NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.2
Class 10 NCERT Solutions Maths Chapter 8 Exercise 8.2 will help extend your expertise in solving trigonometric problems with some known values of trigonometric ratios. The exercise will be focused on understanding these trigonometric ratios with different types of trigonometric problems. These ratios are some of the most important parts of trigonometry in the whole chapter, as well as in advanced levels of mathematical studies. So, let’s get a deeper insight into one of the most crucial sections of trigonometry in class 10th through exercise 8.2.
1.0Download NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.2: Free PDF
2.0Introduction to Trigonometric Ratios:
Before delving into the key concepts of trigonometric ratios, let’s take a quick look at these ratios. Trigonometric ratios refer to the relation between the angles and sides of right-angled triangles. These ratios remain the same for a certain angle and given sides. The trigonometric ratios are used to determine unknown sides and angles for right triangles. They also form the foundation for more advanced trigonometric functions and identities in higher mathematics. Knowing and working with these fundamental ratios is important to continue further studies in trigonometry and to solve practical problems concerning angles, heights, distances, and periodic phenomena.
3.0Key Concepts
Now that we have a basic idea of what trigonometric ratios are let’s go through some of the basic concepts, along with the values of some of these ratios. Which is the basic idea of the whole exercise:
Trigonometric Ratios for Angle A:
As we know, trigonometric ratios can be divided into six different basic ratios. Let’s recall the formulas for these six basic trigonometric ratios for a right-angled triangle ABC, right-angled at B with A as an acute angle :
SinA = PerpendicularHypotenuse=PH
CosA= BaseHypotenuse=BH
tanA= PerpendicularBase =PB
CosecA= HypotenusePerpendicular =HP
SecA= HypotenuseBase=HB
CotA= Base Perpendicular=BP
Here, A is the acute angle, according to which we will be determining the values of different trigonometric ratios.
Values of Trigonometric Ratios for Specific Angles:
In Exercise 8.3, we will be using the values of trigonometric ratios for 0°, 30°, 45°, 60°, and 90°. Although how these values are derived is not as crucial for the exercise itself, it can surely help students memorise and understand these trigonometric ratios of specific angles.
The values of these trigonometric ratios can be found using the specific properties of right-angled triangles and applying trigonometric principles. See the table below for the specific values of the six trigonometric ratios:
It is recommended to memorise the above-mentioned values of trigonometric ratios, as these values will be crucial for solving the questions throughout the exercise.
Relation Between Different Trigonometric Ratios:
In trigonometry, the six basic trigonometric ratios — sine, cosine, tangent, cosecant, secant, and cotangent — are related to each other. Knowing these relationships facilitates simplifying trigonometric expressions and solving complicated problems more effectively. The interconnections among these ratios are derived from their definitions in a right triangle and their reciprocal nature. Let's discuss these connections:
CosecA= 1SinA
SecA= 1CosA
CotA= 1tanA
tanA= SinACosA
CotA= CosASinA
Working on these NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.2 ratios will go a long way in understanding trigonometry and in helping you fetch good marks during examinations.
4.0NCERT Class 10 Maths Chapter 8 introduction to Trigonometry Exercise 8.2 : Detailed Solutions
1. Evaluate:
(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°)
Sol.
(i) sin 60° cos 30° + sin 30° cos 60°
= (√3 / 2)(√3 / 2) + (1 / 2)(1 / 2) = (√3 / 2)² + (1 / 2)²
= 3 / 4 + 1 / 4 = 1
(ii) 2 tan² 45° + cos² 30° - sin² 60°
= 2 × (1)² + (√3 / 2)² - (√3 / 2)²
= 2 + 3 / 4 - 3 / 4 = 2
(iii) cos 45° / (sec 30° + cosec 30°)
= (1 / √2) / (2 / √3 + 2) = (1 / √2) / (2(1 + √3) / √3) = 1(√3) / (2√2(1 + √3))
= √3 / (2√2) × (√3 - 1) / ((√3 + 1)(√3 - 1)) = √3(√3 - 1) / (2√2 × 2)
= (3 - √3) / (4√2) = (3 - √3) / (4√2) × (√2 / √2) = (3√2 - √6) / 8
(iv) (sin 30° + tan 45° - cosec 60°) / (sec 30° + cos 60° + cot 45°)
= (1 / 2 + 1 - 2 / √3) / (2 / √3 + 1 / 2 + 1)
= ((√3 + 2√3 - 4) / (2√3)) / ((4 + √3 + 2√3) / (2√3)) = (3√3 - 4) / (4 + 3√3) × (4 - 3√3) / (4 - 3√3)
= (12√3 - 27 - 16 + 12√3) / (16 - 9 × 3)
= (24√3 - 43) / (-11) = (43 - 24√3) / 11
(v) (5 cos² 60° + 4 sec² 30° - tan² 45°) / (sin² 30° + cos² 30°)
= (5(cos 60°)² + 4(sec 30°)² - (tan 45°)²) / ((sin 30°)² + (cos 30°)²)
= (5(1 / 2)² + 4(2 / √3)² - (1)²) / ((1 / 2)² + (√3 / 2)²) = (5 / 4 + 4 × 4 / 3 - 1) / (1 / 4 + 3 / 4)
= (5 / 4 + 16 / 3 - 1) / (1 / 4 + 3 / 4) = 5 / 4 + 16 / 3 - 1
= (15 + 64 - 12) / 12 = 67 / 12
2. Choose the correct option and justify your choice:
(i) (2 tan 30°) / (1 + tan² 30°) =
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°
(ii) (1 - tan² 45°) / (1 + tan² 45°) =
(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) (2 tan 30°) / (1 - tan² 30°) =
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°
Sol.
(i) Option (A)
(2 tan 30°) / (1 + tan² 30°) = (2(1/√3)) / (1 + (1/√3)²) = (2/√3) / (1 + 1/3)
= (2/√3) × (3/4) = √3/2 = sin 60°
(ii) Option (D)
(1 - tan² 45°) / (1 + tan² 45°) = (1 - 1) / (1 + 1) = 0 / 2 = 0
(iii) Option (A)
By option 0°
sin 2A = sin 0° = 0
2 sin 0° = 2 × 0 = 0
(iv) Option (C)
(2 tan 30°) / (1 - tan² 30°)
= (2 × (1/√3)) / (1 - (1/√3)²) = (2/√3) / (1 - 1/3) = (2/√3) / (2/3) = (2/√3) × (3/2) = 3/√3 = √3
= tan 60°
3. If tan(A+B) = √3 and tan(A-B) = 1/√3; 0° < A+B ≤ 90°; A > B, find A and B.
Sol. tan(A+B) = √3 => A+B = 60° ...(1)
tan(A-B) = 1/√3 => A-B = 30° ...(2)
Adding (1) and (2),
2A = 90° => A = 45°
Then from (1), 45° + B = 60° => B = 15°
4. State whether the following are true or false. Justify your answer.
(i) sin(A+B) = sinA + sinB
(ii) The value of sinθ increases as θ increases.
(iii) The value of cosθ increases as θ increases.
(iv) sinθ = cosθ for all values of θ.
(v) cotA is not defined for A = 0°.
Sol.
(i) False
When A = 60°, B = 30°
LHS = sin(A+B) = sin(60° + 30°) = sin90° = 1
RHS = sinA + sinB = sin60° + sin30° = √3/2 + 1/2 ≠ 1
i.e., LHS ≠ RHS
(ii) True
Note that sin0° = 0, sin30° = 1/2 = 0.5,
sin45° = 1/√2 = 0.7 (approx.),
sin60° = √3/2 = 0.87 (approx.)
and sin90° = 1
i.e., value of sinθ increases as θ increases from 0° to 90°.
(iii) False
Note that cos0° = 1,
cos30° = √3/2 = 0.87 (approx.)
cos45° = 1/√2 = 0.7 (approx.),
cos60° = 1/2 = 0.5 and cos90° = 0
i.e., value of cosθ decreases as θ increases from 0° to 90°.
(iv) False, it is true for only θ = 45°.
(v) True, cotA = 1/0 = not defined.
5.0Benefits of Class 10 Maths Chapter 8 Introduction to Trigonometry
- Strengthens application of trigonometric ratios in varied problems.
- Enhances logical reasoning through systematic use of trigonometric ratios.
- Develops time-management skills by solving problems efficiently under exam conditions.
- Prepares students for real-life applications like measuring heights and distances.
- Improves visualization skills for geometric figures and triangles.