Trigonometry
Class 9th Mathematics Part Ii MHB Solution
- In the Fig. 8.12, ∠R is the right angle of ΔPQR. Write the following ratios. (i) sin P…
- In the right angled ΔXYZ, ∠XYZ = 90^0 and a,b,c are the lengths of the sides as shown…
- In right angled ΔLMN, ∠LMN =90^0 , ∠L = 50^0 and ∠N = 40^0 write the following ratios.…
- In the figure 8.15 ∠PQR = 90^0 , ∠PQS = 90^0 , ∠PRQ = α and ∠QPS = θ Write the…
- In the following table, a ratio is given in each column. Find the remaining two ratios…
- 5sin 30^0 + 3tan45^0 Find the values of -
- 4/5 tan^260^circle + 3sin^260^circle Find the values of -
- 2sin 30^0 + cos 0^0 + 3sin 90^0 Find the values of -
- tan60/sin60+cos60 Find the values of -
- cos^2 45^0 + sin^2 30^0 Find the values of -
- cos 60^0 × cos 30^0 + sin60^0 × sin30^0 Find the values of -
- If sinθ = 4/5 then find cosθ.
- If costheta = 15/17 then find sinθ
- Which of the following statements is true? Choose the correct alternative answer for…
- Which of the following is the value of sin 90°? Choose the correct alternative answer…
- 2tan 45^0 + cos 45^0 - sin 45^0 =? Choose the correct alternative answer for following…
- cos28^circle /sin62^circle = ? Choose the correct alternative answer for following…
- In right angled ΔTSU, TS = 5, ∠S = 90^0 , SU =12 then find sin T, cos T, tan T.…
- In right angled ΔYXZ, ∠X = 90^0 , XZ = 8cm, YZ =17cm, find sin Y, cos Y, tan Y, sin Z,…
- In right angled ΔLMN, if ∠N = θ, ∠M = 90^0 , cosθ = 24/25 find sinθ and tanθ Similarly,…
- Fill in the blanks. i. sin20^circle = cossquare^circle ii. tan30^circle x…
Practice Set 8.1
Question 1.In the Fig. 8.12, ∠R is the right angle of ΔPQR. Write the following ratios.
(i) sin P (ii) cos Q
(iii) tan P (iv) tan Q
Answer:
For any right-angled triangle,
sinθ = Opposite side Side/Hypotenuse
cosθ = Adjacent sideSide/Hypotenuse
tanθ = sinθ/cosθ
= Opposite side Side/Adjacent sideSide
cotθ = 1/tanθ
= Adjacent sideSide/Opposite side Side
secθ = 1/cosθ
= Hypotenuse/Adjacent sideSide
cosecθ = 1/sinθ
= Hypotenuse/Opposite side Side
In the given triangle let us understand, the Opposite side and Adjacent sidesides.
So for ∠ P,
Opposite side Side = QR
Adjacent sideSide = PR
So, for ∠ Q,
Opposite side Side = PR
Adjacent sideSide = QR
In general for the side Opposite side to the 90° angle is the hypotenuse.
So, for Δ PQR, hypotenuse = PQ
(i) sin P = Opposite side Side/Hypotenuse
= QR/PQ
(ii) cos Q = Adjacent sideSide/Hypotenuse
= QR/PQ
(iii) tan P = sinθ/cosθ
= Opposite side Side/Adjacent sideSide
= QR/PR
(iv) tan Q = sinθ/cosθ
= Opposite side Side/Adjacent sideSide
= PR/QR
Question 2.
In the right angled ΔXYZ, ∠XYZ = 900 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.
Answer:
For any right-angled triangle,
sinθ = Opposite side Side/Hypotenuse
cosθ = Adjacent Side/Hypotenuse
tanθ = sinθ/cosθ
= Opposite Side/Adjacent Side
In the given triangle let us understand, the Opposite side and Adjacent side
So for ∠ X,
Opposite Side = YZ = a
Adjacent Side = XY = b
So for ∠ Z,
Opposite Side = XY = b
Adjacent Side = YZ = a
In general for the side Opposite side to the 90° angle is the hypotenuse.
So for Δ XYZ, hypotenuse = XZ = c
(i) sin X = Opposite side Side/Hypotenuse
= YZ/XZ
= a/c
(ii) tan Z = sinθ/cosθ
= Opposite Side/Adjacent Side
= XY/YZ
= b/a
(iii) cos X= Adjacent Side/Hypotenuse
= XY/XZ
= b/c
(iv) tan X = sinθ/cosθ
= Opposite Side/Adjacent Side
= YZ/XY
= a/b
Question 3.
In right angled ΔLMN, ∠LMN =900, ∠L = 500 and ∠N = 400
write the following ratios.
(i) sin 50° (ii) cos 50°
(iii) tan 40° (iv) cos 40°
Answer:
For any right-angled triangle,
sinθ = Opposite side Side/Hypotenuse
cosθ = Adjacent sideSide/Hypotenuse
tanθ = sinθ/cosθ
= Opposite side Side/Adjacent sideSide
cotθ = 1/tanθ
= Adjacent sideSide/Opposite side Side
secθ = 1/cosθ
= Hypotenuse/Adjacent sideSide
cosecθ = 1/sinθ
= Hypotenuse/Opposite side Side
In the given triangle let us understand, the Opposite side and Adjacent sidesides.
So for ∠ 50°,
Opposite side Side = MN
Adjacent sideSide = LM
So for ∠ 40°,
Opposite side Side = LM
Adjacent sideSide = MN
In general, for the side Opposite side to the 90° angle is the hypotenuse.
So, for Δ LMN, hypotenuse = LN
(i) sin 50° = Opposite side Side/Hypotenuse
= MN/LN
(ii) cos 50° = Adjacent sideSide/Hypotenuse
= LM/LN
(iii) tan 40° = sinθ/cosθ
= Opposite side Side/Adjacent sideSide
= LM/MN
(iv) cos 40° = Adjacent sideSide/Hypotenuse
= MN/LN
Question 4.
In the figure 8.15 ∠PQR = 900, ∠PQS = 900, ∠PRQ = α and ∠QPS = θ Write the following trigonometric ratios.
i. sin α, cos α, tan α
ii. sin θ, cos θ, tan θ
Answer:
For any right-angled triangle,
sinθ = Opposite side Side/Hypotenuse
cosθ = Adjacent sideSide/Hypotenuse
tanθ = sinθ/cosθ
= Opposite side Side/Adjacent sideSide
cotθ = 1/tanθ
= Adjacent sideSide/Opposite side Side
secθ = 1/cosθ
= Hypotenuse/Adjacent sideSide
cosecθ = 1/sinθ
= Hypotenuse/Opposite side Side
(i) In the given triangle let us understand, the Opposite side and Adjacent sidesides.
So, for Δ PQR,
So, for ∠ α,
Opposite side Side = PQ
Adjacent sideSide = QR
In general for the side Opposite side to the 90° angle is the hypotenuse.
So, for Δ PQR, hypotenuse = PR
sin α = Opposite side Side/Hypotenuse
= PQ/PR
cos α = Adjacent sideSide/Hypotenuse
= QR/PR
tan α = sinθ/cosθ
= Opposite side Side/Adjacent sideSide
= PQ/QR
(ii) In the given triangle let us understand, the Opposite side and Adjacent sidesides.
So for Δ PQS,
So for ∠θ,
Opposite side Side = QS
Adjacent sideSide = PQ
In general for the side Opposite side to the 90° angle is the hypotenuse.
So for Δ PQS, hypotenuse = PS
sinθ = Opposite side Side/Hypotenuse
= QS/PS
cosθ = Adjacent sideSide/Hypotenuse
= PQ/PS
tanθ = sinθ/cosθ
= Opposite side Side/Adjacent sideSide
= QS/PQ
Practice Set 8.2
Question 1.In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.
Answer:
For first column:
cosθ = 35/37
Adjacent side= 35,
Hypotenuse = 37
By Pythagoras Theorem
Hypotenuse2 = Opposite side2 + Adjacent2
Opposite side2 = Hypotenuse2 - Adjacent2
= 372 - 352
= 1369 – 1225
Opposite side2 = 144
Opposite side = 12
For second column:
Opposite side = 11
Hypotenuse = 61
By Pythagoras Theorem
Hypotenuse2 = Opposite side2 + Adjacent2
Adjacent2 = Hypotenuse2 - Opposite side2
= 612 - 112
= 3721 – 121
Adjacent2 = 3600
Adjacent side= 60
For third column:
Opposite side = 1
Adjacent side= 1
By Pythagoras Theorem
Hypotenuse2 = Opposite side2 + Adjacent2
= 1 + 1
Hypotenuse2 = 2
Hypotenuse = √2
For fourth column:
Opposite side = 1
Hypotenuse = 2
By Pythagoras Theorem
Hypotenuse2 = Opposite side2 + Adjacent2
Adjacent2 = Hypotenuse2 - Opposite side2
= 22 - 12
= 4 – 1
Adjacent2 = 3
Adjacent side= √3
For fifth column:
Adjacent side= 1
Hypotenuse = √3
By Pythagoras Theorem
Hypotenuse2 = Opposite side2 + Adjacent2
Opposite side2 = Hypotenuse2 - Adjacent2
= (√3)2 - 12
= 3 – 1
Opposite side2 = 2
Opposite side = √2
For sixth column:
Opposite side = 21
Adjacent side= 20
By Pythagoras Theorem
Hypotenuse2 = Opposite side2 + Adjacent2
= 212 + 202
Hypotenuse2 = 841
Hypotenuse = 29
For seventh column:
Opposite side = 8
Adjacent side= 15
By Pythagoras Theorem
Hypotenuse2 = Opposite side2 + Adjacent2
= 82 + 152
Hypotenuse2 = 289
Hypotenuse = 17
For eighth column:
Opposite side = 3
Hypotenuse = 5
By Pythagoras Theorem
Hypotenuse2 = Opposite side2 + Adjacent2
Adjacent2 = Hypotenuse2 - Opposite side2
= 52 - 32
= 25 – 9
Adjacent2 = 16
Adjacent side= 4
For ninth column:
Opposite side = 1
Adjacent side= 2√2
By Pythagoras Theorem
Hypotenuse2 = Opposite side2 + Adjacent2
= 12 + (2√2)2
Hypotenuse2 = 9
Hypotenuse = 3
Question 2.
Find the values of –
5sin 300 + 3tan450
Answer:
We know,
sin 30° = 1/2
tan 45° = 1
⟹ 5sin 30° + 3tan 45°
⟹ 
⟹ 2.5 + 3
⟹ 5.5
Question 3.
Find the values of –
Answer:
We know,
tan 60° = √3
sin 60° = √3/2
⟹ 
⟹ 
⟹ 
⟹ 
⟹ 
= 93/20
Question 4.
Find the values of –
2sin 300 + cos 00 + 3sin 900
Answer:
We know,
sin 30° = 1/2
cos 0° = 1
sin 90° = 1
⟹ 
⟹ 
⟹ 1 + 1 + 1
= 3
Question 5.
Find the values of –
Answer:
We know,
tan 60° = √3
sin 60° = √3/2
cos 60° = 1/2
⟹ 
⟹ 
⟹ 
Question 6.
Find the values of –
cos2450 + sin2300
Answer:
We know,
cos 45° = 1/√2
sin 30° = 1/2
⟹ 
⟹ 
⟹ 
Question 7.
Find the values of –
cos 600× cos 300 + sin600 × sin300
Answer:
We know,
sin 30° = 1/2
sin 60° = √3/2
cos 60° = 1/2
cos 30° = √3/2
⟹ 
⟹ 
⟹ 
⟹ 
Question 8.
If sinθ = 4/5 then find cosθ.
Answer:
We know,
sinθ = Opposite side/Hypotenuse
Given:
sinθ = 4/5
Opposite side = 4
Hypotenuse = 5
By Pythagoras Theorem
Hypotenuse2 = Opposite side2 + Adjacent2
Adjacent2 = Hypotenuse2 - Opposite side2
= 52 - 42
= 25 – 16
= 9
Adjacent2 = 9
Adjacent side= 3
cosθ = Adjacent side/Hypotenuse
= 3/5
Question 9.
If 
then find sinθ
Answer:
We know,
cosθ = Adjacent side/Hypotenuse
Adjacent side = 15
Hypotenuse = 17
By Pythagoras Theorem
Hypotenuse2 = Opposite side2 + Adjacent2
Opposite side2 = Hypotenuse2 - Adjacent2
= 172 - 152
= 289 – 225
= 64
Opposite side2 = 64
Opposite side = 8
sinθ = Opposite side /Hypotenuse
= 8/17
Problem Set 8
Question 1.Choose the correct alternative answer for following multiple choice questions.
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)
Answer:
Let us consider the given triangle,
In this Δ PMN,
For ∠θ,
Opposite side = PM
Adjacent side= PN
For ∠ (90 –θ)
Opposite side = MN
Adjacent side = PM
sinθ = Opposite side/Hypotenuse
= PM/PN …………………… (i)
cos (90-θ) = Adjacent/Hypotenuse
= PM/PN ……………………. (ii)
RHS of equation (i) and (ii) are equal
∴ sinθ = cos (90-θ)
So Option A is correct.
Question 2.
Choose the correct alternative answer for following multiple choice questions.
Which of the following is the value of sin 90°?
A. 
B. 0
C. 
D. 1
Answer:
We know that the value of sin 90° = 1
So option D is correct.
Question 3.
Choose the correct alternative answer for following multiple choice questions.
2tan 450 + cos 450 – sin 450 =?
A. 0
B. 1
C. 2
D. 3
Answer:
We know that,
tan 45° = 1
We also know that
cos 45° = sin 45°
So,
⟹ 2 × 1 + cos 45° - cos 45°
= 2
So the correct option is C.
Question 4.
Choose the correct alternative answer for following multiple choice questions.
A. 2
B. -1
C. 0
D. 1
Answer:
We know the identity that,
sinθ = cos (90 –θ)
sin 62° = cos (90 – 62)
= cos 28°
Therefore [cos 28°/cos 28°] = 1
So option D is correct.
Question 5.
In right angled ΔTSU, TS = 5, ∠S = 900, SU =12 then find sin T, cos T, tan T. Similarly find sin U, cos U, tan U.
Answer: By applying Pythagoras theorem to given triangle we have,
TU2=ST2+SU2
TU2= 52+122
TU2= 25+144
TU2=169
TU=13
Now,
sinT
cosT
tanT
Similarly,
Question 6.
In right angled ΔYXZ, ∠X = 900, XZ = 8cm, YZ =17cm, find sin Y, cos Y, tan Y, sin Z, cos Z, tan Z.
Answer:
For any right-angled triangle,
sinθ = Opposite side /Hypotenuse
cosθ = Adjacent side/Hypotenuse
tanθ = sinθ/cosθ
= Opposite side/Adjacent side
cotθ = 1/tanθ
= Adjacent side/Opposite side
secθ = 1/cosθ
= Hypotenuse/Adjacent side
cosecθ = 1/sinθ
= Hypotenuse/Opposite side
In the given triangle let us understand, the Opposite side and Adjacent sides.
So for ∠ Y,
Opposite side = XZ =8
Adjacent side= XY
So for ∠ Z,
Opposite side = XY
Adjacent side = XZ = 8
In general for the side Opposite side to the 90° angle is the hypotenuse.
So for Δ TSU,
By Pythagoras Theorem
YZ2 = XZ2 + XY2
XY2 = 172 - 82
= 289 - 64
= 225
XY = 15
(i) sin Y = Opposite side/Hypotenuse
= XZ/YZ
= 8/17
(ii) cos Y = Adjacent side/Hypotenuse
= XY/YZ
= 15/17
(iii) tan Y = sinθ/cosθ
= Opposite side/Adjacent side
= XZ/XY
= 8/15
(i) sin Z = Opposite side/Hypotenuse
= XY/YZ
= 15/17
(ii) cos Z = Adjacent side/Hypotenuse
= XZ/YZ
= 8/17
(iii) tan Z = sinθ/cosθ
= Opposite side/Adjacent side
= XZ/XY
= 8/15
Question 7.
In right angled ΔLMN, if ∠N = θ, ∠M = 900, cosθ = 24/25 find sinθ and tanθ Similarly, find (sin2θ) and (cos2θ).
Answer:
Give:
cosθ = 24/25
cosθ = Adjacent side/Hypotenuse
Adjacent side = 24
Hypotenuse = 25
By Pythagoras Theorem
Hypotenuse2 = Opposite side2 + Adjacent2
Opposite side2 = Hypotenuse2 - Adjacent2
= 252 - 242
= 625 – 576
= 49
Opposite side2 = 49
Opposite side = 7
sinθ = Opposite side/Hypotenuse
= 7/25
tanθ = sinθ/cosθ
= Opposite side/Adjacent side
= 7/24
sin2θ = (7/25)2
= 49/625
cos2θ = (24/25)2
= 576/625
Question 8.
Fill in the blanks.
i. 
ii. 
iii. 
Answer:
i. We know the following identity,
sinθ = cos (90 -θ)
So sin 20° = cos (90 – 20)
∴ sin 20° = cos 70°
ii. We know that,
Let the unknown angle be θ
tan 30° = 
tanθ = 
= 
tanθ = √3
θ = tan-1(√3)
∴θ = 60°
iii. We know that,
cosθ = sin (90 -θ )
cos 40° = sin (90 – 40)
∴ cos 40° = sin 50°