CBSE Class 10 Maths Chapter-8 Introduction to Trigonometry – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 8 – Introduction to Trigonometry prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Maths Chapter-8 Introduction to Trigonometry Important Questions
CBSE Class 10 Maths Important Questions Chapter 8 – Introduction to Trigonometry
2 Mark Questions
1. In 
ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) 
(ii) 
Ans. Let us draw a right angled triangle ABC, right angled at B.
Using Pythagoras theorem,
AC2 = AB2 + BC2
= 
= 576 + 49 = 625
AC = 25 cm
(i) 
, 
(ii) 
, 
2. In adjoining figure, find 
:
Ans. Using Pythagoras theorem,
PR2 = PQ2 + QR2
QR2
QR2 =169 – 144 = 25
QR = 5 cm
= 
= 
= 0
3. If 
calculate 
and 
Ans. Given: A triangle ABC in which 
B = 
Let BC = 
and AC = 
Then, Using Pythagoras theorem,
AB = 
= 
= 
= 
4. Given 
find 
and 
Ans. Given: A triangle ABC in which B =
Let AB = 
and BC = 
Then using Pythagoras theorem,
AC = 
= 
= 
= 
= 
5. If And B are acute angles such that 
then show that A = B.
Ans. In right triangle ABC,
and 
But 
[Given]
AC = BC
A = B
[Angles opposite to equal sides are equal]
6. State whether the following are true or false. Justify your answer.
(i) The value of 
is always less than 1.
(ii) 
for some value of angle A.
(iii) is the abbreviation used for the cosecant of angle A.
(iv) 
is the product of 
and A.
(v) 
for some angle 
Ans. (i) False because sides of a right triangle may have any length, so may have any value.
(ii) True as 
is always greater than 1.
(iii) False as is the abbreviation of cosine A.
(iv) False as is not the product of ‘cot’ and A. ‘cot’ is separated from A has no meaning.
(v) False as 
cannot be > 1
7. Evaluate:
(i) 
(ii) 
(iii) 
(iv) 
Ans. Solution:
(i) = 
= 
= 1
(ii) = 
= 
= 1
(iii)
= 
= 
= 0
(iv)
= 
= 
=0
8. Show that:
(i) 
(ii) 
Ans. (i) L.H.S. 
= 
= 
= 
= 1 = R.H.S.
(ii) R.H.S. 
= 
= 
= 0 = R.H.S.
9. If 
where 2A is an acute angle, find the value of A.
Ans. Given: 
A = 
10. If 
prove that 
Ans. Given: 
A + B =
11. If 
where 4A is an acute angle, find the value of A.
Ans. Given: 
A = 
12. If A, B and C are interior angles of a ABC, then show that 
Ans. Given: A, B and C are interior angles of aABC.
A + B + C = 
13. Express 
in terms of trigonometric ratios of angles between 
and 
Ans.
= 
= 
14. Express the trigonometric ratios 
and in terms of 
Ans. For 
By using identity 
For 
By using identity 
= 
For 
15. Write the other trigonometric ratios of A in terms of
Ans. For
By using identity, 
= 
For 
For
By using identity 
For 
= 
For 
16. Evaluate:
(i) 
(ii) 
Ans. (i) = 
= 
= 
(ii)
= 
= 
= 
= 1
17. Show that any positive odd integer is of the form6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Ans. Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,
a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3,
Or 6q + 5
18. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Ans. We have to find the HCF(616, 32) to find the maximum number of columns in which they can march.
To find the HCF,we can use Euclid’s algorithm.
616 = 32 × 19 + 8
32 = 8 × 4 + 0
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.
19. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]
Ans. Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
Or,
Where k1, k2, and k3 are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.