Advertisement

PRINTABLE FOR KIDS

XII (12) HSC

XI (11) FYJC
X (10) SSC

Trigonometry Class 9th Mathematics Part Ii MHB Solution

Trigonometry Class 9th Mathematics Part Ii MHB Solution

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°

PDF FILE TO YOUR EMAIL IMMEDIATELY PURCHASE NOTES & PAPER SOLUTION. @ Rs. 50/- each (GST extra)

SUBJECTS

HINDI ENTIRE PAPER SOLUTION

MARATHI PAPER SOLUTION
SSC MATHS I PAPER SOLUTION
SSC MATHS II PAPER SOLUTION
SSC SCIENCE I PAPER SOLUTION
SSC SCIENCE II PAPER SOLUTION
SSC ENGLISH PAPER SOLUTION
SSC & HSC ENGLISH WRITING SKILL
HSC ACCOUNTS NOTES
HSC OCM NOTES
HSC ECONOMICS NOTES
HSC SECRETARIAL PRACTICE NOTES

2019 Board Paper Solution

HSC ENGLISH SET A 2019 21st February, 2019

HSC ENGLISH SET B 2019 21st February, 2019

HSC ENGLISH SET C 2019 21st February, 2019

HSC ENGLISH SET D 2019 21st February, 2019

SECRETARIAL PRACTICE (S.P) 2019 25th February, 2019

HSC XII PHYSICS 2019 25th February, 2019

CHEMISTRY XII HSC SOLUTION 27th, February, 2019

OCM PAPER SOLUTION 2019 27th, February, 2019

HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019

HSC MATHS PAPER SOLUTION SCIENCE 2nd, March, 2019

SSC ENGLISH STD 10 5TH MARCH, 2019.

HSC XII ACCOUNTS 2019 6th March, 2019

HSC XII BIOLOGY 2019 6TH March, 2019

HSC XII ECONOMICS 9Th March 2019

SSC Maths I March 2019 Solution 10th Standard11th, March, 2019

SSC MATHS II MARCH 2019 SOLUTION 10TH STD.13th March, 2019

SSC SCIENCE I MARCH 2019 SOLUTION 10TH STD. 15th March, 2019.

SSC SCIENCE II MARCH 2019 SOLUTION 10TH STD. 18th March, 2019.

SSC SOCIAL SCIENCE I MARCH 2019 SOLUTION20th March, 2019

SSC SOCIAL SCIENCE II MARCH 2019 SOLUTION, 22nd March, 2019

XII CBSE - BOARD - MARCH - 2019 ENGLISH - QP + SOLUTIONS, 2nd March, 2019

HSC Maharashtra Board Papers 2020

(Std 12th English Medium)

HSC ECONOMICS MARCH 2020

HSC OCM MARCH 2020

HSC ACCOUNTS MARCH 2020

HSC S.P. MARCH 2020

HSC ENGLISH MARCH 2020

HSC HINDI MARCH 2020

HSC MARATHI MARCH 2020

HSC MATHS MARCH 2020

SSC Maharashtra Board Papers 2020

(Std 10th English Medium)

English MARCH 2020

HindI MARCH 2020

Hindi (Composite) MARCH 2020

Marathi MARCH 2020

Mathematics (Paper 1) MARCH 2020

Mathematics (Paper 2) MARCH 2020

Sanskrit MARCH 2020

Important-formula

THANKS