# Maharashtra Board Class 9 Math Part 2 Chapter 8 Trigonometry Solution

## Maharashtra Board Class 9 Math Part 2 Solution Chapter 8 – Trigonometry

Balbharati Maharashtra Board Class 9 Math Part 2 Solution Chapter 8: Trigonometry. Marathi or English Medium Students of Class 9 get here Trigonometry full Exercise Solution.

 Std Maharashtra Class 9 Subject Math Part 2 Solution Chapter Trigonometry

### Practice Set – 8.1

(1) In the Fig. 8.12, R is the right angle of PQR. Write the following ratios.

(i) sin P

(ii) cos Q

(iii) tan P

(iv) tan Q

Solution:

(i) sin P = RQ/PQ

(ii) cos Q = QR/PQ

(iii) tan P = QR/PR

(iv) tan Q = PR/QR

(2) In the right angled XYZ, XYZ = 90° and a, b, c are the lengths of the sides as shown in the figure. Write the following ratios,

(i) sin X

(ii) tan Z

(iii) cos X

(iv) tan X

Solution:

(i) sin X = a/c

(ii) tan Z = b/a

(iii) cos X = b/c

(iv) tan X = a/b

(3) In right angled LMN, LMN = 90° L = 50° and N = 40°,

Write the following ratios.

(i) sin 50°

(ii) cos 50°

(iii) tan 40°

(iv) cos 40°

Solution:

(i) sin 50° = MN/LN

(ii) cos 50° = LM/LN

(iii) tan 40° = LM/MN

(iv) cos 40° = MN/LN

(4) In the figure 8.15, PQR = 90°, PQS = 90°, PRQ = α and QPS = θ

Write the following trigonometric ratios.

(i) sin α, cos α, tan α

(ii) sin θ, cos θ, tan θ

Solution:

(i) sin α = PQ/RP

cos α = RQ/RP

tan α = PQ/QR

(ii) sin θ = QS/PS, cos θ = PQ/PS, tan θ = QS/PQ

### Practice Set – 8.2

(1) In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.

 sin θ 11/61 1/2 3/2 cos θ 35/37 1/√3 tan θ 1 21/20 8/15 1/2√2

Solution:

Given cos θ = 35/37 ∴ b = 35, h = 37

∴ P = √372 – 352 = 12

∴ sin θ = 12/37, tan θ = 12/35

Given, sin θ = 11/61 ∴ P = 11, h = 61

∴ b = √612 – 112 = 60

cos θ = 60/61, tan θ = 11/60

Given, tan θ = 1 => tan 45° = 1

∴ θ = 45°

∴ sin θ = 1/√2, cos θ = 1/√2

Given, sin θ = ½ => sin 30° = ½ ∴ θ = 30°

∴ cos θ = √3/2, tan θ = 1/√3

Given, cos θ = 1/√3 ∴ b = 1, h = √3

∴ P = √3 – 1 = √2

∴ sin θ = √2/√3, tan θ = √2/1

Given, tan θ = 21/20 P = 21, b = 20

∴ h = √212 + 202 = 29

sin θ = 21/29, cos θ = 20/29

Given, tan θ = 8/15, P = 8, b = 15

h = √82 + 152 = 17

sin θ = 8/17, cos θ = 15/17

Given, sin θ = 3/5, P = 3, h = 5

b = √52 – 32

= √25 – 9 = 4

cos θ = 4/5, tan θ = ¾

Given, tan θ = 1/2√2 P = 1, b = 2√2

∴ h = √12 + (2√2)2

= 3

∴ sin θ = 1/3 cos θ = 2√2/3

(2) Find the values of –

(i) 5 sin 30° + 3 tan 45°

= 5/2 + 3

= 11/2

(ii) 4/5 tan2 60° + 3 sin2 60°

= 4/5 × 3 + 3 × ¾

= 12/5 + 9/4

= 48+45/20

= 93/20

(iii) 2 sin 30° + cos 0° + 3sin 90°

= 2 × ½ + 1+ 3 ×1

= 5

(iv) tan 60°/sin 60° + cos 60°

= √3/(√3/2 + ½)

= 2√3/√3 +1

(v) cos2 45° + sin2 30°

= ½ + 1/4  = 3/4

(vi) cos 60° × cos 30° + sin 60° × sin 60°

= ½ × √3/2 + √3/2 × 1/2

= √3/2

(3) If sin θ = 4/5 then find cos θ

Solution:

Sin θ = 4/5

P = 4, h = 5

b = √52 – 42 = 3

∴ cos θ = 3/5

(4) If cos θ = 15/17 then find sin θ

Solution:

Cos θ = 15/17

b = 15 h = 17

p = √172 – 152 = 8

∴ sin θ = 8/17

### Problem set – 8

(1) Choose the correct alternative answer for following multiple choice questions.

(i) Which of the following statements is true?

(A) sin θ = cos (90 – θ)

(B) cos θ = tan (90 – θ)

(C) sin θ = tan (90 – θ)

(D) tan θ = tan (90 – θ)

Ans:

(A) sin θ = cos (90 – θ).

(ii) Which of the following is the value of sin 90°?

(A) √3/2

(B) 0

(C) ½

(D) 1

Ans:

(D) 1.

(iii) 2 tan 45° + cos45° – sin45° =?

(A) 0

(B) 1

(C) 2

(D) 3

Ans:

(C) 2.

(iv) cos 28°/sin 62° =?

(A) 2

(B) -1

(C) 0

(D) 1

Ans:

Cos28°/sin62° = cos28°/cos(90-62) = cos28°/cos28° = 1 (D) 1.

(2) In right angled TSU, TS = 5, S = 90°, SU = 12 then find sin T, cos T, tan T. Similarly find sin U, cos U, tan U.

Solution:

T = 5, SU = 12

∴ TU = √122 + 52 = 13

∴ sin T = 12/13, cos T = 5/13, tan T = 12/5

Sin U = 5/13, cos U = 12/13, tan U = 5/12

(3) In right angled YXZ, X = 90°, XZ = 8 cm, YZ = 17 cm, find sin Y, cos Y, tan Y, sin Z, cos Z, tan Z.

Solution:

XZ = 8, YZ = 17

∴ XY = √172 – 82 = 15 cm

Sin Y = 8/17, cos Y = 15/17, tan Y = 8/15

Sin Z = 15/17, cos Z = 8/17, tan Z = 15/8

(4) In right angled LMN, if N = θ, M = 90°, cos θ = 24/25, find sinθ and tanθ Similarly, find (sin2 θ) and (cos2 θ).

Solution:

cos θ = 24/25

∴ b = 24, h = 25

P = √252 – 242 = 7

∴ sin θ = 7/25, tan θ = 7/24

sin2 θ = 49/625, cos2 = 576/625

(5) Fill in the blanks.

(i) sin 20° = cos 70°

(ii) tan 30° × tan 60° = 1

(iii) cos 40° = sin 50°

Updated: September 11, 2021 — 11:15 pm