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°
