Exercise 8.2 (NCERT) - Trigonometry, Class 10
| Chapter 8 | Exercise 8.2 | Class 10 |
|---|
Below are the Quick links for all questions of Exercise 8.2, Trigonometry, Class 10.
Click a link to view the solution of corresponding question.
Below are the solutions of every question of Exercise 8.2 of Chapter 8, Introduction to Trigonometry, NCERT of Class 10. YouTube video tutorial of every question is also embedded along with the written solution.
This exercise is primarily about the standard values of trigonometric ratios and that’s why the questions are quite easy to solve.
Very Important: Before you start solving the questions of this exercise, it is very important that you remember the table displayed below.
Trigonometric ratios of standard angles
| ∠A | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
|
sin A |
0 |
\(\begin{aligned} \frac {1}{2}\end{aligned} \) |
\(\begin{aligned} \frac {1}{\sqrt{2}}\end{aligned} \) |
\(\begin{aligned} \frac {\sqrt{3}}{2}\end{aligned} \) |
1 |
|
cos A |
1 |
\(\begin{aligned} \frac {\sqrt{3}}{2}\end{aligned} \) |
\(\begin{aligned} \frac {1}{\sqrt{2}}\end{aligned} \) |
\(\begin{aligned} \frac {1}{2}\end{aligned} \) |
0 |
|
tan A |
0 |
\(\begin{aligned} \frac {1}{\sqrt {3}}\end{aligned} \) |
1 |
\(\begin{aligned} \sqrt {3} \end{aligned} \) |
N.D. |
|
cosec A |
N.D. |
2 |
\(\begin{aligned} \sqrt {2} \end{aligned} \) |
\(\begin{aligned} \frac {2}{\sqrt{3}}\end{aligned} \) |
1 |
|
sec A |
1 |
\(\begin{aligned} \frac {2}{\sqrt{3}}\end{aligned} \) |
\(\begin{aligned} \sqrt{2}\end{aligned} \) |
2 |
N.D. |
|
cot A |
N.D. |
\(\begin{aligned} \sqrt{3}\end{aligned} \) |
1 |
\(\begin{aligned} \frac {1}{\sqrt{3}}\end{aligned} \) |
1 |
N.D. – Not Defined
How to quickly make this table in exams
Let’s start using these trigonometric ratios of angles 0°, 30°, 45°, 60° and 90° to solve the questions.
Question 1
Evaluate the following :
Note: As per my understanding, this should be treated as the final answer, but in your book, the denominator has been rationalized. To achieve that, we need to further solve it as follows.
Rationalizing the denominator
Rationalizing the denominator
Question 2
Choose the correct option and justify your choice :
Hence (A) is the answer.
Hence (D) is the answer.
(iii) sin 2A = 2 sin A is true when A =
Let us take A = 0°
Hence (A) is the answer.
Hence (C) is the answer.
Question 3
; 0° < A + B ≤ 90°; A > B, find A and B.
Put the value of A in (1)
Answer: A = 45° and B = 15°
Question 4
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
Answer: False
Consider A = 30° and B = 60°
From (1) and (2)
Hence the statement of not true in general.
(ii) The value of sin θ increases as θ increases.
Answer: Actually it’s answer is False but it is given as True in your book.
The reason is, the value of angle θ can be more than 90 degrees as well. The value of sin θ increases as θ increases from 0 to 90 degrees but sin θ starts decreasing as the value of θ increases from 90 to 180 degrees.
Now talk to your class teacher to know what to write in exam 😉
Ideally the question should be as follows:
The value of sin θ increases as θ increases from 0 to 90 degrees.
(iii) The value of cos θ increases as θ increases.
Answer: False
Reason: cos 0° = 1 and cos 90° = 0
(iv) sin θ = cos θ for all values of θ.
Answer: False
Reason: sin 0° = 0 and cos 0° = 1
(v) cot A is not defined for A = 0°.
Answer: True
Reason: Just remember the value of cot 0° or try to evaluate cos 0°/sin 0° 😉
Some general tips for solving trigonometry questions easily and quickly are:
- Prepare a table of all standard values of sin, cos, tan, cosec, sec and cot. This will help you memorize them to be applied quickly to your questions.
- Another important tip is that if you get stuck somewhere, try to change everything into sin and cos. In some questions it may help you progress further towards the solution.
- Some important relations are sin(x) = cos(90-x) , tan(x) = cot(90-x) , cosec(x) = sec(90-x) and vice versa.
- It is important to know which side of the triangle is base and which is perpendicular. The one opposite to the angle under consideration is known as perpendicular and the one adjacent is the base. (the longest one is the hypotenuse)
- In case you are not able to prove LHS equal to RHS, then progress both of them separately and bring them to a common point. This will be considered a valid solution to the question.