Chapter 3:Pair of Linear Equations in Two Variables

Chapter 5: Arithmetic Progression

Chapter 8: Introduction to Trigonometry

Chapter 9:Some Applications of Trigonometry

Chapter 12:Area Related to Circles

<

Introduction to Trigonometry

(i) sin A, cos A

(ii) sin C, cos C

∠B = 90°

Given: AB = 24 cm and BC = 7 cm By applying Pythagoras theorem, we get

AC

AC

AC

AC

AC = √625 = 25

Therefore, AC = 25 cm

(i) Sin A, Cos A We know that Sin (A) =

Opposite side
/
Hypotenuse

=
BC
/
AC

=
7
/
25

Cos (A) =

side adjacent to angle A
/
Hypotenuse

=
AB
/
AC

=
24
/
25

(ii) Sin C, Cos C

Sin C =

AB
/
AC

=
24
/
25

Cos C =

BC
/
AC

=
7
/
25

PR = 13cm, PQ = 12cm

apply the Pythagorean theorem

PR

13

169 = QR

QR

QR

QR = √25 = 5

Therefore, the side QR = 5 cm

So tan P – cot R:

tan P =

side opposite to angle p
/
side adjacent to angle p

=
QR
/
QP

=
5
/
12

Cot R =

side adjacent to angle R
/
side opposite to angle R

=
QR
/
QP

=
5
/
12

Therefore, tan P – cot R =

5
/
12

–
5
/
12

= 0 Therefore, tan P – cot R = 0

3
/
4

, Calculate cos A and tan A.Given: Sin A =

3
/
4

Therefore, Sin A =

Opposite side
/
Hypotenuse

=
3
/
4

Let BC be 3k and AC be 4k

where k is a positive real number.

AC

(4k)

16k

AB

Therefore, AB = √7k

Cos A =

Adjacent side
/
Hypotenuse

AB
/
AC

=
√7k
/
4k

=
√7
/
4

Therefore, cos A =

√7
/
4

tan A =

Opposite side
/
Adjacent side

=
BC
/
AB

=
3k
/
√7k

=
3
/
√7

Therefore, tan A =

3
/
√7

Given: 15 cot A = 8

So, Cot A =

8
/
15

Therefore, cot A =

Adjacent side
/
Opposite side

=
AB
/
BC

=
8
/
15

Let AB be 8k and BC be 15k

Where, k is a positive real number.

According to the Pythagoras theorem,

AC

AC

AC

AC

Therefore, AC = 17k

We know that,

Sin A =

Opposite side
/
Hypotenuse

Sin A =

BC
/
AC

=
15k
/
17k

=
15
/
17

Therefore, sin A =

15
/
17

Sec A =

Hypotenuse
/
Adjacent side

we get,

AC
/
AB

=
17k
/
8k

=
17
/
8

Therefore sec A =

17
/
8

13
/
12

Calculate all other trigonometric ratios
Let us assume a right angled triangle ABC, right angled at B

sec θ =

13
/
12

=
Hypotenuse
/
Adjacent side

=
AC
/
AB

Let AC be 13k and AB be 12k

Where, k is a positive real number.

According to the Pythagoras theorem,

AC

(13k)

169k

169k

BC

BC

Therefore, BC = 5k

Sin θ =

Opposite Side
/
Hypotenuse

=
BC
/
AC

=
5
/
13

Cos θ =

Adjacent Side
/
Hypotenuse

=
AB
/
AC

=
12
/
13

tan θ =

Opposite Side
/
Adjacent Side

=
BC
/
AB

=
5
/
12

Cosec θ =

Hypotenuse
/
Opposite Side

=
AC
/
BC

=
13
/
5

cot θ =

Adjacent Side
/
Opposite Side

=
AB
/
BC

=
12
/
5

Give that the angles A and B are acute angles, such that

Cos A = cos B

Cos

AD
/
AC

=
BD
/
BC

Now, interchange the terms, we get

AD
/
BD

=
AC
/
BC

Let take a constant value

AD
/
BD

=
AC
/
BC

= k
Now consider the equation as

AD = k BD …(1)

AC = k BC …(2)

By applying Pythagoras theorem in △CAD and △CBD we get,

CD

CD

From the equations (3) and (4) we get,

AC

Now substitute the equations (1) and (2) in (3) and (4)

K2(BC

Putting this value in equation, we obtain

AC = BC

∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)

(i)

(1 + sin θ)(1 – sin θ)
/
(1+cos θ)(1-cos θ)

(ii) cot2 θ

Given: cot θ = BC/AB = 7/8

Let BC = 7k and AB = 8k, where k is a positive real number

According to Pythagoras theorem in △ABC we get.

AC

AC

AC

AC

AC = √113 k

sin θ = AB/AC = Opposite Side/Hypotenuse

= 8k/√113 k = 8/√113 and

cos θ = Adjacent Side/Hypotenuse

= BC/AC = 7k/√113 k = 7/√113

sssss

cot(A) = AB/BC = 4/3

Let AB = 4k an BC =3k, where k is a positive real number.

According to the Pythagorean theorem,

AC

AC

AC

AC

AC=5k

tan(A) = BC/AB = 3/4

sin (A) = BC/AC = 3/5

cos (A) = AB/AC = 4/5

Now compare the left hand side(LHS) with right hand side(RHS)

Since, both the LHS and RHS = 7/25

R.H.S. =L.H.S.

Hence,

(1-tan2 A)
/
(1+tan2 A)

= cos2 A – sin 2 A is proved(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

tan A = BC/AB = 1/√3

Let BC = 1k and AB = √3 k,

Where k is the positive real number of the problem

By Pythagoras theorem in ΔABC we get:

AC

AC

AC

AC

AC = 2k

Sin A = BC/AC = 1/2

Cos A = AB/AC = √3/2

Sin C = AB/AC = √3/2

Cos C = BC/AC = 1/2

Now, substitute the values in the given problem

(i) sin A cos C + cos A sin C

= (1/2) ×(1/2 )+ √3/2 ×√3/2 = 1/4 + 3/4 = 1

(ii) cos A cos C – sin A sin C

= (√3/2 )(1/2) – (1/2) (√3/2 ) = 0

Solution:

In a given triangle PQR

PQ = 5 cm

PR + QR = 25 cm

Now let us assume, QR = x

PR = 25-QR

PR = 25- x

According to the Pythagorean Theorem,

PR2 = PQ2 + QR2

Substitute the value of PR as x

(25- x)

25

625 + x

-50x = -600

x= -600/-50

x = 12 = QR

Now, find the value of PR

PR = 25- QR

Substitute the value of QR

PR = 25-12

PR = 13

Now, substitute the value to the given problem

(1) sin p = Opposite Side/Hypotenuse

= QR/PR = 12/13

(2) Cos p = Adjacent Side/Hypotenuse

= PQ/PR = 5/13

(3) tan p =Opposite Side/Adjacent side

= QR/PQ = 12/5

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 θ = 4/3 for some angle θ.

(i) The value of tan A is always less than 1.

Proof: In ΔMNC in which ∠N = 90∘,

MN = 3, NC = 4 and MC = 5

tan M = 4/3.

Pythagoras theorem.

MC

5