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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…
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°


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