CBSE NCERT Solutions for Class 10 Mathematics Chapter 8

Class- X-CBSE-Mathematics

Introduction to Trigonometry

CBSE NCERT Solutions for Class 10 Mathematics Chapter 8

Back of Chapter Questions

1. In ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A , cos A (ii) sin C , cos C Solution: In ABC, apply Pythagoras theorem AC2 = AB2 + BC2 = (24)2 + (7)2 = 625

AC = 625 = 25 cm

(i)

sin A = Side opposite to A = BC

hypotenuse

AC

7 =

25

Side adjacent to A AB cos A = hypotenuse = AC

24 = 25

(ii)

Practice more on Introductions to Trigonometry

Page - 1



Class- X-CBSE-Mathematics

Introduction to Trigonometry

Side opposite to C AB 24 sin C = hypotenuse = AC = 25

Side adjacent to C BC 7

cos C =

==

hypotenuse

AC 25

2. In Fig., find tan P - cot R .

Solution: Apply Pythagoras theorem in PQR PR2 = PQ2 + QR2 (13)2 = (12)2 + QR2 169 = 144 + QR2 25 = QR2 QR = 5

Side opposite to P QR tan P = Side adjacent to P = PQ

5 =

12 Side adjacent to R QR

cot R = Side opposite to R = PQ 5

= 12 55

tan P - cot R = 12 - 12 = 0

Practice more on Introductions to Trigonometry

Page - 2



Class- X-CBSE-Mathematics

Introduction to Trigonometry

3. If sin A = 3, calculate cos A and tan A.

4

Solution: Let us assume that ABC is a right triangle, right angled at vertex B.

Given that 3

sin A = 4 sin A = Side opposite to A

hypotenuse

BC 3 AC = 4 Let BC and AC be 3R and 4R respectively, where R is any positive number. In ABC, apply Pythagoras theorem AC2 = AB2 + BC2 (4 R)2 = AB2 + (3 R)2 16 R2 - 9 R2 = AB2 7 R2 = AB2

AB = 7R Side adjacent to A

cos A = hypotenuse

= AB = 7R = 7 AC 4R 4 Side opposite to A

tan A = Side adjacent to A

BC 3R 3 = AB = 7R = 7 4. Given 15 cot A = 8, find sin A and sec A. Solution: Let us assume that ABC is a right triangle, right angled at vertex B.

Practice more on Introductions to Trigonometry

Page - 3



Class- X-CBSE-Mathematics

Introduction to Trigonometry

Side adjacent to A cot A = Side opposite to A

AB =

BC cot A = 8 (given)

15

AB 8 BC = 15 Let AB and BC be 8R and 15R respectively, where R is a positive number.

Now applying Pythagoras theorem in ABC AC2 = AB2 + BC2

= (8R)2 + (15R)2

= 64 R2 + 225 R2

= 289 R2

AC = 17 R

Side opposite to A BC

sin A =

=

hypotenuse

AC

15 R 15 = 17 R = 17

hypotenuse sec A =

Side adjacent to A

AC 17 = AB = 8

5.

Given

sec

=

13,

12

calculate

all

other

trigonometric

ratios.

Solution:

Let us assume that ABC is a right angled triangle, right angled at vertex B.

Practice more on Introductions to Trigonometry

Page - 4



Class- X-CBSE-Mathematics

Introduction to Trigonometry

hypotenuse sec = Side adjacent to B

13 AC 12 = AB

Let AC and AB be 13R and 12R, where R is a positive number.

Now applying Pythagoras theorem in ABC

(AC)2 = (AB)2 + (BC)2

(13 R)2 = (12 R)2 + BC2

169 R2 = 144 R2 + BC2

25 R2 = BC2

BC = 5R

Side opposite to BC 5R 5

sin =

hypotenuse

== = AC 13R 13

Side adjacent to AB 12R 12 cos = hypotenuse = AC = 13R = 13

Side opposite to BC 5R 5 tan = Side adjacent to = AB = 12R = 12

Side adjacent to AB 12R 12 cot = Side opposite to = BC = 5R = 5

hypotenuse

AC 13R 13

coses = Side opposite to = BC = 5R = 5

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

Solution:

Let us assume a right triangle, right angled at vertex C.

Practice more on Introductions to Trigonometry

Page - 5



 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download