##### Trigonometry Class 9th Mathematics Part Ii MHB Solution

**Practice Set 8.1**

- 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…

**Practice Set 8.2**

- 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

**Problem Set 8**

- 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 = 90^{0} 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 =90^{0}, ∠L = 50^{0} and ∠N = 40^{0}

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 = 90^{0}, ∠PQS = 90^{0}, ∠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

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

Opposite side^{2} = Hypotenuse^{2} - Adjacent^{2}

= 37^{2} - 35^{2}

= 1369 – 1225

Opposite side^{2} = 144

Opposite side = 12

__For second column:__

Opposite side = 11

Hypotenuse = 61

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

Adjacent^{2} = Hypotenuse^{2} - Opposite side^{2}

= 61^{2} - 11^{2}

= 3721 – 121

Adjacent^{2} = 3600

Adjacent side= 60

__For third column:__

Opposite side = 1

Adjacent side= 1

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

= 1 + 1

Hypotenuse^{2} = 2

Hypotenuse = √2

__For fourth column:__

Opposite side = 1

Hypotenuse = 2

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

Adjacent^{2} = Hypotenuse^{2} - Opposite side^{2}

= 2^{2} - 1^{2}

= 4 – 1

Adjacent^{2} = 3

Adjacent side= √3

__For fifth column:__

Adjacent side= 1

Hypotenuse = √3

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

Opposite side^{2} = Hypotenuse^{2} - Adjacent^{2}

= (√3)^{2} - 1^{2}

= 3 – 1

Opposite side^{2} = 2

Opposite side = √2

__For sixth column:__

Opposite side = 21

Adjacent side= 20

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

= 21^{2} + 20^{2}

Hypotenuse^{2} = 841

Hypotenuse = 29

__For seventh column:__

Opposite side = 8

Adjacent side= 15

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

= 8^{2} + 15^{2}

Hypotenuse^{2} = 289

Hypotenuse = 17

__For eighth column:__

Opposite side = 3

Hypotenuse = 5

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

Adjacent^{2} = Hypotenuse^{2} - Opposite side^{2}

= 5^{2} - 3^{2}

= 25 – 9

Adjacent^{2} = 16

Adjacent side= 4

__For ninth column:__

Opposite side = 1

Adjacent side= 2√2

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

= 1^{2} + (2√2)^{2}

Hypotenuse^{2} = 9

Hypotenuse = 3

**Question 2.**

Find the values of –

5sin 30^{0} + 3tan45^{0}

**Answer:**

We know,

sin 30° = 1/2

tan 45° = 1

⟹ 5sin 30° + 3tan 45°

⟹

⟹ 2.5 + 3

⟹ 5.5

**Question 3.**

**Answer:**

We know,

tan 60° = √3

sin 60° = √3/2

⟹

⟹

⟹

⟹

⟹

= 93/20

**Question 4.**

Find the values of –

2sin 30^{0} + cos 0^{0} + 3sin 90^{0}

**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 –

cos^{2}45^{0} + sin^{2}30^{0}

**Answer:**

We know,

cos 45° = 1/√2

sin 30° = 1/2

⟹

⟹

⟹

**Question 7.**

Find the values of –

cos 60^{0}× cos 30^{0} + sin60^{0} × sin30^{0}

**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

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

Adjacent^{2} = Hypotenuse^{2} - Opposite side^{2}

= 5^{2} - 4^{2}

= 25 – 16

= 9

Adjacent^{2} = 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

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

Opposite side^{2} = Hypotenuse^{2} - Adjacent^{2}

= 17^{2} - 15^{2}

= 289 – 225

= 64

Opposite side^{2} = 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 45^{0} + cos 45^{0} – sin 45^{0} =?

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 = 90^{0}, 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,

TU

^{2}=ST

^{2}+SU

^{2}

TU

^{2}= 5

^{2}+12

^{2}

TU

^{2}= 25+144

TU

^{2}=169

TU=13

Now,

sinT

cosT

tanT

Similarly,

**Question 6.**

In right angled ΔYXZ, ∠X = 90^{0}, 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

YZ^{2} = XZ^{2} + XY^{2}

XY^{2} = 17^{2} - 8^{2}

= 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 = 90^{0}, cosθ = 24/25 find sinθ and tanθ Similarly, find (sin^{2}θ) and (cos^{2}θ).

**Answer:**

Give:

cosθ = 24/25

cosθ = Adjacent side/Hypotenuse

Adjacent side = 24

Hypotenuse = 25

By Pythagoras Theorem

Hypotenuse^{2} = Opposite side^{2} + Adjacent^{2}

Opposite side^{2} = Hypotenuse^{2} - Adjacent^{2}

= 25^{2} - 24^{2}

= 625 – 576

= 49

Opposite side^{2} = 49

Opposite side = 7

sinθ = Opposite side/Hypotenuse

= 7/25

tanθ = sinθ/cosθ

= Opposite side/Adjacent side

= 7/24

sin^{2}θ = (7/25)^{2}

= 49/625

cos^{2}θ = (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°