Trigonometry Class 9th Mathematics Part Ii MHB Solution

Trigonometry Class 9th Mathematics Part Ii MHB Solution

Practice Set 8.1

1. In the Fig. 8.12, ∠R is the right angle of ΔPQR. Write the following ratios. (i) sin P…
2. In the right angled ΔXYZ, ∠XYZ = 90^0 and a,b,c are the lengths of the sides as shown…
3. In right angled ΔLMN, ∠LMN =90^0 , ∠L = 50^0 and ∠N = 40^0 write the following ratios.…
4. In the figure 8.15 ∠PQR = 90^0 , ∠PQS = 90^0 , ∠PRQ = α and ∠QPS = θ Write the…

Practice Set 8.2

1. In the following table, a ratio is given in each column. Find the remaining two ratios…
2. 5sin 30^0 + 3tan45^0 Find the values of -
3. 4/5 tan^260^circle + 3sin^260^circle Find the values of -
4. 2sin 30^0 + cos 0^0 + 3sin 90^0 Find the values of -
5. tan60/sin60+cos60 Find the values of -
6. cos^2 45^0 + sin^2 30^0 Find the values of -
7. cos 60^0 × cos 30^0 + sin60^0 × sin30^0 Find the values of -
8. If sinθ = 4/5 then find cosθ.
9. If costheta = 15/17 then find sinθ

Problem Set 8

1. Which of the following statements is true? Choose the correct alternative answer for…
2. Which of the following is the value of sin 90°? Choose the correct alternative answer…
3. 2tan 45^0 + cos 45^0 - sin 45^0 =? Choose the correct alternative answer for following…
4. cos28^circle /sin62^circle = ? Choose the correct alternative answer for following…
5. In right angled ΔTSU, TS = 5, ∠S = 90^0 , SU =12 then find sin T, cos T, tan T.…
6. In right angled ΔYXZ, ∠X = 90^0 , XZ = 8cm, YZ =17cm, find sin Y, cos Y, tan Y, sin Z,…
7. In right angled ΔLMN, if ∠N = θ, ∠M = 90^0 , cosθ = 24/25 find sinθ and tanθ Similarly,…
8. Fill in the blanks. i. sin20^circle = cossquare^circle ii. tan30^circle x…

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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

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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

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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°