AD is perpendicular to BC, BD = 15 cm, sin B = 4/5 and tan C = 1.

(i) Calculate the lengths of AD, AB, DC and AC.

(ii) Show that tan2 B â€“ 1/cos2 B = – 1.

Solution:

(a) It is given that

âˆ†ABC is isosceles with AB = AC = 5 cm and BC = 6 cm

D is the mid point of BC

So BD = CD

Here

BD = CD = 6/2 = 3 cm

In right angled âˆ†ABD

Using Pythagoras theorem

We can write it as

Substituting the values

By further calculation

AD2 = 25 â€“ 9 = 16

So we get

(i) In right angled âˆ†ACD

sin C = perpendicular/hypotenuse

sin C = AD/AC = 4/5

(ii) In right angled âˆ†ABD

tan B = perpendicular/base

tan B = AD/BD = 4/3

(iii) In right angled âˆ†ACD

tan C = perpendicular/base

tan C = AD/CD = 4/3

In right angled âˆ†ABD

cot B = base/perpendicular

cot B = BD/AD = Â¾

Here

tan C â€“ cot B = 4/3 â€“ Â¾

Taking LCM

tan C â€“ cot B = (16 â€“ 9)/ 12 = 7/12

(b) It is given that

âˆ†ABC is right-angled at B

AB = 2 units and BC = 1 unit

In right angled âˆ†ABC

Using Pythagoras theorem

AC2 = AB2Â + BC2

Substituting the values

AC2 = 22 + 12

AC2 = 4 + 1 = 5

So we get

AC2 = 5

AC = âˆš5 units

In right angled âˆ†ABC

sin Î¸ = perpendicular/hypotenuse

sin Î¸ = AB/AC = 2/âˆš5

In right angled âˆ†ABC

tan Î¸ = perpendicular/base

tan Î¸ = AB/BC = 2/1

We know that

sin2 Î¸ + tan2 Î¸ = (2/âˆš5)2 + (2/1)2

By further calculation

= 4/5 + 4/1

Taking LCM

= (4 + 20)/ 5

= 24/5

= 4 4/5

(c) (i) In âˆ†ABC

BD = 15 cm

sin B = 4/5

tan C = 1

In âˆ†ABD

sin B = perpendicular/hypotenuse

sin B = AD/AB = 4/5

Consider AD = 4x and AB = 5x

Using Pythagoras theorem

In right angled âˆ†ABD

We can write it as

Substituting the values

(15)2 = (5x)2 â€“ (4x)2

225 = 25x2 â€“ 16x2

By further calculation

225 = 9x2

x2 = 225/9 = 25

So we get

x = âˆš25 = 5

Here

AD = 4 Ã— 5 = 20

AB = 5 Ã— 5 = 25

In right angled âˆ†ACD

tan C = perpendicular/base

So we get

tan C = AD/CD = 1/1

Consider AD = X then CD = x

Using Pythagoras theorem

Substituting the values

AC2 = x2 + x2 â€¦..(1)

So the equation becomes

AC2 = 202 + 202

AC2 = 400 + 400 = 800

So we get

AC = âˆš800 = 20âˆš2

Length of AD = 20 cm

Length of AB = 25 cm

Length of DC = 20 cm

Length of AC = 20âˆš2 cm

(ii) In right angled âˆ†ABD

tan B = perpendicular/base

So we get

Substituting the values

tan B = 20/15 = 4/3

In right angled âˆ†ABD

cos B = base/hypotenuse

So we get

cos B = BD/AB

Substituting the values

cos B = 15/25 = 3/5

Here

LHS = tan2 B â€“ 1/cos2 B

Substituting the values

= (4/3)2 â€“ 1/(3/5)2

By further calculation

= (4)2/(3)2 â€“ (5)2/(3)2

= 16/9 â€“ 25/9

So we get

= (16 â€“ 25)/9

= -9/9

= – 1

= RHS

Hence, proved.

9. If sin Î¸ =3/5 and Î¸ is acute angle, find

(i) cos Î¸

(ii) tan Î¸.

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 + BC2

BC2 = (5x)2 â€“ (3x)2

BC2 = (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

10. Given that tan Î¸ = 5/12 and Î¸ is an acute angle, find sin Î¸ and cos Î¸.

Solution:

Consider âˆ† ABC be right angled at B and âˆ ACB = Î¸

It is given that

tan Î¸ = 5/12

AB/BC = 5/12

Consider AB = 5x and BC = 12x

In right angled âˆ† ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = (5x)2 + (12x)2

By further calculation

AC2 = 25x2 + 144x2 = 169x2

So we get

AC2 = (13x)2

AC = 13x

In right angled âˆ† ABC

sin Î¸ = perpendicular/hypotenuse

So we get

sin Î¸ = AB/AC = 5x/13x = 5/13

In right angled âˆ† ABC

cos Î¸ = base/hypotenuse

So we get

cos Î¸ = BC/AC

Substituting the values

cos Î¸ = 12x/13x = 12/13

11. If sin Î¸ = 6/10, find the value of cos Î¸ + tan Î¸.

Solution:

Consider âˆ† ABC be right angled at B and âˆ ACB = Î¸

It is given that

sin Î¸ = AB/AC

sin Î¸ = 6/10

Take AB = 6x then AC = 10x

In right angled âˆ† ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

(10x)2 = (6x)2 + BC2

By further calculation

BC2 = 100x2 â€“ 36x2 = 64x2

So we get

BC2 = (8x)2

BC = 8x

In right angled âˆ† ABC

cos Î¸ = base/hypotenuse

cos Î¸ = BC/AC

Substituting the values

cos Î¸ = 8x/10x = 4/5

In right angled âˆ† ABC

tan Î¸ = perpendicular/base

tan Î¸ = AB/BC

Substituting the values

tan Î¸ = 6x/8x = Â¾

Here

cos Î¸ + tan Î¸ = 4/5 + Â¾

Taking LCM

= (16 + 15)/ 20

= 31/20

= 1 11/20

12. If tan = 4/3, find the value of sin Î¸ + cos Î¸ (both sin Î¸ and cos Î¸ 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

AC2 = AB2 + BC2

AC2 = AB2 + BC2

AC2 = AB2 + BC2

AC2 = AB2 + BC2

(AC2 = (4x)2 + (3x)2

AC2 = 16x2 + 9x2

AC2 = 25x2

AC2 = (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

13. 1f cosec = âˆš5 and Î¸ is less than 900, find the value of cot Î¸ â€“ cos Î¸.

Solution:

Given cosec Î¸ = âˆš5/1 = OP/PM

OP = âˆš5 and PM = 1

Now OP2 = OM2 + PM2 using Pythagoras theorem

(âˆš5)2 = OM2 + 12

5 = OM2 + 1

OM2 = 5 â€“ 1

OM2 = 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)/ âˆš5

14. Given sin Î¸ = p/q, find cos Î¸ + sin Î¸ in terms of p and q.

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

AC2 = AB2 + BC2

BC2 = AC2 â€“ AB2

BC2 = q2x2 â€“ p2x2

BC2 = (q2 â€“ p2)x2

BC = âˆš( q2 â€“ p2)x

In right angled triangle ABC,

Cos Î¸ = base/ hypotenuse

= BC/AC

= âˆš( q2 â€“ p2)x/qx

= âˆš( q2 â€“ p2)/ q

Now,

Sin Î¸ + cos Î¸ = p/q + âˆš( q2 â€“ p2)/ q

= [p + âˆš( q2 â€“ p2)]/ q

15. If Î¸ is an acute angle and tan = 8/15, find the value of sec Î¸ + cosec Î¸.

Solution:

Given tan Î¸ = 8/15

Î¸ is an acute angle

in the figure triangle OMP is a right angled triangle,

âˆ M = 90o and âˆ Q = Î¸

TanÎ¸ = PM/OL = 8/15

Therefore, PM = 8, OM = 15

But OP2 = OM2 + PM2 using Pythagoras theorem,

= 152 + 82

= 225 + 64

= 289

= 172

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

= 3 31/120

16. 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

AC2 = AB2 + BC2

BC2 = AC2 â€“ BC2

BC2 = (13x)2 â€“ (5x)2

BC2 = 169x2 â€“ 25x2

BC2 = 144x2

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

17. Given A is an acute angle and cosec A = âˆš2, find the value of

(2 sin2 A + 3 cot2 A)/ (tan2 A â€“ cos2 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

AC2 = AB2 + BC2

(âˆš2x)2 = AB2 + x2

AB2 = 2x2 â€“ x2

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 sin2 A + 3 cot2 A/ (tan2 A â€“ cos2 A) = 8

18. The diagonals AC and BD of a rhombus ABCD meet at O. If AC = 8 cm and BD = 6 cm, find sin âˆ OCD.

Solution:

It is given that

Diagonals AC and BD of rhombus ABCD meet at O

AC = 8 cm and BD = 6 cm

O is the mid point of AC

We know that

AO = OC = AC/2 = 8/2 = 4 cm

O is the mid point of BD

BO = OD = BD/2 = 6/2 = 3 cm

In right angled âˆ†COD

CD2 = OC2 + OD2

Substituting the values

CD2 = 42 + 32

So we get

CD2 = 16 + 9 = 25

CD2 = 52

CD = 5 cm

In right angled âˆ†COD

sin âˆ OCD = perpendicular/ hypotenuse

So we get

sin âˆ OCD = OD/CD = 3/5

19. If tan Î¸ = 5/12, find the value of (cos Î¸ + sin Î¸)/ (cos Î¸ â€“ sin Î¸).

Solution:

Consider âˆ†ABC be right angled at B and âˆ ACB = Î¸

It is given that

tan Î¸ = AB/BC = 5/12

Take AB = 5x then BC = 12x

In right angled âˆ†ABC,

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = (5x)2 + (12x)2

By further calculation

AC2 = 25x2 + 144x2 = 169x2

So we get

AC2 = (13x)2

AC = 13x

In right angled âˆ†ABC

cos Î¸ = base/hypotenuse

cos Î¸ = BC/AC

Substituting the values

cos Î¸ = 12x/13x = 12/13

In right angled âˆ†ABC

sin Î¸ = perpendicular/hypotenuse

sin Î¸ = AB/AC

Substituting the values

sin Î¸ = 5x/13x = 5/13

Here

(cos Î¸ + sin Î¸)/ (cos Î¸ â€“ sin Î¸) = [12/13 + 5/13]/ [12/13 â€“ 5/13]

Taking LCM

= [(12 + 5)/ 13]/[(12 â€“ 5)/ 13]

So we get

= 17/13/ 7/13

= 17/13 Ã— 13/7

= 17/7

Therefore, (cos Î¸ + sin Î¸)/ (cos Î¸ â€“ sin Î¸) = 17/7 = 2 3/7.

20. Given 5 cos A â€“ 12 sin A = 0, find the value of (sin A + cos A)/ (2 cos A â€“ sin A).

Solution:

It is given that

5 cos A â€“ 12 sin A = 0

We can write it as

5 cos A = 12 sin A

So we get

sin A/ cos A = 5/12

We know that sin A/ cos A = tan A

tan A = 5/12

Consider âˆ†ABC right angled at B and âˆ A is acute angle

Here

tan A = BC/AB = 5/12

Take BC = 5x then AB = 12x

In right angled âˆ†ABC

Using Pythagoras theorem

AC2 = BC2 + AB2

Substituting the values

AC2 = (5x)2 + (12x)2

AC2 = 25x2 + 144x2 = 169x2

So we get

AC2 = (13x)2

AC = 13x

In right angled âˆ†ABC

sin A = perpendicular/hypotenuse

So we get

sin A = BC/AC = 5x/13x = 5/13

In right angled âˆ†ABC

cos A = base/hypotenuse

So we get

cos A = AB/AC = 12x/13x = 12/13

Here

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

By further calculation

= [(5 + 12)/13]/ [24/13 â€“ 5/13]

So we get

= [(5 + 12)/13]/[(24 â€“ 5)/13]

= 17/13/ 19/13

= 17/13 Ã— 13/19

= 17/19

Therefore, (sin A + cos A)/ (2 cos A â€“ sin A) = 17/19

21. If tan Î¸ = p/q, find the value of (p sin Î¸ â€“ q cos Î¸)/ (p sin Î¸ + q cos Î¸).

Solution:

It is given that

tan Î¸ = p/q

Consider âˆ†ABC be right angled at B and âˆ BCA = Î¸

tan Î¸ = BC/AB = p/q

BC = px then AB = qx

In right angled âˆ†ABC

Using Pythagoras theorem

AC2 = BC2 + AB2

Substituting the values

AC2 = (px)2 + (qx)2

AC2 = p2x2 + q2x2

AC2 = x2 (p2 + q2)

So we get

AC = âˆšx2 (p2 + q2)

AC = x(âˆšp2 + q2)

In right angled âˆ†ABC

sin Î¸ = perpendicular/hypotenuse

sin Î¸ = BC/AC

Substituting the values

sin Î¸ = px/ x(âˆšp2 + q2)

So we get

sin Î¸ = p/ (âˆšp2 + q2)

In right angled âˆ†ABC

cos Î¸ = base/hypotenuse

cos Î¸ = AB/AC

Substituting the values

cos Î¸ = qx/ x(âˆšp2 + q2)

So we get

cos Î¸ = q/ (âˆšp2 + q2)

Here

22. If 3 cot Î¸ = 4, find the value of (5 sin Î¸ â€“ 3 cos Î¸)/ (5 sin Î¸ + 3 cos Î¸).

Solution:

It is given that

3 cot Î¸ = 4

cot Î¸ = 4/3

Consider âˆ†ABC be right angled at B and âˆ ACB = Î¸

cot Î¸ = BC/AB = 4/3

Take BC = 4x then AB = 3x

In right angled âˆ†ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = (3x)2 + (4x)2

AC2 = 9x2 + 16x2 = 25x2

So we get

AC2 = (5x)2

AC = 5x

In right angled âˆ†ABC

sin Î¸ = perpendicular/hypotenuse

sin Î¸ = AB/AC

Substituting the values

sin Î¸ = 3x/5x = 3/5

In right angled âˆ†ABC

cos Î¸ = base/hypotenuse

cos Î¸ = BC/AC

Substituting the values

cos Î¸ = 4x/5x = 4/5

23. (i) If 5 cos Î¸ â€“ 12 sin Î¸ = 0, find the value of (sin Î¸ + cos Î¸)/(2 cos Î¸ – sin Î¸).

(ii) If cosec Î¸ = 13/12, find the value of (2 sin Î¸ â€“ 3 cos Î¸)/(4 sin Î¸ â€“ 9 cos Î¸).

Solution:

(i) It is given that

5 cos Î¸ â€“ 12 sin Î¸ = 0

We can write it as

5 cos Î¸ = 12 sin Î¸

sin Î¸/ cos Î¸ = 5/12

tan Î¸ = 5/12

Dividing both numerator and denominator by cos Î¸

(ii) It is given that

cosec Î¸ = 13/12

We know that cosec Î¸ = 1/ sin Î¸

1/sin Î¸ = 13/12

sin Î¸ = 12/13

Here cos2 Î¸ = 1 â€“ sin2 Î¸

Substituting the values

= 1 â€“ (12/13)2

By further calculation

= 1 â€“ 144/169

Taking LCM

= (169 â€“ 144)/ 169

= 25/169

So we get

= (5/13)2

cos Î¸ = 5/13

Here

24. If 5 sin Î¸ = 3, find the value of (sec Î¸ â€“ tan Î¸)/ (sec Î¸ + tan Î¸).

Solution:

Consider âˆ†ABC be right angled at B and âˆ ACB = Î¸

It is given that

5 sin Î¸ = 3

sin Î¸ = AB/AC = 3/5

Take AB = 3x then AC = 5x

In right angled âˆ†ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

BC2 = AC2 â€“ AB2

Substituting the values

BC2 = (5x)2 â€“ (3x)2

So we get

BC2 = 25x2 â€“ 9x2 = 16x2

BC2 = (4x)2

BC = 4x

In right angled âˆ†ABC

sec Î¸ = hypotenuse/base

sec Î¸ = AC/BC = 5x/4x = 5/4

In right angled âˆ†ABC

tan Î¸ = perpendicular/base

tan Î¸ = AB/BC = 3x/4x = Â¾

25. If Î¸ is an acute angle and sin Î¸ = cos Î¸, find the value of 2 tan2 Î¸ + sin2 Î¸ â€“ 1.

Solution:

Consider âˆ†ABC be right angled at B and âˆ ACB = Î¸

It is given that

sin Î¸ = cos Î¸

sin Î¸/cos Î¸ = 1

tan Î¸ = AB/BC = 1

Take AB = x then BC = x

In right angled âˆ†ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

AC2 = x2 + x2 = 2x2

So we get

AC = âˆš2x2

AC = (âˆš2)x

In right angled âˆ†ABC

sin Î¸ = perpendicular/hypotenuse

sin Î¸ = AB/AC = x/âˆš2x = 1/âˆš2

Here

2 tan2 Î¸ + sin2 Î¸ â€“ 1 = 2 Ã— (1)2 + (1/âˆš2)2 â€“ 1

By further calculation

= 2 Ã— 1 + Â½ – 1

= 2 + Â½ – 1

= 1+ Â½

Taking LCM

= (2 + 1)/2

= 3/2

Therefore, 2 tan2 Î¸ + sin2 Î¸ â€“ 1 = 3/2.

26. Prove the following:

(i) cos Î¸ tan Î¸ = sin Î¸

(ii) sin Î¸ cot Î¸ = cos Î¸

(iii) sin2 Î¸/ cos Î¸ + cos Î¸ = 1/ cos Î¸.

Solution:

(i) cos Î¸ tan Î¸ = sin Î¸

LHS = cos Î¸ tan Î¸

We know that tan Î¸ = sin Î¸/cos Î¸

= cos Î¸ (sin Î¸/cos Î¸)

So we get

= 1 Ã— sin Î¸/1

= sin Î¸

= RHS

Therefore, LHS = RHS.

(ii) sin Î¸ cot Î¸ = cos Î¸

LHS = sin Î¸ cot Î¸

We know that cot Î¸ = cos Î¸/sin Î¸

= sin Î¸ (cos Î¸/sin Î¸)

= 1 Ã— cos Î¸/1

= cos Î¸

= RHS

Therefore, LHS = RHS.

(iii) sin2 Î¸/cos Î¸ + cos Î¸ = 1/cos Î¸

LHS = sin2 Î¸/cos Î¸ + cos Î¸/1

Taking LCM

= (sin2 Î¸ + cos2 Î¸)/cos Î¸

We know that sin2 Î¸ + cos2 Î¸ = 1

= 1/cos Î¸

= RHS

Therefore, LHS = RHS.

27. If in âˆ†ABC, âˆ C = 900 and tan A = Â¾, prove that sin A cos B + cos A sin B = 1.

Solution:

It is given that

tan A = BC/AC = Â¾

Using Pythagoras theorem

AB2 = AC2 + BC2

Substituting the values

= 42 + 32

= 16 + 9

= 25

= 52

So we get AB = 5

Here

sin A = BC/AC = 3/5

cos A = AC/AB = 4/5

cos B = BC/AB = 3/5

sin B = AC/AB = 4/5

LHS = sin A cos B + cos A sin B

Substituting the values

= 3/5 Ã— 3/5 + 4/5 Ã— 4/5

By further calculation

= 9/25+ 16/25

= (9 + 16)/ 25

= 25/25

= 1

= RHS

Therefore, LHS = RHS.

28. (a) In figure (1) given below, âˆ†ABC is right-angled at B and âˆ†BRS is right-angled at R. If AB = 18 cm, BC = 7.5 cm, RS = 5 cm, âˆ BSR = x0 and âˆ SAB = y0, then find:

(i) tan x0

(ii) sin y0.

(b) In the figure (2) given below, âˆ†ABC is right angled at B and BD is perpendicular to AC. Find

(i) cos âˆ CBD

(ii) cot âˆ ABD.

Solution:

(a) âˆ†ABC is right-angled at B, âˆ†BSC is right-angled at S and âˆ†BRS is right-angled at R

It is given that

AB = 18 cm, BC = 7.5 cm, RS = 5 cm, âˆ BSR = x0 and âˆ SAB = y0

By Geometry âˆ†ARS and âˆ†ABC are similar

AR/AB = RS/BC

Substituting the values

AR/18 = 5/7.5

By further calculation

AR = (5 Ã— 18)/7.5 = (1 Ã— 18)/1.5

Multiply both numerator and denominator by 10

AR = (18 Ã— 10)/15

AR = (10 Ã— 6)/5

AR = (2 Ã— 6)/1 = 12

So we get

RB = AB â€“ AR

RB = 18 â€“ 12 = 6

In right angled âˆ†ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = 182 + 7.52

By further calculation

AC2 = 324 + 56.25 = 380.25

AC = âˆš380.25 = 19.5 cm

(i) In right angled âˆ†BSR

tan x0 = perpendicular/base

tan x0 = RB/RS = 6/5

(ii) In right angled âˆ†ASR

sin y0 = perpendicular/hypotenuse

Using Pythagoras theorem

AS2 = 122 + 52

By further calculation

AS2 = 144 + 25 = 169

AS = âˆš169 = 13 cm

So we get

sin y0 = RS/AS = 5/13

(b) We know that

âˆ†ABC is right angled at B and BD is perpendicular to AC

In right angled âˆ†ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = 122 + 52

By further calculation

AC2 = 144 + 25 = 169

So we get

AC2 = (13)2

AC = 13

By Geometry âˆ CBD = âˆ A and âˆ ABD = âˆ C

(i) cos âˆ CBD = cos âˆ A = base/hypotenuse

In right angled âˆ†ABC

cos âˆ CBD = cos âˆ A = AB/AC = 12/13

(ii) cos âˆ ABD = cos âˆ C = base/perpendicular

In right angled âˆ†ABC

cos âˆ ABD = cos âˆ C = BC/AB = 5/12

29. In the adjoining figure, ABCD is a rectangle. Its diagonal AC = 15 cm and âˆ ACD = Î±. If cot Î± = 3/2, find the perimeter and the area of the rectangle.

Solution:

cot Î± = CD/AD = 3/2

Take CD = 3x then AD = 2x

Using Pythagoras theorem

Substituting the values

(15)2 = (3x)2 + (2x)2

By further calculation

13x2 = 225

x2Â = 225/13

So we get

x = âˆš225/13 = 15/âˆš13

Length of rectangle (l) = 3x = (3 Ã— 15)/ âˆš13 = 45/âˆš13 cm

Breadth of rectangle (b) = 2x = (2 Ã— 15)/ âˆš13 = 30/âˆš13 cm

(i) Perimeter of rectangle = 2 (l + b)

Substituting the values of l and b

= 2 (45/âˆš13 + 30/âˆš13)

So we get

= 2 Ã— 75/âˆš13

= 150/âˆš13 cm

(ii) Area of rectangle = l Ã— b

Substituting the values of l and b

= 45/âˆš13 Ã— 30/âˆš13

So we get

= 1350/13

= 103 11/13 cm2

30. Using the measurements given in the figure alongside,

(a) Find the values of:

(i) sin Ï•

(ii) tan Î¸.

(b) Write an expression for AD in terms of Î¸.

Solution:

From the figure

BC = 12, BD = 13

In right angled âˆ†BCD

Using Pythagoras theorem

BD2 = BC2 + CD2

It can be written as

CD2 = BD2 â€“ BC2

Substituting the values

CD2 = (13)2 â€“ (12)2

CD2 = 169 â€“ 144 = 25

So we get

CD = âˆš25 = 5

Construct BE perpendicular to AB

CD = BE = 5 and EA = AE = 14 â€“ 5 = 9

(a) (i) sin Ï• = perpendicular/hypotenuse

In right angled âˆ†BCD

sin Ï• = CD/BD = 5/13

(ii) tan Î¸ = perpendicular/hypotenuse

In right angled âˆ†AED

tan Î¸ = ED/AE = BC/AE = 12/9 = 4/3 (Since ED = BC)

(b) In right angled âˆ†AED

sin Î¸ = perpendicular/hypotenuse

cos Î¸ = base/perpendicular

We can write it as

Substituting the values

31. Prove the following:

(i) (sin A + cos A)2 + (sin A â€“ cos A)2 = 2

(ii) cot2 A â€“ 1/sin2 A + 1 = 0

(iii) 1/(1 + tan2 A) + 1/(1 + cot2 A) = 1

Solution:

(i) (sin A + cos A)2 + (sin A â€“ cos A)2 = 2

LHS = (sin A + cos A)2 + (sin A â€“ cos A)2

Using the formula

(a + b)2 = a2 + b2 + 2ab and (a â€“ b)2 = a2 + b2 â€“ 2ab

= [(sin A)2 + (cos A)2 + 2 sin A cos A] + [(sin A)2 + (cos A)2 â€“ 2 sin A cos A]

By further calculation

= sin2Â A + cos2 A + 2 sin A cos A + sin2 A + cos2 A â€“ 2 sin A cos A

= sin2 A + cos2 A + sin2 A + cos2 A

= 2 sin2 A + 2 cos2 A

We know that sin2 A + cos2Â A = 1

= 2 (sin2 A + cos2 A)

= 2 (1)

= 2

= RHS

Therefore, LHS = RHS.

(ii) cot2 A â€“ 1/sin2 A + 1 = 0

LHS = cot2 A â€“ 1/sin2 A + 1

We know that

1/sin A = cosec A

= cot2 A â€“ cosec2 A + 1

= (1 + cot2 A) â€“ cosec2 A

We know that 1 + cot2 A = cosec2 A

= cosec2 A â€“ cosec2 A

= 0

= RHS

Therefore, LHS = RHS.

(iii) 1/(1 + tan2 A) + 1/(1 + cot2 A) = 1

LHS = 1/(1 + tan2 A) + 1/(1 + cot2 A)

We know that

sec2 A â€“ tan2 A = 1

sec2 A = 1 + tan2 A

cosec2 A â€“ cot2 A = 1

cosec2 A = 1 + cot2 A

So we get

= 1/sec2 A + 1/cosec2 A

Here 1/sec A = cos A and 1/cosec A = sin A

= cos2 A + sin2 A

= 1

= RHS

Therefore, LHS = RHS.

32. Simplify

Solution:

We know that 1 = sin2 Î¸ + cos2 Î¸

= âˆšcos2 Î¸/sin2 Î¸

= cos Î¸/ sin Î¸

Here cos Î¸/sin Î¸ = cot Î¸

= cot Î¸

Therefore,
= cot Î¸.

33. If sin Î¸ + cosec Î¸ = 2, find the value of sin2 Î¸ + cosec2 Î¸.

Solution:

It is given that

sin Î¸ + cosec Î¸ = 2

sin Î¸ + 1/sin Î¸ = 2

By further calculation

sin2 Î¸ + 1 = 2 sin Î¸

sin2 Î¸ â€“ 2 sin Î¸ + 1 = 0

So we get

(sin Î¸ â€“ 1)2 = 0

sin Î¸ â€“ 1 = 0

sin Î¸ = 1

Here

sin2 Î¸ + cosec2 Î¸ = sin2 Î¸ + 1/sin2 Î¸

Substituting the values

= 12 + 1/12

= 1 + 1/1

= 1 + 1

= 2

34. If x = a cos Î¸ + b sin Î¸ and y = a sin Î¸ â€“ b cos Î¸, prove that x2 + y2 = a2 + b2.

Solution:

It is given that

x = a cos Î¸ + b sin Î¸ â€¦. (1)

y = a sin Î¸ â€“ b cos Î¸ â€¦. (2)

By squaring and adding both the equations

x2 + y2 = (a cos Î¸ + b sin Î¸)2 + (a sin Î¸ â€“ b cos Î¸)2

Using the formula

(a + b)2 = a2 + b2 + 2ab and (a â€“ b)2 = a2 + b2 â€“ 2ab

= [(a cos Î¸)2 + (b sin Î¸)2 + 2 (a cos Î¸) (b sin Î¸)] + [(a sin Î¸)2 + (b cos Î¸)2 â€“ 2 (a sin Î¸) (b cos Î¸)]

By further calculation

= a2 cos2 Î¸ + b2 sin2 Î¸ + 2 ab sin Î¸ cos Î¸ + a2 sin2 Î¸ + b2 cos2Â Î¸ â€“ 2 ab sin Î¸ cos Î¸

= a2 cos2 Î¸ + b2 sin2 Î¸ + a2 sin2 Î¸ + b2 cos2Â Î¸

So we get

= a2 (cos2 Î¸ + sin2 Î¸) + b2 (sin2 Î¸ + cos2Â Î¸)

Here sin2 Î¸ + cos2Â Î¸ = 1

= a2 (1) + b2 (1)

= a2 + b2

Therefore, x2Â + y2 = a2 + b2.

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

AC2 = AB2 + BC2

AB2 = AC2 – BC2

AB2 = (3)2 â€“ (2)2

AB2 = 9 â€“ 4

AB2 = 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¸

Therefore, x = 10 cot Î¸ and y = 10 cosec Î¸.

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

(i) sin âˆ ABC

(ii) tan x â€“ cos x + 3 sin x.

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

(i) 5 sin x

(ii) 7 tan x

(iii) 5 cos x â€“ 17 sin y â€“ tan x.

Solution:

(a) From the figure

BC = 12, CD = 9 and BC = 20

In right angled âˆ†ABC,

Using Pythagoras theorem

AB2 = AC2 + BC2

It can be written as

AC2 = AB2 â€“ BC2

Substituting the values

AC2 = (20)2 â€“ (12)2

By further calculation

AC2 = 400 â€“ 144 = 256

So we get

AC2 = (16)2

AC = 16

In right angled âˆ†BCD

Using Pythagoras theorem

BD2 = BC2 + CD2

Substituting the values

BD2 = 122 + 92

By further calculation

BD2 = 144 + 81 = 225

So we get

BD2 = (15)2

BD = 15

(i) In right angled âˆ†BCD

sin âˆ ABC = perpendicular/hypotenuse

So we get

sin âˆ ABC = AC/AB = 16/20 = 4/5

(ii) In right angled âˆ†BCD

tan x = perpendicular/base

So we get

tan x = BC/CD = 12/9 = 4/3

In right angled âˆ†BCD

cos x = base/hypotenuse

So we get

cos x = CD/BD = 9/15 = 3/5

In right angled âˆ†BCD

sin x = perpendicular/hypotenuse

So we get

sin x = BC/BD = 12/15 = 4/5

tan x â€“ cos x + 3 sin x = 4/3 â€“ 3/5 + 3 Ã— 4/5

By further calculation

= 4/3 â€“ 3/5 + 12/5

Taking LCM

= (4 Ã— 5 â€“ 3 Ã— 3 + 12 Ã— 3)/ 15

So we get

= (20 â€“ 9 + 36)/ 15

= (56 â€“ 9)/ 15

= 27/15

= 3 2/15

Therefore, tan x â€“ cos x + 3 sin x = 3 2/15.

(b) In the figure

AC = 17, AB = 25, AD = 15

In right angled âˆ†ACD

Using Pythagoras theorem

Substituting the values

(17)2 = (15)2 + (CD)2

By further calculation

CD2 = (17)2 â€“ (15)2

CD2 = 289 â€“ 225 = 64

So we get

CD2 = 82

CD = 8

In right angled âˆ†ABD

Using Pythagoras theorem

Substituting the values

(25)2 = (15)2 + BD2

By further calculation

BD2 = (25)2 â€“ (15)2

BD2 = 625 â€“ 225 = 400

So we get

BD2 = (20)2

BD = 20

(i) In right angled âˆ†ABD

5 sin x = 5 (perpendicular/hypotenuse)

So we get

= 5 Ã— 15/25

= 15/5

= 3

(ii) In right angled âˆ†ABD

7 tan x = 7 (perpendicular/base)

So we get

= 7 Ã— 15/20

= 7 Ã— Â¾

= 21/4

= 5 Â¼

(iii) In right angled âˆ†ABD

cos x = base/hypotenuse

So we get

cos x = BD/AB = 20/25 = 4/5

In right angled âˆ†ACD

sin y = perpendicular/hypotenuse

So we get

sin y = CD/AC = 8/17

In right angled âˆ†ABD

tan x = perpendicular/base

So we get

tan x = AD/BD = 15/20 = Â¾

5 cos x â€“ 17 sin y â€“ tan x = 5 Ã— 4/5 â€“ 17 Ã— 8/17 â€“ Â¾

It can be written as

= 4/1 â€“ 8/1 â€“ Â¾

Taking LCM

= (16 â€“ 32 â€“ 3)/4

= (16 â€“ 35)/4

So we get

= – 19/4

= – 4 Â¾

Therefore, 5 cos x â€“ 17 sin y â€“ tan x = – 4 Â¾.

3. If q cos Î¸ = p, find tan Î¸ â€“ cot Î¸ in terms of p and q.

Solution:

Consider ABC as a triangle right angled at B and âˆ ACB = Î¸

It is given that

q cos Î¸ = p

cos Î¸ = BC/AC = p/q

Take BC = px then AC = qx

In right angled âˆ†ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

It can be written as

AB2 = AC2 â€“ BC2

Substituting the values

AB2 = (qx)2 â€“ (px)2

AB2 = q2x2 â€“ p2x2

Taking out the common terms

AB2 = (q2 â€“ p2)x2

So we get

AB = âˆš(q2 â€“ p2) x2

AB = (âˆšq2 â€“ p2)x

In right angled âˆ†ABC

tan Î¸ = perpendicular/base

So we get

tan Î¸ = AB/BC = [(âˆšq2 â€“ p2)x]/px

tan Î¸ = (âˆšq2 â€“ p2)/p

In right angled âˆ†ABC

cot Î¸ = base/perpendicular

So we get

cot Î¸ = BC/AB = px/[(âˆšq2 â€“ p2)x]

cot [(âˆšq2 â€“ p2)x] = p/(âˆšq2 â€“ p2)

4. Given 4 sin Î¸ = 3 cos Î¸, find the values of:

(i) sin Î¸

(ii) cos Î¸

(iii) cot2 Î¸ â€“ cosec2 Î¸.

Solution:

It is given that

4 sin Î¸ = 3 cos Î¸

sin Î¸/cos Î¸ = Â¾

tan Î¸ = Â¾

Consider âˆ†ABC right angled at B and âˆ ACB = Î¸

tan Î¸ = perpendicular/base

Substituting the values

Â¾ = AB/BC

AB/BC = Â¾

Take AB = 3x then BC = 4x

In right angled âˆ†ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

AC2 = (3x)2 + (4x)2

By further calculation

AC2 = 9x2 + 16x2 = 25x2

So we get

AC2Â = (5x)2

AC = 5x

(i) In right angled âˆ†ABC

sin Î¸ = perpendicular/hypotenuse

So we get

sin Î¸ = AB/AC = 3x/5x = 3/5

(ii) In right angled âˆ†ABC

cos Î¸ = base/hypotenuse

So we get

cos Î¸ = BC/AC = 4x/5x = 4/5

(iii) In right angled âˆ†ABC

cot Î¸ = base/perpendicular

So we get

cot Î¸ = BC/AB = 4x/3x = 4/3

In right angled âˆ†ABC

cosec Î¸ = hypotenuse/perpendicular

So we get

cosec Î¸ = AC/AB = 5x/3x = 5/3

Here

cot2 Î¸ â€“ cosec2 Î¸ = (4/3)2 â€“ (5/3)2

By further calculation

= 16/9 â€“ 25/9

= (16 â€“ 25)/9

= -9/9

= – 1

Therefore, cot2 Î¸ â€“ cosec2 Î¸ = -1.

5. If 2 cos Î¸ = âˆš3, prove that 3 sin Î¸ â€“ 4 sin3Â Î¸ = 1.

Solution:

It is given that

2 cos Î¸ = âˆš3

cos Î¸ = âˆš3/2

We know that

sin2 Î¸ = 1 â€“ cos2Â Î¸

Substituting the values

= 1 â€“ (âˆš3/2)2

= 1 â€“ Â¾

= Â¼

sin Î¸ = âˆš Â¼ = Â½

Consider

LHS = 3 sin Î¸ â€“ 4 sin3Â Î¸

It can be written as

= sin Î¸ (3 â€“ 4 sin2 Î¸)

Substituting the values

= Â½ (3 â€“ 4 Ã— Â¼)

= Â½ (3 â€“ 1)

= Â½ Ã— 1

= 1

= RHS

Therefore, proved.

6. If (sec Î¸ â€“ tan Î¸)/ (sec Î¸ + tan Î¸) = Â¼, find sin Î¸.

Solution:

We know that

By cross multiplication

4 â€“ 4 sin Î¸ = 1 + sin Î¸

We get

4 â€“ 1 = sin Î¸ + 4 sin Î¸

3 = 5 sin Î¸

sin Î¸ = 3/5

7. If sin Î¸ + cosec Î¸ = 3 1/3, find the value of sin2 Î¸ + cosec2 Î¸.

Solution:

It is given that

sin Î¸ + cosec Î¸ = 3 1/3 = 10/3

By squaring on both sides

(sin Î¸ + cosec Î¸)2Â = (10/3)2

Expanding using formula (a + b)2 = a2 + b2 + 2ab

sin2Â Î¸ + cosec2Â Î¸ + 2 sin Î¸ cosec Î¸ = 100/9

We know that sin Î¸ = 1/cosec Î¸

sin2Â Î¸ + cosec2Â Î¸ + 2 sin Î¸ Ã— 1/ sin Î¸ = 100/9

By further calculation

sin2Â Î¸ + cosec2Â Î¸ + 2 = 100/9

sin2Â Î¸ + cosec2Â Î¸ = 100/9 â€“ 2

Taking LCM

sin2Â Î¸ + cosec2Â Î¸ = (100 â€“ 18)/9 = 82/9

So we get

sin2Â Î¸ + cosec2Â Î¸ = 9 1/9

8. In the adjoining figure, AB = 4 m and ED = 3 m.

If sin Î± = 3/5 and cos Î² = 12/13, find the length of BD.

Solution:

It is given that

sin Î± = AB/AC = 3/5

AB = 3 and AC = 5

Using Pythagoras theorem

AC2Â = AB2 + BC2

Substituting the values

52 = 32 + BC2

By further calculation

25 = 9 + BC2

BC2Â = 25 â€“ 9 = 16

So we get

BC2 = 42

BC = 4

We know that

tan Î± = AB/BC = 4/5

cos Î² = CD/CE = 12/13

CD = 12 and CE = 13

Using Pythagoras theorem

CE2 = CD2 + ED2

Substituting the values

132Â = 122 + ED2

By further calculation

ED2 = 132 â€“ 122

ED2 = 169 â€“ 144 = 25

So we get

ED2 = (5)2

ED = 5

tan Î² = ED/CD = 5/12

From the figure

tan Î± = AB/BC = 4/BC

So we get

Â¾ = 4/BC

BC = (4 Ã— 4)/3 = 16/3 m

tan Î² = ED/CD = 3/CD

5/12 = 3/CD

So we get

CD = (12 Ã— 3)/5 = 36/5 m

Here

BD = BC + CD

Substituting the values

= 16/3 + 36/5

Taking LCM

= (80 + 108)/15

= 188/15 m

= 12 8/15 m