Introduction to Trigonometry Important Questions: Applications of Trigonometry

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Trigonometry is a branch of mathematics that studies the relation between the side and angles of a triangle and this concept was given by the Greek Mathematician Hipparchus.

Chapter 8 Introduction to Trigonometry falls under Unit 5 Trigonometry who carries a total weightage of 12 marks according to the current CBSE Class 10 Mathematics Exam Pattern.

Table of Content

Applications of Trigonometry
Basic Formulas

Applications of Trigonometry

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Trigonometry has different applications such as :

It is used extensively in the aviation industry
It is used for cartography for the creation of maps.
It is used for light and sound waves.

There are basically six functions of an angle commonly used in trigonometry. The names are :

Sine (sin)
Cosine (cos)
Tangent(tan)
Cotangent(cot)
Secant(sec)
Cosecant(cosec)

Basic Formulas

sinA = Opposite/Hypotenuse

cosA = Adjacent/Hypotenuse

tanA = Opposite/Adjacent = sinA/cosA

cosA = 1/sinA

secA = 1/cosA

cotA = 1/tanA = cosA/sinA

Basic Identities

Sin²A + cos²A = 1

Tan²A + 1 = sec²A

1 + cot²A = cosec²A

Important Questions

Ques 1. Given that, In a right angle triangle ABC, tan B =12/5, then find sin B? (2 marks)

Ans: 1st Method

tanB = 12/5

cotB = 5/12

Cosec²B = 1 + cot²B = 1+[(5/12)2]

= 1 + [25/144]

= (144 + 25 )/144

=169/144

cosecB= 13/12

sinB=12/13

2nd Method

tanB = 12/5

tanB= AC/BC

let AC=12k, BC= 5k

In right angle triangle ΔACB,

AB² = AC² + BC² …..[Pythagoras theorem]

AB² = (12k)² + (5k)²

AB² = 144k² + 25k² = 169k²

AB = 13K

sinB = AC/AB = 12k/13k = 12/13

Ques 2. If sinA = 3/4 calculate cosA and tanA ? (3 marks)

Ans: Let ABC is a right angle triangle, right-angled at B.

sinA = 3/4

sinA = Opposite side/ Hypotenuse side = 3/4

Now, let BC = 3k and AC = 4k

Where k is a positive real number.

According to pythagoras theorem,

H² = P² + B²

AC² = AB² + BC²

Substitute the values in to the equation

(4k)² = (AB)² + (BC)²

16k² – 9k² = AB²

AB² = 7k²

Hence, AB = 7k

Now, We need to find the value of cosA and tanA.

cosA = Adjacent side / Hypotenuse side = AB/AC

cosA = 7k/4k = 3/7.

Ques 3. If 3cotA = 4, check whether (1 – tan²A)(1+tan²A) = cos²A – sin²A or not ? (4 marks)

Ans: Given,

3cotA = 4

cotA = 4/3

since, tanA = 1/cotA

tanA = 1/(4/3)=3/4

BC/AB = 3/4

Let BC = 3k and AB = 4k

By Pythagoras theorem:

H² = P² + B²

AC² = AB² + BC²

AC² = (4k)² + (3k)²

AC² = 16k² + 9k²

AC = 25k2 = 5k

sinA = Opposite side/ Hypotenuse

= BC/AC

= 3k/5k

= 3/5

In the same way,

cosA = Adjacent side/ Hypotenus

=AB/AC

= 4k/5k

=4/5

To check: (1-tan²A)/(1+tan²A) = cos²A – sin²A or not

L.H.S

(1-tan²A)/(1+tan²A) = [1 – (3/4)²]/ [1+ (3/4)²]

= [1-(9/16)]/[1+(9/16)] = 7/25

R.H.S. = cos²A – sin²A = (4/5)² – (3/5)²

(16/25) – (9/25) – 7/25

Since, L.H.S =R.H.S.

Hence, Proved

Ques 4. In a triangle, right angled at Q, PR + QR = 25 cm and PQ =5 cm. Determine the value of sinP ? (3 marks)

Ans:

In this triangle PQR,

PQ = 5cm

PQ + QR = 25cm

Let say, QR = X

Then PR = 25 – QR = 25 –X

Using Pythagoras theorem

PR² = PQ² + QR²

Now, Substituting the values

(25 – X)² = (5)² + (X)²

25² +X²-50X = 25 + X²

625 – 50X = 25

50X = 600

X = 12

So, QR = 12cm

PR = 25 – QR = 25 – 12 = 13 cm

Therefore,

sinP = QR/PR = 12/13

cosP = PQ/PR = 5/13

tanP= QR/PQ = 12/5

Ques 5. If tan²A = cot (A-18⁰), where 2A is an acute angle, Find the value of A. (2 marks)

Ans: Given,

tan²A = cot (A-18⁰)

tan²A – cot (90⁰ – 2A)

Substituting the values,

Cot(90⁰ – 2A ) = cot (A-18⁰ )

Therefore,

90⁰ – 2A = A-18⁰

108⁰ = 3A

A= 108⁰ /3

Hence, The value of A = 36⁰

Ques 6. In triangle ABC, right- angled at B, AB = 24 cm, BC = 7 cm.
SinA, cosA
sicC, cosC (3 marks)

Ans: In a given triangle ABC, right-angled a B=900

Given: AB = 24 cm and BC = 7 cm

Hence AC = Hypotenuse

According to the theorem,

H² = P² + B²

AC² = AB² + BC²

AC² = (24)² + 7²

AC² = (576 + 49)

AC² = 625 cm²

Therefore, AC= 25 cm.

We need to find sinA and cosA

As we know, sine of the angle is equal to the ratio of perpendicular and hypotenuse of the triangle.

sinA = BC/AC = 7/25

cosA = AB/AC= 24/25

We need to find sinC and cosC

sinC=AB/AC= 24/25

cosC =BC/AC = 7/25

Ques 7. If A,B,C are interior angles of a triangle ABC, then show that sin[(B + C)/2] = cos A/2. (2 marks)

Ans: The sum of all its interior angles is equals to 180⁰

A + B + C = 180⁰

B + C = 180⁰ – A

Divide the equation by 2

(B + C)/2 = (180⁰ – A)/2

(B +C)/2 = 90⁰ – A/2

Now, put sin function on both the sides

sin( B + C)/2 = sin (90⁰ – A/2)

since,

sin(90⁰ –A/2) = cosA/2

sin(B+C)/2 =cosA/2

Ques 8. Evaluate 2 tan²45 + cos²30 – sin²60. (1 mark)

Ans:

Tan 45 =1

cos30 = 3/2

sin 60 = 3/2

put the value in the equation

2(1)² + (3/2)² – (3/2)²

=2+0

Question 9. If sec 4A = cosec ( A – 20⁰ ) where 4A is an aute angle, Find the value of A. (1 mark)

Ans:

Sec 4A = cosec (A-20⁰ )

Cosec (90⁰-4A) = cosec(A-20⁰ )

90⁰ - 4A = A-20⁰

110⁰ = 5A

A=22⁰

Ques 10. If tanA + cotA = 5 , Find the value of tan²A + cotA? (1 mark)

Ans:

Given: tanA + cotA = 5

Squaring both the sides

tan²A+cot²A+2tanAcotA = 25

tan²A+cot²A +2=25

Hence, tan²A+cot²A =23

Question 11. Prove the following identity:
sin³A+cos³A/sinA+cosA = 1-sinA.cosA (1 mark)

Ans:

L.H.S. = sin³A+cos³A/sinA+cosA

= (sinA + cosA)(sin²A + cos²A - sinAcosA)/(sinA + cosA)

= 1 - sinAcosA = R.H.S. ………[sin²A + cos²A = 1]

Question 12. Prove that
sinA – cosA/sinA+cosA + sinA+cosA/sinA-cosA = 2/2sin²-1 (2 marks)

Ans:

L.H.S. = sinA – cosA/sinA+cosA + sinA+cosA/sinA-cosA

=(sinA - cosA)² + (sinA + cosA)²/(sinA + cosA)(sinA - cosA)

=sin²A + cos²A - 2sinAcosA + sin²A+cos²A + 2sinAcosA/sin²A - cos²A

=1 + 1/sin²A - (1 - sin²A)

= 2/sin²A – 1 + sin²A

= 2/2sin²A - 1

Ques 13. Evaluate 4(sin⁴30 + cos⁴60)- 3(cos²450 – sin²900) (2 marks)

Ans:

Given : 4(sin⁴30 + cos⁴60)- 3(cos²450 – sin²900)

=4[(1/2)⁴ + [(1/2)⁴] – 3[(1/2)² -1 ]

=4[1/16 + 1/16] – 3[1/2 - 1]

= 4*2/16-3(-1/2)

= 1/2 + 3/2 = 4/2 =2

Ques 14. Prove that tan³A/1+tan²A + cot³A/1+cot²A = secAcosecA -2sinAcosA. (3 marks)

Ans:

tan³A/1+tan²A + cot³A/1+cot²A = secAcosecA -2sinAcosA.

=sin³A/cos³A *cos²A + cos³A/sin³A*sin²A

=sin³A/cosA + cos³A/sinA

= sin⁴A + cos⁴A/sinAcosA

=(sin²A)² + (cos²A)²/sinAcosA

= (sin²A + cos²A) – 2sin²Acos²A/sinAcosA

=1-2sin²Acos²A/sinAcosA

=1/sinAcosA – 2 sin²Acos²A/sinAcosA

=secAcosecA – 2sinAcosA

Ques 15. If cosecA + cotA = m , show that m²-1/m²+1 =cosA (3 marks)

Ans:

L.H.S. = m²-1/m²+1

= (cosecA + cotA)² – 1/(cosecA - cotA)² + 1

=cosec²A+cot²A+2cosecAcotA-1/ cosec²A+cot²A+2cosecAcotA+1

= (cosec²A - 1) + cot²A + 2cosecAcotA/ cosec²A + (1 + cot²A)+ 2cosecAcotA

= cot²A +cot²A+2cosecAcotA/cosec²A+cosec²A+2cosecAcotA

=2cot²A + 2cosecAcotA/2cosec²A +2cosecAcotA

= 2cotA(cotA+cosecA)/2cosecA(cosecA + cotA)

=cotA/cosecA

=cosA/sinAcosecA

=cosA = R.H.S.

Hence,

cosA = m² -1/m²+1

Ques 16. In a triangle, right angled at B, BC= 7 cm and AC – AB = 1 cm. Find the value of cosA +sinA (2 marks)

Ans:

Let AC = x cm , AB = y cm

Therefore, x-y=1 ……………….(1)

Now, B is a right angle

So, AC² = AB² + BC²

x²-y² =72

(x - y)(x + y) = 49

1*(x + y) = 49

x+y = 49 ……………..(2)

Solving eq 1 and 2,

X= 25 and y= 24

Now, cosA + sinA = AB/AC + BC/AC

= 24/25 + 7/25

= 31/25

Ques 17. Prove that (1 + cotA – cosecA) (1 + tanA + secA) = 2 (2 marks)

Ans:

L.H.S = (1 + cotA – cosecA) (1 + tanA + secA)

= ( 1+ cosA/sinA – 1/sinA)(1 + sinA/cosA + 1/cosA)

=(sinA + cosA – 1/sinA) (cosA+sinA+1/cosA)

= (sinA + cosA)²-1/sinAcosA

= sin²A +cos²A + 2sinAcosA -1 /sinAcosA

= 1+2sinAcosA -1/sinAcosA =2sinAcosA/sinAcosA=2

Ques 18. Prove that sinA ( 1 + tanA) + cosA(1 + cotA) = secA + cosecA (2 marks)

Ans:

L.H.S. sinA ( 1 + tanA) + cosA(1 + cotA)

= sinA ( cosA + sinA)/cosA + cosA(sinA + cosA)/sinA

=(cosA + sinA)[sinA/cosA + cosA/sinA]

= (cosA + sinA)/cosAsinA (sin²A + cos²A)

=cosA/cosAsinA + sin/cosAsinA

=cosecA + secA

Ques 19. sin³A + cos³A/sinAcosA + sinAcosA (1 mark)

Ans:

sin³A + cos³A/sinAcosA + sinAcosA

= (sinA + cosA) ( sin²A + cos²A – sinAcosA)/(sinA + cosA) + sinA.cosA

= 1- sinAcosA + sinAcosA =1

Ques 20. cosA/1+sinA + 1+sinA/cosA = 2secA (2 marks)

Ans:

L.H.S. cosA/1+sinA + 1+sinA/cosA

= cos²A + ( 1 + sinA)²/(1+sinA)cosA

= cos²A + sin²A + 1+ 2sinA/(1+sinA)cosA

= 2(1+sinA)/(1+sinA)cosA

= 2/cosA

= 2secA

= R.H.S.

Ques 21. If sinA = 1/3, then find the value of ( 2cot²A+2) (1 mark)

Ans:

2(cot²A+1) =2.cosec²A

= 2/sin²A

= 2/(1/9)

=18

Length (in mm)	Number of leaves
118 - 126	3
127 - 135	5
136 - 144	9
145 - 153	12
154 - 162	5
163 - 171	4
172 - 180	2

Introduction to Trigonometry Important Questions: Applications of Trigonometry

Applications of Trigonometry

Basic Formulas

CBSE X Related Questions

1.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

4.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Similar Mathematics Concepts

Comments

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Introduction to Trigonometry Important Questions: Applications of Trigonometry

Applications of Trigonometry

Basic Formulas

CBSE X Related Questions

1. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

4. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

4.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.