207-NCERT New Syllabus Grade 10 Introduction to Trigonometry Ex-8.1

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NCERT New Syllabus Mathematics

Class: 10

Exercise 8.1

Topic: Introduction to Trigonometry

Introduction to Trigonometry: Unlocking the Secrets of Angles

Welcome to the world of Trigonometry, where we explore the hidden relationships between angles and lengths that have fascinated mathematicians for centuries! Trigonometry, derived from the Greek words "trigonon" (triangle) and "metron" (measure), takes us on an exciting journey, unraveling the geometry of triangles and helping us solve complex real-world problems — from calculating the heights of mountains to navigating through the stars!

Class 10 introduces the basic trigonometric ratios, which form the foundation of everything from surveying to advanced physics. This chapter will empower you to solve problems involving right-angled triangles and lead you toward a deeper understanding of the angles that shape the world around us.

Are you ready to decode the mystery of angles and embrace the power of trigonometry? Let’s get started!

Click here to explore the Basics of Trigonometry - Part 1
Click here to explore the Basics of Trigonometry - Part 2

Details of trigonometric ratios:
1) sin A  ------  sine of angle A
2) cos A ------  co-sine of angle A
3) tan A  ------  tangent of angle A
4) csc A ------  co-secant of angle A (generally written as cosec A)
5) sec A ------  secant of angle A
6) cot A ------  co-tangent of angle A

Now we will define all six trigonometric ratios as follows:

1) sin A   =  BC/AC  =  opposite / hypotenuse
2) cos A  =  AB/AC  =  adjacent / hypotenuse
3) tan A   =  BC/AB  =  opposite / adjacent
4) csc A  =  AC/BC  =  hypotenuse / opposite
5) sec A  =  AC/AB  =  hypotenuse / adjacent
6) cot A  =  AB/BC  =  adjacent /  opposite

Trigonometric Ratios of an Angle A	Trigonometric Ratios of Complimentary Angle of A.  (90°– A)	Ratio (In the form of the length of the sides of a triangle)
sin A	cos (90°– A)	BC/AC
cos A	sin (90°– A)	AB/AC
csc A	sec (90°– A)	AC/BC
sec A	csc (90°– A)	AC/AB
tan A	cot (90°– A)	BC/AB
cot A	tan (90°– A)	AB/BC

EXERCISE 8.1

Q1. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :

(i) sin A, cos A

(ii) sin C, cos C

Solution:

(i) sin A, cos A

1) By the theorem of Pythagoras, we have,

(AC)² = (AB)² + (BC)²

(AC)² = (24)² + (7)²

(AC)² = 576 + 49
(AC)² = 625
AC = ± 5

2) As distance is always positive, AC = 5 ------- equation 1.

3) Now we will find sin A, cos A.

4) We know that:

a) First we will find sin A 

sin A = (side opposite)/hypotenuse

sin A = 7/25

b) Now we will find cos A

cos A = (adjacent side)/hypotenuse

cos A = 24/25

5) Therefore,

sin A = 7/25

cos A = 24/25 

(ii) sin C, cos C

 

6) Now we will find sin C, cos C.

7) We know that:

a) First we will find sin C 

sin C = (side opposite)/hypotenuse

sin C = 24/25

b) Now we will find cos C

cos C = (adjacent side)/hypotenuse

cos C = 7/25

8) Therefore, 

sin C = 24/25

cos C = 7/25 

 

Q 2. In the following fig., find tan P – cot R.

Solution:

1) By the theorem of Pythagoras, we have,

(QR)² = (PR)² – (PQ)²

(QR)² = (13)² – (12)²

(QR)² = (13 – 12) (13 + 12)
(QR)² = (1) (25)
QR = ± 5

2) As distance is always positive, QR = 5 ------- equation 1.

3) First we will find tan P, cot R.

4) We know that:

a) First we will find tan P 

tan P = opposite side/adjacent side

tan P = 5/12 ------- equation 2

b) Now we will find cot R

cot R = adjacent side/opposite side

cot R = 5/12 ------- equation 3

5) Therefore,

tan P – cot R =  (5/12) – (5/12)

tan P – cot R =  0.

Q 3. If sin A = 3/4 calculate cos A and tan A.

Solution:

1) Here, sin A = 3/4, so by definition,

Opposite side BC = 3 and hypotenuse AC = 4.

2) By the theorem of Pythagoras, we have,

(AB)² = (AC)² – (BC)²

(AB)² = (4)² – (3)²

(AB)² = (4 – 3) (4 + 3)

(AB)² = (7)
AB = ± √7

3) As distance is always positive, AB = √7 ------- equation 1.

4) Now we will find cos A and tan A.

5) We know that:

a) First we will find cos A 

cos A = (adjacent side)/hypotenuse
cos A = √7/4

b) Now we will find tan A

tan A = opposite side/adjacent side

tan A = 3/√7

6) Therefore, cos A = √7/4 and tan A = 3/√7.

Q 4. Given 15 cot A = 8, find sin A and sec A.

Solution:

1) Here, 15 cot A = 8, so cot A = 8/15, so by definition,

adjacent side AB = 8, opposite side BC = 15.

2) By the theorem of Pythagoras, we have,

(AC)² = (AB)² + (BC)²

(AC)² = (8)² + (15)²

(AC)² = (64) + (225)

(AC)² = (289)
AC= ± 17

3) As distance is always positive, AC = 17 ------- equation 1.

4) Now we will find sin A and sec A.

5) We know that:

a) First we will find sin A 

sin A = (opposite side)/hypotenuse
sin A = 15/17

b) Now we will find sec A

sec A = hypotenuse/adjacent side

sec A = 17/8

6) Therefore, sin A = 15/17 and sec A = 17/8.

Q 5. Given sec 𝛳 = 13/12, calculate all other trigonometric ratios.

Solution:

1) Here, sec 𝛳 = 13/12, so by definition,

hypotenuse AC = 13, adjacent side AB = 12.

2) By the theorem of Pythagoras, we have,

(BC)² = (AC)² – (AB)²

(BC)² = (13)² – (12)²

(BC)² = (13 – 12) (13 + 12)

(BC)² = (1) (25) 

(BC)² = (25)
BC= ± 5

3) As distance is always positive, BC = 5 ------- equation 1.

4) Now we will find sin A, cos A, tan A, cosec A, and cot A.

5) We know that:

a) First we will find sin A 

sin A = opposite side/hypotenuse
sin A = 5/13

b) Now we will find cos A

cos A = adjacent side/hypotenuse

cos A = 12/13

c) Now we will find tan A 

tan A = opposite side/adjacent side
tan A = 5/12

d) Now we will find cosec A 

cosec A = hypotenuse/opposite side

cosec A = 13/5

e) Now we will find cot A 

cot A = adjacent side/opposite side

cot A = 12/5

6) Therefore, sin A = 5/13, cos A = 12/13, tan A = 5/12, cosec A = 13/5, and 

cot A = 12/5.

Q 6. If ∠ A and ∠ B are acute angles such that cos A = cos B, then show

that ∠ A = ∠ B.

Solution:

1) cos A = cos B (given) ------- equation 1

2) Now we will find cos A and cos B.

3) Here,

a) First we will find cos A 

cos A = (adjacent side)/hypotenuse

cos A = AC/AB ------- equation 2

b) Now we will find cos B

cos B = (adjacent side)/hypotenuse

cos B = CB/AB ------- equation 3

4) From equations 1, 2, and 3, we have

AC/AB = CB/AB

AC = CB ------- equation 4

5) From equation 4, we have

∠ A = ∠ B, hence proved.

Q 7. If cot 𝛳 = 7/8, evaluate 

(i) [(1 + sin 𝛳) (1 – sin 𝛳)]/[(1 + cos 𝛳) (1 – cos 𝛳)], 

(ii) cot² 𝛳

Solution:

1) cot 𝛳 = 7/8, in above diagram,

2) By the theorem of Pythagoras, we have,

(AB)² = (AC)² + (BC)²

(AB)² = (8)² + (7)²

(AB)² = (64) + (49)

(AB)² = (113) 

AB= ± √113

3) As distance is always positive, AB = √113 ------- equation 1.

4) Now we will find sin 𝛳, cos 𝛳.

5) We know that:

a) First we will find sin 𝛳 

sin 𝛳 = opposite side/hypotenuse
sin 𝛳 = 8/√113

b) Now we will find cos 𝛳

cos 𝛳 = adjacent side/hypotenuse

cos 𝛳 = 7/√113

6) Now we will find the value of:

i) [(1 + sin 𝛳) (1 – sin 𝛳)]/[(1 + cos 𝛳) (1 – cos 𝛳)] 

= [(1 + sin 𝛳) (1 – sin 𝛳)]/[(1 + cos 𝛳) (1 – cos 𝛳)]

= [(1 – sin²𝛳)]/[(1 – cos²𝛳)]

= [(1 – (8/√113)²]/[(1 – (7/√113)²]

= [(1 – (64/113)]/[(1 – (49/113)]
= [(113 – 64)/113]/[(113 – 49)/113]
= [(49)/113]/[(64)/113]
= [(49)/(64)] 

 (ii) cot² 𝛳

= cot² 𝛳
= (7/8)²
= (49/64)

7) Therefore,

(i) [(1 + sin 𝛳) (1 – sin 𝛳)]/[(1 + cos 𝛳) (1 – cos 𝛳)] = (49)/(64)

(ii) cot² 𝛳 = (49)/(64) 

Q 8. If 3 cot A = 4, check whether (1 – tan² A)/(1 + tan² A) = cos² A – sin² A or not.

Solution:

1) 3 cot A = 4, so cot A = 4/3. 

2) By the theorem of Pythagoras, we have,

(AC)² = (AB)² + (BC)²

(AC)² = (4)² + (3)²

(AC)² = (16) + (9)

(AC)² = (25) 

AC= ± 5

3) As distance is always positive, AC = 5 ------- equation 1.

4) We will find sin A, cos A, tan A.

5) We know that:

a) First we will find sin A 

sin A = opposite side/hypotenuse
sin A = 3/5

b) Now we will find cos A

cos A = adjacent side/hypotenuse

cos A = 4/5

c) Now we will find tan A

tan A = opposite side/adjacent side

tan A = 3/4 

6) Now we will check for (1 – tan² A)/(1 + tan² A) = cos² A – sin² A is true or not.

(1 – tan² A)/(1 + tan² A) = cos² A – sin² A

LHS = (1 – tan² A)/(1 + tan² A)
LHS = (1 – (3/4)²)/(1 + (3/4)²)
LHS = (1 – (9/16))/(1 + (9/16))
LHS = [(16 – 9)/16]/[(16 + 9)/16]
LHS = [(7)/16]/[(25)/16]
LHS = 7/25 ------- equation 2
RHS = cos² A – sin² A
RHS = (4/5)² – (3/5)²
RHS = (16/25) – (9/25)
RHS = (16 – 9)/25
RHS = 7/25 ------- equation 3

7) From equations 2 and 3, we have

(1 – tan² A)/(1 + tan² A) = cos² A – sin² A is true.

Q 9. In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of:

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Solution:

1) tan A = 1/√3, so according to the diagram, we have,

2) By the theorem of Pythagoras, we have,

(AC)² = (AB)² + (BC)²
(AC)² = (√3)² + (1)²
(AC)² = (3) + (1)
(AC)² = (4) 
AC= ± 2

3) As distance is always positive, AC = 2 ------- equation 1.

4) Now we will find sin A, sin C, cos A, and cos C.

5) We know that:

a) First we will find sin A 

sin A = opposite side/hypotenuse
sin A = 1/2

b) First we will find sin C 

sin C = opposite side/hypotenuse
sin C = √3/2

c) Now we will find cos A

cos A = adjacent side/hypotenuse

cos A = √3/2

d) Now we will find cos C

cos C = adjacent side/hypotenuse

cos C = 1/2 

6) Now we will find the following:

(i) sin A cos C + cos A sin C

= sin A cos C + cos A sin C

= (1/2) (1/2) + (√3/2) (√3/2)

= (1/4) + (3/4)
= (1 + 3)/4
= 4/4
= 1 ------- equation 2

(ii) cos A cos C – sin A sin C

= cos A cos C – sin A sin C

= (√3/2) (1/2) – (1/2) (√3/2)

= (√3/4) – (√3/4)
= (√3 – √3)/4
= 0/4
= 0 ------- equation 3

7) Therefore,

(i) sin A cos C + cos A sin C = 1

(ii) cos A cos C – sin A sin C = 0

Q 10. In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P, and tan P.

Solution:

1) In ∆ PQR, PR + QR = 25, let QR be x, so PR will be (25 – x) we have,

2) By the theorem of Pythagoras, we have,

(PR)² = (PQ)² + (QR)²
(25 – x)² = (5)² + (x)²

(25 – x)² – (x)² = (5)²      applying (a² – b²) = (a – b) (a + b) we get,
(25 – x - x) (25 – x + x) = (5)²
(25 – 2x) (25) = 25
(25 – 2x) = 25/25
(25 – 2x) = 1
– 2x = 1 – 25

– 2x = – 24
x = (– 24)/(– 2)
x = 12
QR = 12 ------- equation 1

3) We know that 

PR + QR = 25

PR + 12 = 25
PR = 25 – 12
PR = 13 ------- equation 2

4) So our diagram will be: 

5) Now we will find sin P, cos P, and tan P.

6) We know that:

a) First we will find sin P 

sin P = opposite side/hypotenuse
sin P = 12/13

b) Now we will find cos P

cos P = adjacent side/hypotenuse

cos P = 5/13

c) Now we will find tan P

tan P = opposite side/adjacent side

tan P = 12/5

7) Therefore, 

sin P = 12/13,
cos P =5/13,
tan P = 12/5. 

Q 11. State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin 𝛳 = 4/3 for some angle 𝛳.

Solution:

(i) The value of tan A is always less than 1.

Ans: We know that tan 60 = √3, so this statement is false.

 

(ii) sec A = 12/5 for some value of angle A.

Ans: sec A = hypotenuse/adjacent side, and hypotenuse is the largest

side, so sec A = 12/5 for some value of angle A is true.

 

(iii) cos A is the abbreviation used for the cosecant of angle A.

Ans: false. cos A is the abbreviation used for cosine of angle A.

(iv) cot A is the product of cot and A.

Ans: false. cot A is not the product of cot and A, it is the abbreviation used for cotangent of angle A.

(v) sin 𝛳 = 4/3 for some angle 𝛳. 

Ans: sin 𝛳 = opposite side/hypotenuse, and hypotenuse is the largest

side, so sin 𝛳 = 4/3 for some value of angle A is false. The value of sin 𝛳 is always less than 1.

Wrapping Up: Your Journey into Trigonometry Begins Here!

As we conclude this introduction to trigonometry, you’ve taken the first steps toward mastering a concept that extends far beyond the classroom. From calculating distances to understanding the intricacies of wave patterns and even space exploration, trigonometry is your gateway to a world of endless possibilities. Keep exploring, keep practicing, and soon you'll find yourself confidently solving real-world problems with the power of trigonometric ratios.

Stay curious and remember — this is just the beginning! The more you engage with trigonometry, the clearer its magic becomes.

Let’s keep learning together and unravel the mysteries of triangles, one angle at a time!

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#TrigonometryBasics #MathForClass10 #NCERTMath #TrigonometryMadeEasy #LearnWithTriangles #MathMagic #MathInEverydayLife #Class10Mathematics #MathematicalThinking #STEMLearning

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NCERT New Syllabus Class 10 - Introduction to Trigonometry Exercise 8.2

Anil Satpute