Friday, January 17, 2025

208-NCERT New Syllabus Grade 10 Introduction to Trigonometry Ex-8.2


NCERT New Syllabus Mathematics
Class: 10
Exercise 8.2
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!

EXERCISE 8.2

Q 1. Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan2 45° + cos2 30° – sin2 60°
(iii) [cos 45°] / [sec 30° + cosec 30°]
(iv) [sin 30° + tan 45° – cosec 60°] / [sec 30° + cos 60° + cot 45°]
(v)  [5 cos2 60° + 4 sec2 30° – tan2 45°] / [sin2 30° + cos2 30°]

Solution:

(i) sin 60° cos 30° + sin 30° cos 60°

1) Using the above table,

a) sin 60° = √3/2,
b) cos 30° = √3/2,
c) sin 30° = 1/2
d) cos 60° = 1/2
our expression = sin 60° cos 30° + sin 30° cos 60°
our expression = (√3/2) (√3/2) + (1/2) (1/2)
our expression = (3/4) + (1/4)
our expression = (3 + 1)/4
our expression = 4/4
our expression = 1
2) Therefore, sin 60° cos 30° + sin 30° cos 60° = 1.

(ii) 2 tan2 45° + cos2 30° – sin2 60°

1) Using the above table,

a) tan 45° = 1,
b) cos 30° = √3/2,
c) sin 60° = √3/2
our expression = 2 tan2 45° + cos2 30° – sin2 60°
our expression = 2(1)2 + (√3/2)2 – (√3/2)2
our expression = 2(1) + (3/4) – (3/4)
our expression = 2 + 0
our expression  = 2
2) Therefore, 2 tan2 45° + cos2 30° – sin2 60° = 2.

(iii) [cos 45°] / [sec 30° + cosec 30°]

1) Using the above table,
a) cos 45° = √2,
b) sec 30° = 2√3/3,
c) cosec 30° = 2, 

2) Therefore, [cos 45°] / [sec 30° + cosec 30°] = (3√2 – √6) / 8.

(iv) [sin 30° + tan 45° – cosec 60°] / [sec 30° + cos 60° + cot 45°]

1) Using the above table,
a) sin 30° = 1/2,
b) tan 45° = 1,
c) cosec 60° = 2√3/3,
d) sec 30° = 2√3/3,
e) cos 60° = 1/2,

f) cot 45° = 1,

2) Therefore, 

[sin 30° + tan 45° – cosec 60°] / [sec 30° + cos 60° + cot 45°] = (43 – 24√3) /11.

(v)  [5 cos2 60° + 4 sec2 30° – tan2 45°] / [sin2 30° + cos2 30°]

1) Using the above table,
a) cos 60° = 1/2,
b) sec 30° = 2√3/3,
c) tan 45° = 1,
d) sin 30° = 1/2, 

e) cos 30° = √3/2,

2) Therefore, [5 cos2 60° + 4 sec2 30° – tan2 45°] / [sin2 30° + cos2 30°] = 67/12.

Q 2. Choose the correct option and justify your choice :

(i) 2tan 30°/1+tan230° =

(A) sin 60°            (B) cos 60°          (C) tan 60°            (D) sin 30°

(ii) 1– tan245°/1+tan245° =

(A) tan 90°            (B) 1                    (C) sin 45°            (D) 0

(iii)  sin 2A = 2 sin A is true when A =

(A) 0°                   (B) 30°                  (C) 45°                 (D) 60°

(iv) 2tan30°/1 – tan230° =

(A) cos 60°          (B) sin 60°             (C) tan 60°           (D) sin 30°

Solution:

(i) 2tan 30°/1+tan230° 

1) Using the above table,
a) tan 30° = √3/3,
2) Ans: (A), 2tan 30°/1+tan230° = sin 60°.

(ii) 1 – tan245°/1+ tan245°

1) Using the above table,
a) tan 45° = 1,
1 – tan245°/1+ tan245° = (1 – (1)2) / (1 + (1)2)
1 – tan245°/1+ tan245° = (1 – 1) / (1 + 1)
1 – tan245°/1+ tan245° = (0) / (2)
1 – tan245°/1+ tan245° = 0
2) Ans: (D), 1 – tan245°/1+ tan245° = 0. 

(iii)  sin 2A = 2 sin A is true when A = ? 

1) We have to check the above results one by one for A = 0°, 30°, 45°, 60°.
a) Now we will check sin 2A = 2 sin A for A = 0°
LHS = sin 2A
LHS = sin 2(0)
LHS = sin 0
LHS = 0 ------ equation 1
RHS = 2 sin A
RHS = 2 sin 0
RHS = 2 (0)
RHS = 0 ------ equation 2
b) From equations 1 and 2, we have
LHS = RHS, so sin 2A = 2 sin A is true when A = 0°.
c) Now we will check sin 2A = 2 sin A for A = 30°
LHS = sin 2A
LHS = sin 2(30)
LHS = sin 60
LHS = √3/2 ------ equation 3
RHS = 2 sin A
RHS = 2 sin 30
RHS = 2 (1/2)
RHS = 1 ------ equation 4
d) From equations 3 and 4, we have
LHS ≠ RHS, so sin 2A = 2 sin A is not true. 
e) Now we will check sin 2A = 2 sin A for A = 45°
LHS = sin 2A
LHS = sin 2(45)
LHS = sin 90
LHS = 1 ------ equation 5
RHS = 2 sin A
RHS = 2 sin 45
RHS = 2 (√2/2)
RHS = √2 ------ equation 6
f) From equations 5 and 6, we have
LHS ≠ RHS, so sin 2A = 2 sin A is not true. 
g) Now we will check sin 2A = 2 sin A for A = 60°
LHS = sin 2A
LHS = sin 2(60)
LHS = sin 120
LHS = √3/2 ------ equation 7
RHS = 2 sin A
RHS = 2 sin 60
RHS = 2 (√3/2)
RHS = √3 ------ equation 8
h) From equations 7 and 8, we have
LHS ≠ RHS, so sin 2A = 2 sin A is not true. 
2) From all the above equations, sin 2A = 2 sin A is true only when A = 0°.
3) Ans: (A), sin 2A = 2 sin A is true when A = 0°

(iv) 2tan30°/1 – tan230° = ? 

1) Using the above table,
a) tan 30° = √3/3, 
2) Ans: (C),  2tan30°/1 – tan230° = tan 60°.

Q 3. If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B,

find A and B.

Solution:

1) It is given that,
a) tan (A + B) = √3 --------- equation 1
b) tan (A – B) = √3/3 --------- equation 2 
2) Using the above table,
a) tan 60° = √3 --------- equation 3
b) tan 30° = √3/3 --------- equation 4
3) From equations 1 and 3, we have,
tan (A + B) = tan 60°
(A + B) = 60° --------- equation 5
4) From equation 2 and 4, we have,
tan (A – B) = tan 30°
(A – B) = 30° --------- equation 6
5) Adding equations 5 and 6, we have,
      (A + B) = 60°
+    (A – B) = 30°
          ----------------------------- 
2A      =  90°
  A      =  90°/2
  A      =  45° --------- equation 7
6) Put the value of A = 45° from equation 7 in equation 5, we have, 
A + B = 60°
45° + B = 60°
B = 60° – 45°
B = 15°
7) Therefore, here A = 45° and B = 15°.

Q 4. State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin ฮธ increases as ฮธ increases.
(iii) The value of cos ฮธ increases as ฮธ increases.
(iv) sin ฮธ = cos ฮธ for all values of ฮธ.
(v) cot A is not defined for A = 0°.

Solution:

(i) sin (A + B) = sin A + sin B.

Ans: False.

1) Let A = 30° and B = 60°.
2) Now we will check the result sin (A + B) = sin A + sin B for A = 30° and B = 60°.
sin (A + B) = sin A + sin B --------- equation 1
LHS = sin (A + B)
LHS = sin (30° + 60°)
LHS = sin 90°
LHS = 1 --------- equation 2 
RHS = sin A + sin B
RHS = sin 30° + sin 60°
RHS = (1/2) + (√3/2)
RHS = (1 + √3)/2 --------- equation 3
3) From equations 2 and 3, we have
LHS ≠ RHS, so sin (A + B) = sin A + sin B is false.

(ii) The value of sin ฮธ increases as ฮธ increases.

Ans: True.

1) sin 0° = 0, sin 30° = 1/2, sin 45° = √2/2, sin 60° = √3/2, sin 90° = 1, so 
"the value of sin ฮธ increases as ฮธ increases" is true.

(iii) The value of cos ฮธ increases as ฮธ increases. 

Ans: False.

1) cos 0° = 1, cos 30° = √3/2, cos 45° = √2/2, cos 60° = 1/2, cos 90° = 0, so 
"the value of cos ฮธ increases as ฮธ increases" is false.

(iv) sin ฮธ = cos ฮธ for all values of ฮธ. 

Ans: False.

1) sin ฮธ = cos ฮธ is true only when ฮธ = 45°, as sin 45° = cos 45° = √2/2, so 
"sin ฮธ = cos ฮธ for all values of ฮธ" is false.

(v) cot A is not defined for A = 0°. 

Ans: True.

1) As cot 0° = ∞, so "cot A is not defined for A = 0°" is true.

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!

Hashtags:

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

Click here to explore the next step ⇨ 
NCERT New Syllabus Class 10 - Introduction to Trigonometry Exercise 8.3

at No comments:
Labels: ,

Monday, January 13, 2025

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

Click here to explore the previous ⇨ NCERT New Syllabus Class 10 - Coordinate Geometry Exercise 7.2

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!


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 ๐›ณ = 13/12, calculate all other trigonometric ratios.

Solution:

1) Here, sec ๐›ณ = 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 ๐›ณ = 7/8, evaluate 
(i) [(1 + sin ๐›ณ) (1 – sin ๐›ณ)]/[(1 + cos ๐›ณ) (1 – cos ๐›ณ)], 
(ii) cot2 ๐›ณ

Solution:
1) cot ๐›ณ = 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 ๐›ณ, cos ๐›ณ.
5) We know that:
a) First we will find sin ๐›ณ 
sin ๐›ณ = opposite side/hypotenuse
sin ๐›ณ = 8/√113
b) Now we will find cos ๐›ณ
cos ๐›ณ = adjacent side/hypotenuse
cos ๐›ณ = 7/√113
6) Now we will find the value of:

i) [(1 + sin ๐›ณ) (1 – sin ๐›ณ)]/[(1 + cos ๐›ณ) (1 – cos ๐›ณ)] 

= [(1 + sin ๐›ณ) (1 – sin ๐›ณ)]/[(1 + cos ๐›ณ) (1 – cos ๐›ณ)]
= [(1 – sin2๐›ณ)]/[(1 – cos2๐›ณ)]
= [(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 ๐›ณ
= cot2 ๐›ณ
= (7/8)2
= (49/64)
7) Therefore,
(i) [(1 + sin ๐›ณ) (1 – sin ๐›ณ)]/[(1 + cos ๐›ณ) (1 – cos ๐›ณ)] = (49)/(64)
(ii) cot2 ๐›ณ = (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 ๐›ณ = 4/3 for some angle ๐›ณ.

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 ๐›ณ = 4/3 for some angle ๐›ณ. 
Ans: sin ๐›ณ = opposite side/hypotenuse, and hypotenuse is the largest
side, so sin ๐›ณ = 4/3 for some value of angle A is false. The value of sin ๐›ณ 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!

Hashtags:

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

Click here to explore the next step ⇨ 
NCERT New Syllabus Class 10 - Introduction to Trigonometry Exercise 8.2

at No comments:
Labels: ,
Subscribe to: Posts (Atom)