x-CH5 Trigonometry

CHAPTER 5
TRIGONOMETRY

Prepared By
Cecilia Joseph
HST
St.John De Britto’s A.I.H.S
Fortkochi

Trignomerty Chapter -5 1
Trigonometry
Online class – 49 23 /10 / 2020
Previous Knowledge
To view class

If 3 sides of a triangle are equal to 3 sides of another
triangle, then the angles opposite to equal sides are equal.

Such triangles are called equal triangles.

If 3 angles of a triangle are equal to 3 angles of another
triangle, then the side opposite to equal angles are in the
same ratio.

Such triangles are called similar triangles.

The word trigonometry is derived from the Greek words
‘Trigon’ (meaning triangles) and ‘Metron’ (meaning measures).

Trigon + Metron

(Triangles) (Measures)

Trigonometry
Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 2

A triangle has 2 measures , angles and length of the sides ,
so we can say

Trigonometry is the study of the relation between
angles and sides of triangles.

Angles and sides
a)Consider the given right triangle.

3 cm We know that
sides opposite
What are the lengths of its other sides? to equal
* The perpendicular side opposite angles are
also equal.
to the 45 o angle = 3cm
√32 = 3
* According to Pythagoras Theorem,

(Hypotenuse)2 = (Base)2 + (Altitude)2

=32 + 32

∴ Hypotenuse = √2× 32 = 3√2 cm

3 cm 3√2 cm

3 cm

Measure of the angles are 45° , 45° , 90°

Sides are in the ratio 3 : 3 : 3√2

Reduced form 1 : 1 : √2

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 3

b)Consider the given right triangle.

5 cm √52 = 5

What are the lengths of its other sides?
* The perpendicular side opposite

to the 45 o angle = 5cm

* According to Pythagoras Theorem,

(Hypotenuse)2 = (Base)2 + (Altitude)2

=52 + 52

∴ Hypotenuse = √2× 52 = 5√2 cm

5 cm 5√2 cm

5 cm

Measure of the angles are 45° , 45° , 90°

Sides are in the ratio 5 : 5 : 5√2

Reduced form 1 : 1 : √2

In general,

Considering a right triangle with one perpendicular side ‘x’cm
we have,

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 4

x √ x2 = x

* The perpendicular side opposite
to the 45 o angle = x

*(Hypotenuse)2 = (Base)2 + (Altitude)2

= x2 + x2

∴ Hypotenuse = √2× x2 = x√2 cm

x x√2

x

Measure of the angles are 45° , 45° , 90°

Sides are in the ratio x : x : x√2

Reduced form 1 : 1 : √2

The sides of any triangle of angles 45° , 45° , 90°
are in the ratio 1 : 1 : √2

1 √2
1

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 5

Note: From the above we can see that

If the angles of the right triangle are 45° , 45° , 90°
then, hypotenuse is √2 times the perpendicular sides
and, perpendicular sides are √2 part of the hypotenuse.

Q) Find other sides of given triangles.

a)

Given, BC = 6cm

..... ..... So, AB = 6cm
AC = 6√2 cm

6cm
b)

Given, PR = 11cm

11cm ..... So, PQ = 11 cm

√2

RQ = 11 cm

√2

.....

c)

Given, AB = 9 cm
9cm ..... So, BC = 9cm

AC = 9√2 cm

.....

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 6

Assignment
Find the other two sides of the triangles given
a) b)

8 cm

8 cm

c)

Diameter = AB =7√2 cm
AC = BC

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 1

Chapter -5
Trigonometry

Online class – 50 27 /10 / 2020

To view class

Answer of last class assignment

a)
Given,∠A=900,∠B=450

∴∠C=1800 −(900 + 450)=450

Given, AC = 8cm 45° , 45° , 90°
8 cm So, AB = 8cm 1 : 1 : √2
8 8 8√2
BC = 8√2 cm

b)
Given,∠Q=900,∠P=450
∴∠R=1800 −(900 + 450)=450

Given, PR = 8cm
8
8 cm So, PQ = cm 45° , 45° , 90°
√2 1 : 1 : √2
8
QR = cm 8 / √2 8 / √2 8
√2

c) Given AB is diameter, So ∠C = 90°

Given AC = BC

So, ∠A =∠B = 180 °−90° = 90° =45°
2 2

Given, AB =7√2 cm 45° , 45° , 90°
7√2 1 : 1 : √2
∴ AC = √2 = 7cm, 7 7 7√2

BC = 7√2 = 7 cm
√2

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 2

Angles and sides
a) Consider the given right triangle.

A

D 3cm C

Visualizing it as half of an equilateral triangle we have,

A

3√3cm 6cm

B 3cm 6cDm 3cm C

Since all angles are 60°, ∆ABC is an equilateral triangle.
So BC = AC = AB = 6cm

Consider ∆ADC , Hypotenuse = 6cm, Base = 3cm

∴ Altitude(AD) = √(Hypotenuse)2−(Base)2

= √(6)2−(3)2

= √(6+ 3)(6−3)

= √9x3

= 3√3 cm

Measure of the angles are 30° , 60° , 90°

Sides are in the ratio 3 : 3√3 : 6 Dividing
Reduced form 1 : √3 : 2 each by 3

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 3

b) Consider the given right triangle.

A

D 5cm C

Visualizing it as half of an equilateral triangle we have,

A

5√3cm 10cm

B 5cm 10Dcm 5cm C

Since all angles are 60°, ∆ABC is an equilateral triangle.
So BC = AC = AB = 10cm

Consider ∆ADC , Hypotenuse = 10cm, Base = 5cm

∴ Altitude(AD) = √(Hypotenuse)2−(Base)2

= √(10)2−(5)2

= √(10+5)(10−5)

= √15 x 5

= √3x5x5

= 5√3 cm

Measure of the angles are 30° , 60° , 90°

Sides are in the ratio 5 : 5√3 : 10 Dividing
Reduced form 1 : √3 : 2 each by 5

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 4

In general,

Considering a right triangle with side opposite to 300 angle as ‘x’

we have, A

Dx C

Visualizing it as half of an equilateral triangle we have,

A

√3x 2x

B x 2xD x C

Since all angles are 60°, ∆ABC is an equilateral triangle.

So BC = AC = AB = 2x

Consider ∆ADC , Hypotenuse = 2x , Base = x

∴ Altitude(AD) = √(Hypotenuse)2−(Base)2

= √(2 x)2−(x)2

= √(2 x+x)(2 x−x)

= √3x×x

= √3x

Measure of the angles are 30° , 60° , 90°

Sides are in the ratio x : √3x : 2x Dividing
Reduced form 1 : √3 : 2 each by x

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 5

In any triangle of angles 30o , 60o , 90o the sides
are in the ratio 1 : √3 : 2

√3 2

1

Note:
From the above we can see that

If the angles of a right triangle are 30° , 60° , 90° and
a) If side opposite to 300 angle is given, then
Hypotenuse = 2 × side opposite to 300 angle
Side opposite to 600angle = √3 × side opposite to 300 angle

b) If side opposite to 900 angle(Hypotenuse) is given, then
Side opposite to 300angle = Hypotenuse
2
Side opposite to 600angle= √3 × side opposite to 300 angle

c) If side opposite to 600 angle is given, then
Side opposite to 300angle = side opposite to 600 angle
√3
Hypotenuse = 2 × side opposite to 300 angle

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 6

Q) Find the area of an equilateral triangle of side 4cm.

Ans) A

2√3 cm Given ABC is an equilateral triangle.

∴ ∠A = ∠B = ∠C = 600
4 cm Draw AD perpendicular to BC

∠DAC =1800 −(900+600)=1800 −1500= 300

B D 2 cm C

Consider ∆ADC,

AC = 4cm
4
∴ DC = 2 = 2cm DC=Hypotenuse
2
AD = 2√3 cm
AD = √3xDC

∴ Base = 2 + 2 = 4cm, Height = 2√3 cm

Area of the triangle = 1 Χ base Χ height
2

= 1 Χ 4 Χ 2√3
2

= 4√3 cm2

Assignment
Q) Find the area of an equilateral triangle of side 7cm.

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 1

Online class – 51 Chapter -5
Trigonometry

30 /10 / 2020

To view class

Answer of last class assignment

Q) Find the area of an equilateral triangle of side 7cm.

Ans) A

Given ABC is an equilateral triangle.

3.5√3 cm ∴ ∠A = ∠B = ∠C = 600
7 cm Draw AD perpendicular to BC

∠DAC =1800 −(900+600)=1800 −1500= 300

B D 3.5 cm C
7 cm

Consider ∆ADC,

AC = 7cm
7
∴ DC = 2 = 3.5cm DC=Hypotenuse
2
AD = 3.5√3 cm
AD = √3xDC

∴ Base = 7cm, Height = 3.5√3 cm

Area of the triangle = 1 Χ base Χ height
2

= 1 Χ 7 Χ 3.5√3
2

= 12.25√3 cm2

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 2

Computing the ratios of the sides of some non-right
triangles

Consider the given triangle

A

BC

Find the ratio of the sides of this triangle.

Ans) Drawing the perpendicular from the top vertex ‘A’
to the base BC, we can split the triangle into two right
triangles ΔADB & ΔADC

A

√2x 2x
x

B xD √3x C

∠BAD = 180° − (90° + 45°) = 45°

∠CAD =105° − 45° = 60° (Given ∠A=105°)

Let AD = x

Consider ΔADB ,

Angles are 45° , 45° , 90° 45° , 45° , 90°
1 : 1 : √2
So sides are in the ratio 1 : 1 : √2 x x √2x

Since AD = x , BD = x , AB = √2x 30° , 60° , 90°
1 : √3 : 2
Consider ΔADC , x √3x 2x
Angles are 30° , 60° , 90°

So sides are in the ratio 1 : √3 : 2
Since AD = x , DC = √3x , AC = 2x

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 3

Sides of the triangle are, AB= √2x , AC = 2x , BC = x + √3x
= x(1 + √3)

Angles of given triangle are 30°, 45° , 105° Dividing
Sides are in the ratio √2x : 2x : x(1 + √3) each by x
= √2 : 2 : (1 + √3)

T B Page 103 4 cm 4 cm
Q1)In the triangle shown, what

is the perpendicular distance
from the top vertex to the
bottom side? What is the area
of the triangle?

Ans)

A

4 cm 4 cm
2cm

B 2√3cm D 2√3cm C

ΔABC is an isosceles triangle. 30° , 60° , 90°
1 : √3 : 2
Given ∠A = 120° , AB = AC 2 2√3 4
180−120 60
∴ ∠B = ∠C = 2 = 2 = 30°

Draw AD perpendicular to BC.

In Δ ADC,

∠DAC = 180° −(30° + 90°) = 60°

Given AC =4cm,
4
∴ AD = 2 = 2cm , DC= 2√3cm

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 4

In ΔADB,

∠DAB = 180° −(30° + 90°) = 60° 30° , 60° , 90°
1 : √3 : 2
Given AB =4cm, 2 2√3 4
4
∴ AD = 2 = 2cm , BD = 2√3cm

In ΔABC, Height (AD)= 2cm , Base (BC) = 2√3 + 2√3

1 = 4√3cm
2
Area of ΔABC = x base x height

= 1 x 4√3 x 2
2

= 4√3 cm2

Q2)a) In the following parallelogram, find the distance between
the top and bottom side?
Calculate the area of the
parallelogram.

Ans) Draw DE perpendicular to AB.
D C In parallelogram ABCD ,

2cm √2cm Distance between
parallel sides = DE

A √2cm E 4cm B

In △ AED

∠ A = 45°, ∠AED = 90° , ∠ADE = 180° − (90° + 45°) = 45°

Angles of the triangle are 45° , 45° , 90°

So sides are in the ratio 1 : 1 : √2 45° , 45° , 90°
1 : 1 : √2
Since AD = 2cm , √2 √2 2

AE = 2 = √2 x√2 = √2cm , DE = √2cm
√2
√2

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 5

So, Distance between parallel sides = DE = √2cm

∴ Area of = One parallel x Distance between
parallelogram ABCD side parallel sides

= 4 x √2
= 4√2 cm2

Assignment

T B Page 103

Q2) b)In the following parallelogram, find the distance between
the top and bottom side? Calculate the area of the
parallelogram.

2cm

4cm

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty Chapter -5 1
Trigonometry
Online class – 52 02 /11 / 2020

To view class

Answer of last class assignment
T B Page 103

Q2) b)In the following parallelogram, find the distance between
the top and bottom side?
Calculate the area of the
parallelogram.
2cm
2cm
Ans) 4cm
C
Draw DE perpendicular to AB.
In parallelogram ABCD ,
D

√3cm Distance between = DE
parallel sides

A 1cm E B
4cm

In △ AED
∠ A = 60°, ∠AED = 90° , ∠ADE = 180° − (90° + 60°) = 30°
Angles of the triangle are 30° , 60° , 90°

So sides are in the ratio 1 : √3 : 2
Since AD = 2cm ,
AE = 1cm , DE = √3cm
So,

Distance between parallel sides = DE = √3cm

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 2

∴ Area of = One parallel x Distance between
parallelogram ABCD side parallel sides

= 4 x √3
= 4√3 cm2

Q3) A rectangular board is to be cut along the diagonal and the
pieces rearranged to form an equilateral triangle as shown
below. The sides of the triangle must be 50 centimetres.
What should be the length
and breadth of the rectangle?

Ans) A

25√3cm 50cm Given Δ ABC is
equilateral
25√3cm

∴ Each angle = 60°

25cm B 25cm D C

Consider ΔADB,
∠B = 60°, ∠D = 90° So, ∠A = 30°
The angles of Δ ADB are 30° , 60° , 90°

So sides are in the ratio 1 : √3 : 2

Given each side of Δ ABC is 50cm 30° , 60° , 90°
1 : √3 : 2
So AB = 50cm 25 25√3 50

∴ BD = 50 = 25cm , AD = 25√3cm
2

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 3

So, Length of rectangle = 25√3 cm
Breadth of rectangle = 25 cm

Q4) Two rectangles are cut along the diagonal and the triangles
got are to be joined to another rectangle to make a regular
hexagon as shown below:

If the sides of the hexagon are to be 30 centimetres, what

would be the length and breadth of the rectangles?

Ans)

E 30cm D

15√3cm 15√3cm C
30√3cm F 15cm G
30cm
15√3cm
30cm

15cm 30cm AB

We know each angle of a regular hexagon is 120°
∴ ∠FED = 120°

All angles of a rectangle are 90°
∴ ∠GED = 90°
So, ∠FEG = 120°− 90° = 30°
In Δ FGE , ∠E = 30° , ∠G = 90°, ∠F = 180°−(90°+30° )= 60°

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 4

The angles of Δ FGE are 30° , 60° , 90° 30° , 60° , 90°
1 : √3 : 2
So sides are in the ratio 1 : √3 : 2 15 15√3 30

Given,

each side of the regular hexagon is 30cm
30
∴ FE = 30 cm , FG = 2 = 15cm , GE = 15√3cm

∴ Length of small rectangle = 15 √ 3cm
Width of small rectangle = 15cm

Length of the large rectangle = 15√3 + 15√3
= 30 √ 3 cm

Width of the large rectangle = 30cm

Assignment

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty Chapter -5 1
Trigonometry
Online class – 53 03 /11 / 2020

To view class

Answer of last class assignment
T B Page 103

Ans) A
Consider △ ABC
x √3y
Draw AD perpendicular to BC
Dy
∴ ∠ADB = ∠ADC = 90° 4cm
Consider △ ADB
∠ADB = 90° , ∠B = 45° (given)

∴ ∠BAD = 180° −( 90°+ 45°) = 45° B x C
Let AD = x

The angles of Δ ADB are 45° , 45° , 90°

So sides are in the ratio 1 : 1 : √2
∴ Since AD = x, BD = x

Consider △ ADC
∠ADC = 90° , ∠C = 60° (given)
∴ ∠CAD = 180° −( 90°+ 60°) = 30°

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 2

Let DC = y 30° , 60° , 90°
The angles of Δ ADC are 30° , 60° , 90°
1 : √3 : 2
So sides are in the ratio 1 : √3 : 2 y √3y
∴ Since DC = y, AD = √3y

Since AD = x and AD = √3y,

x = √3y ..............(1)

From figure , BC = x + y. Also given BC = 4cm

∴ x + y =4

So, √3y + y = 4 ( Substituting √3y for x

y ( √3 + 1 ) = 4 ( Taking ‘y’ outside )

∴ y = 4 .............(2)
√3 + 1

From (1) we have, x = √3y

x = √3 × 4 = 4√3
√3 + 1
√3 + 1

So , Height of the triangle = 4√3 cm
√3 + 1

Given , Base = 4 cm

Area of triangle = 1 × base × height
2
2
= 1 4√3
2 ×4×
√3 + 1

= 8√3 cm2
√3 + 1

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 3

New measure of angles
Consider the given triangles

bc p
r

a
q

Let’the sides of the small triangle be a, b, c
and those of the large triangle be p, q, r

We know “The sides of triangles with the same angles, taken in
the order of size, are in the same ratio”

So, a = b = c =k Assume, let
p q r the ratio be
a b c equal to ‘k’
p = k ⇒ a = pk , q = k⇒ b = qk , r =k ⇒ c = rk

Using ratios, this means.

a:b:c=p:q:r

In triangles of the same angles drawn in different

sizes, the lengths of the sides are different, but

their ratios are the same.

In other words.

The angles of a triangle determine the ratio of its

sides.

It is not easy in general to compute the ratios of sides from the
angles. However, mathematicians from very early times have
found techniques to compute such ratios for right triangles
and arranged them in special tables .

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 4

For example, these tables show that,

In a right triangle of hypotenuse 5 5 × 0.766 5 ×0.6428
metres and one angle 40o , 5m
The side opposite to angle 40o= 5 ×0.6428

= 3.214 m

The third side = 5 × 0.766
= 3.83 m

The numbers in these tables have special names.

In the example above the number 0.6428 shows how much of the
hypotenuse is the side opposite the 40 o angle.

It is called the sine of 40 o and is written sin 40 o .

Sin 40° = 0.6428 = Opposite side of 400 angle
Hypotenuse

The other number 0.7660 shows what part of the hypotenuse is
the other side of the 40 o angle. (It is called the adjacent side of
the angle).

It is called the cosine of 40 o and is written cos 40 o .

Cos 40° = 0.7660 = Adjacent side of 400 angle
Hypotenuse

Tables of sine and cosine of angle
differing by 1o are available.
A part of such a table is shown on the
right.
From this table, we find for example,

sin 35 o ≈ 0.5736
cos 35 o ≈ 0.8192

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 5

Assignment
Write down the Sine and Cosine values of the following angles

from the table given in your text book.

0°, 30°, 45°, 60 °, 90°

Angle Sin Cos

0°

30°

45°

60°

90°

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 1

Chapter -5
Trigonometry

Online class – 54 05 /11 / 2020

Answer of last class assignment To view class

Angle Sin Cos
1
0° 0
0.8660
30° 0.5 0.7071

45° 0.7071 0.5
0
60° 0.8660

90° 1

Important Points

Opposite Side Longest Side
Hypotenuse

Side opposite to <A

Adjacent Side
Side adjacent to <A

Sin A = Opposite Side Cos A = Adjacent Side
Hypotenuse Hypotenuse

Opposite Side = Hypotenuse × Sin A
Adjacent Side = Hypotenuse × Cos A

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 2

Writing the ratios for two types of right triangles, in terms of

sine and cosine

Consider Δ ABC ,

Let BC = x, then, AB =x, AC= √2x A
x
Consider ∠C Opposite Side AB x
Hypotenuse AC
Sin 45° = = = √2 x

∴ Sin 45° = 1 √2x

√2

Cos 45° = Adjacent Side = BC = x
Hypotenuse AC
√2 x
1 B xC
∴ Cos 45° =
√2

Consider Δ PQR ,
Let QR = x, then, PQ = √3x, PR= 2x

Sin 30° = Opposite Side = QR = x P
Hypotenuse PR 2x √3x 2x

∴ Sin 30° = 1
2

Cos 30° = Adjacent Side = PQ = √3 x
Hypotenuse PR
2x

∴ Cos 30° = √3 Qx R

2

Sin 60° = Opposite Side = PQ = √3 x
Hypotenuse PR
2x

∴ Sin 60° = √3

2

Cos 60° = Adjacent Side = QR = x
Hypotenuse PR 2x
1
∴ Cos 60° = 2

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 3

45° √2 30° 2
1 √3

45° 60°
1
1

Sin 45° = 1 Sin 30° = 1 Sin 60° = √3
2
√2 2

Cos 45° = 1 Cos 30° = √3 Cos 60° = 1
2
√2 2

Q1) Find the area of given triangle.

A

4 cm B 6 cm C
4 cm
Ans) Draw AD perpendicular to BC Opposite Side =
A Consider Δ ADB Hypotenuse × Sin50°
AD = AB × Sin 50°
= 4 × 0.7660 Area of triangle=
= 3.064 cm ½× base × height

B 6Dcm C

Given BC = 6cm 1
2
∴ Area of triangle = × BC × AD

= 1 ×6× 3.064
2

= 3 × 3.064

= 9.192 cm2

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 4

Q2) Find the area of given triangle.

A

B C

Ans) Draw the height AD of Δ ABC4 cm

A

∠ABD = 180°−130°
(Linear Pair)

∠ABD = 50°

Consider Δ ADB D B 6 cm C

AD = AB × Sin 50° Opposite Side = Hypotenuse ×Sin50°

= 4 × 0.7660

= 3.064 cm

Given BC = 6cm 1
2
∴ Area of triangle = × BC × AD Area of triangle=
½× base × height
= 1 ×6× 3.064
2

= 3 × 3.064

= 9.192 cm2

T. B Page 109

Q3) The sides of a parallelogram are 8 centimetres and 12

centimetres and the angle between them is 50o.Calculate

its area. A D
Ans) Draw AE perpendicular to BC

So, 8 cm

Height = AE

B 12Ecm C

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 5

Consider Δ AEB,

AE = AB × Sin 50° Opposite Side = Hypotenuse ×Sin50°

= 8 × 0.7660

= 6.128 cm

Given BC = 12 cm

∴ Area of parallelogram = base × height

= BC × AE

= 12 × 6.128

= 73.536 cm2

Assignment
T. B Page 109

Q4) Angles of 50 o and 65 o are drawn at the ends of a 5
centimetres long line, to make a triangle. Calculate its area.

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 1

Chapter -5
Trigonometry

Online class – 55 06 /11 / 2020

To view class

Answer of last class assignment

T. B Page 109
Q4) Angles of 50 o and 65 o are drawn at the ends of a 5

centimetres long line, to make a triangle. Calculate its area.

Ans)

Consider Δ ABC, A
Given ∠B = 50°, ∠C = 65°
∴ ∠A = 180° −( 50° + 65°) = 65°

Given BC = 5cm, 5 cm
Since sides opposite to equal
angles are also equal, AB = 5 cm

Draw AD perpendicular to BC. B 5 cm D C
So height (h) = AD

Consider right triangle Δ ADB,

AD = AB × Sin 50° Opposite Side = Hypotenuse ×Sin50°

= 5 × 0.7660

= 3.830 cm
1
∴ Area of triangle = 2 × BC × AD
= 1
2 1.915

× 5× 3.830

= 5 × 1.915

= 9.575 cm2

T. B Page 109
Q1) The lengths of two sides of a triangle are 8 centimetres

and 10 centimetres and the angle between them is 40o .
Calculate its area. What is the area of the triangle with

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 2
C
sides of the same length, but angle between them140 o ? C

Ans)

a) Draw AD perpendicular to BC. A

AD is the height of the triangle.

Consider Δ ADB, 10 cm
AD = Hypotenuse × Sin 40°

= 10 × 0.6428

= 6.428 cm 8 cm D

Given, base = 8cm B

Area of triangle = 1 × base × height
= 2
1 4
2
× 8 × 6.428

= 4 × 6.428

= 25.712 cm2

b) A
Draw the height AD of Δ ABC

∠ABD = 180°−140° h 10 cm
D
(Linear Pair) B 8 cm
∠ABD = 40°

Consider Δ ADB

AD = Hypotenuse × Sin 40°

= 10 × 0.6428

= 6.428 cm

Given base = 8cm 1
2
∴ Area of triangle = 1 × base × height
= 2
4

× 8 × 6.428

= 4 × 6.428

=25.712 cm2

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 3

Note: For any two triangles if the two sides are equal and angles
between the equal sides are supplementary, then their area
are equal.

Q2) The sides of a rhombus are 5 centimetres long and one of
its angles is 100 o . Compute its area.

Ans)

* A rhombus is a quadrilateral

with four sides equal.

* Area of rhombus = ½ × d1 × d2 d1 d2
where, d1 & d2 are diagonals .

* In a rhombus, diagonals bisect

each other at right angles.

* Diagonals bisect the angles of a rhombus.

Consider rhombus ABCD,

Given one side is 5cm long, A
So, AB = BC = CD = DA = 5cm

Diagonals are AC & BD. Since 5 cm

diagonals bisect each other at right B O D

angles , OA = OC, OB = OD , ∠AOB=90°

Given one angle =100°,∠B =100°,

since diagonals bisect the angles of a C

rhombus, ∠ABO = 100 ° = 50°
2

Consider Δ AOB,

OA =Hypotenuse × sin 50° OB =Hypotenuse × Cos 50°

= 5 × 0.7660 = 5 × 0.6428

= 3.83cm = 3.214cm

∴ Diagonal , AC = 2 × OA ∴ Diagonal , BD = 2 × OB

= 2 × 3.83 = 2 × 3.214

= 7.66 cm = 6.428 cm

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 4

∴ Area of rhombus = 1 × AC × BD
= 2
1 3.214
2
× 7.66 × 6.428

= 7.66 × 3.214

= 24.62 cm2

Assignment

T. B Page 109
Q5) A triangle is to be drawn with one side 8 centimetres and

an angle on it 40 o . What should be the minimum length of
the side opposite this angle?

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 1

Chapter -5
Trigonometry

Online class – 56 09 /11 / 2020

To view class

Answer of last class assignment

T. B Page 109

Q5) A triangle is to be drawn with one side 8 centimetres and
an angle on it 40 o . What should be the minimum length of

the side opposite this angle?

Ans)

The triangle given in the

question can be drawn in many

ways as given in the figure. C

Among these triangles we get h

minimum length of the side

opposite to 40° , when we draw

the third side BC perpendicular

to AC. A 8 cm B

∴ Minimum length , h = 8 × Sin40°

= 8 × 0.6428

= 5.1424cm

Triangle and circle
Length of arc
Q1) In the given figure an arc makes an angle

60° at the centre of the circle. What is the
length of the arc?

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 2

Ans) 60 ° 1
360 ° 6
=

So , arc length is 1 part of circumference
6

of the circle. 1
6
∴ Arc length = 2πr ×

Q2) In the given figure an arc makes an angle
120° at the centre of the circle. What is the
length of the arc?

Ans) 120 ° 1
360 ° 3
=

So , arc length is 1 part of circumference
3

of the circle. 1
3
∴ Arc length = 2πr ×

In general,

If an arc makes an angle x°

at the centre of the circle,

then, r

Arc length = 2πr × x°
360 °

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 3

Length of chord O
a) Length of a chord of central angle 60o
r 60° r
Here, OA = OB = r (radius of circle)
Given ∠O = 60° 60° 60°
∴ ∠A = ∠B = 60°
Since Δ OAB is equilateral,
AB = r

Ar B

∴ Length of a chord of =r
central angle 60°

b) Length of a chord of central angle 120o

Draw the perpendicular OP from the

centre to the chord AB . OP bisects the

chord AB and its central angle ∠O.
120 °
∠POB = 2 = 60° O
r
The angles of ΔOPB are 30o , 60o , 90o
r
so sides are in the ratio 1 : √3 : 2 A 2 B

So, if we take the radius as r , P √3r

2

then , OP = r & PB = √3 r
2 2

Length of chord AB = 2 × PB = 2 × √3 r= √3 r
2

∴ Length of a chord of = √3 r
central angle 120°

r 120° r
√3r

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 4

c) Length of a chord of central angle 90o Or B
Δ AOB is right triangle.
Since OA =OB, r √2r
∠A = 45° & ∠B = 45° .
The angles of ΔAOB are 45o , 45o , 90o A
so sides are in the ratio 1 : 1 : √2
Since OA = OB = r, AB = √2r

Length of a chord of
∴ central angle 90° = √2r

Q1) What the length of the chord in this picture?

3cm

Ans)

Draw OP perpendicular to the chord AB .

OP bisects the chord AB and its

central angle ∠O.
100 °
∴ ∠POB = 2 = 50°

O Δ OPB is right triangle.
PB
3cm Sin50° = OB

A P B ∴ PB = OB × Sin 50°
= 3 × 0.7660

= 2.298 cm

∴ Length of chord AB = 2 × PB

= 2 × 2.298

= 4.596 cm

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 5

Q2) What is the length of a chord of central angle xo ?

Ans) Chord AB makes an angle x° at the

centre of the circle.

Draw OP perpendicular to AB.
x°
Now, ∠POB = 2 O
Xr
Radius of the circle is ‘r’
2

In Δ OPB , A P B
x°
∠POB = 2 , ∠OPB = 90° , OB = r cm

Sin( x° )= PB
2 OB
x° x°
∴ PB = OB × Sin( 2 )= r × Sin( 2 )

AB = 2 × PB x° x°
2 2
=2× r × Sin( )= 2r Sin( )

That is,

In a circle, the length of any chord is double the product of

the radius and the sine of half the central angle

In a circle of radius ‘r’,

Length of a chord of = 2r Sin( x° ) r
central angle x° 2

Assignment O 20 m B
Q) Raju and Babu are standing at the
A
starting point A of a circular track of
radius 20 metres. Raju walks through
the arc AB and Babu walks through
the chord AB , and reach at B. If the
central angle is 160° , how much
distance did Raju walk more than
Babu?

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 1

Chapter -5
Trigonometry

Online class – 57 10 /11 / 2020

Answer of last class assignment To view class
Q) Raju and Babu are standing at the
O 20 m B
starting point A of a circular track of A
radius 20 metres. Raju walks through
the arc AB and Babu walks through
the chord AB , and reach at B. If the
central angle is 160° , how much
distance did Raju walk more than
Babu?
Ans)

Given ∠AOB = 160°, radius = 20m
x°
O 20 m Length of arc AB = 2πr × 360 °
=
Babu B 2π × 20 × 160 °
360 °
A Raju 4
= 40 × 3.14 × 9

= 55.822

Raju walk along the arc AB.

So distance travelled by Raju = 5.82m
x°
Length of chord AB = 2r Sin( 2 )

= 2 × 20 × Sin( 160 ° )
2

= 2 × 20 × Sin 80°

= 40 × 0.9848 = 39.392

Babu walk along the chord AB.

So distance travelled by Babu = 39.392m

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 2

Difference in distance = 55.822 − 39.392 55.822
= 16.43 m 39.392
16.430
Raju walks 16.43 meters more than Babu.

Computing the ratio of the sides of a triangle from its angles

a) When all the angles of a triangle are less than 90° .

Consider triangle Δ ABC .

If we draw the circumcircle of the triangle, all its sides become

chords of the circle and the central angle of each of these arcs

is double the angle opposite to it in the triangle. A
∠A = 65° ∴ ∠BOC = 130°
∠B = 55° ∴ ∠AOC = 110°
∠C = 60° ∴ ∠AOB = 120°

We know, O

Length of chord of = 2r Sin( x° ) B C
central angle x° 2

A

∴ BC = 2r Sin( 130 ° )= 2r Sin65°
2
110 °
2r Sin60° 2r Sin55° AC = 2r Sin( 2 )= 2r Sin55°

AB = 2r Sin( 120 ° )= 2r Sin60°
2

B 2r Sin65° C

Ratio of length of = 2r Sin65° : 2r Sin55° : 2r Sin60°
sides of triangle

= Sin65° : Sin55° : Sin60°

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 3
C
b) When one angle of triangle is greater than 90°

Consider Δ ABC A
Here ∠A = 110° is an obtuse angle.

If we draw the circumcircle of the

triangle, all its sides become B
chords of the circle.

The circumcircle and the central angles of the chords which

are sides of the triangle are like this, A

AA

B CB CB C

O OO

AB = 2r Sin( 80 ° ) AC = 2r Sin( 60 ° ) BC = 2r Sin( 140 ° )
2 2 2

= 2r Sin 40° = 2r Sin 30° = 2r Sin 70°

A

2r Sin 40° 2r Sin 30°

B 2r Sin 70° C

Ratio of length of
sides of triangle = 2r Sin30° : 2r Sin40° : 2r Sin70°
= Sin30° : Sin40° : Sin70°

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 4
A
c) When one angle of triangle is 90°
Consider Δ ABC
Here ∠A = 90° .

If we draw the circumcircle of the

triangle, all its sides become B C

chords of the circle.

A

2r Sin 50° 2r Sin 40° AC = 2r Sin( 80 ° ) = 2r Sin40°
2
100 °
AB = 2r Sin( 2 )= 2r Sin 50°

B O 2r C

Side BC of the triangle is the

diameter of the circle.

∴ BC = 2r

Ratio of length of = 2r : 2r Sin40° : 2r Sin50°
sides of triangle = 1 : sin 40o : sin50o

In general we can say

The ratio of sides of acute and right triangle is the
ratio of Sine of angles opposite to it.

If one angle is greater than 90 ° we have to take the
supplementary angle’s measure to find the length of

side opposite to the 90° angle.

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 5

Q1) The three angles of a triangle are 50° , 60° and 70° and the

circum radius is 5cm,

a) Find the ratio of the sides of the triangle?

b) Find the measure of the three sides of the triangle?

Ans) 2r Sin 60° A 2r Sin 50°
a)
Ratio of length of = Sin50° : Sin60° : Sin70° ˘70°
sides of triangle

b) Given r = 5cm B )50° 60°( C
Length of AC = 2r Sin50°
= 2 x 5 x Sin 50° 2r Sin 70°
= 10 x 0.7660
= 7.66 cm
Length of AB = 2r Sin60°
= 2 x 5 x Sin 60°
= 10 x 0.8660
= 8.66 cm
Length of BC = 2r Sin70°
= 2 x 5 x Sin 70°
= 10 x 0.9397
= 9.397 cm

Assignment
Q2) If one side of a triangle is 4cm and angles on

both ends of that side are 80° and 70°
respectively.
a) Find the measure of the third angle?
b) Find the circum radius?
c) Find the measure of other two sides?

4 cm

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 1

Chapter -5
Trigonometry

Online class – 58 12 /11 / 2020
Answer of last class assignment
To view class

Q2) If one side of a triangle is 4cm and angles on 4 cm
both ends of that side are 80° and 70°
respectively.

a) Find the measure of the third angle?
b) Find the circumradius?
c) Find the measure of other two sides?
Ans)

A

˘ a) ∠A = 180°−(80° + 70° )

30° = 180° − 150° = 30°

b) If circumradius is ‘r’ then

BC = 2r sin30°

Given, BC = 4cm 1
2
So, 2r sin30° = 4 sin30°=

B 4 cm C 2r × 1 =4
2

∴ r = 4 cm

c) AB = 2r sin70° AC = 2r sin80°
= 2 × 4 × 0.9397 = 2 × 4 × 0.9848
= 7.5176 cm = 7.8784 cm

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 2

If three angles of a triangle are A, B, C, then A
Ratio between 3 sides = SinA : SinB: SinC

If the cirumradius is ‘r’ then 3 sides are 2r SinC 2r SinB

BC = 2r SinA

AC = 2r SinB B 2r SinA C
AB = 2r SinC

Another observation( Sine Rule)

Consider Δ ABC , let the sides be a, b, c
A
If ‘r’ is the cSiirncuAmra∴diuSs,ianA
a = 2r
= 2r

c b b = 2r Sin B ∴ b = 2r

SinB

B a C c = 2r Sin C ∴ c = 2r

SinC

∴ a =b = c = 2r
SinA SinB
SinC

This rule can be used to compute the remaining sides
of a triangle when one side and two angles on it are known.

Q) The figure shows a triangle. Compute the lengths of the
other two sides of the triangle.

10
Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 3

Ans) ˘A

∠A = 180° − (40° + 80° ) cb

= 180° − 120°

= 60°

We know, c
a b sin C
sin A = sin B =

B 10 C

Given a = 10, ∠B = 40° , ∠C = 80° a
a b sin A c
sin A = sin B = sin80 °

10 = b 10 = c
sin 60 ° sin 40 ° sin 60 ° sin80 °

∴ b= 10 sin 40° ∴ c= 10 sin 80 °
sin60 ° sin 60 °
10 × 0.6428 10 × 0.9848
= 0.8660 = 0.8660

= 7.42 cm = 11.37cm

T.B Page 114 A
Q1) The figure shows a triangle and its 60°

circumcircle:
What is the radius of the circle?

Ans)

Given , ∠ A = 60°, BC = 3 cm B 3 cm C

2r sin60° = 3

2r × √3 =3

2

√3 r = 3

r= 3 = 3 × √3 = 3√3 = √3 cm
√3
√3 √3 3

∴ Radius of the circle = √3 cm

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 4

Q4) A circle is to be drawn, passing through the ends of a line, 5

centimetres long; and the angle on the circle on one side of

the line should be 80° . What should be the radius of the

circle? A
Ans) Given ∠A = 80° ,

BC = 2r sin80° 80°
BC
Given, BC = 5
5 cm
∴ 2r sin80° = 5
5
r= 2 × sin80 °

= 5
2 × 0.9848
5
= 1.9696 = 2.54

ie, radius of circle = 2.54 cm

Q5) The picture below shows part of a circle:

8 cm

What is the radius of the circle? If one angle is greater
Ans) Given ∠A =140° than 90 ° we have to take
the supplementary
A C angle’s measure .

B 8 cm

∴ BC = 2r sin(180° – 140° )= 2r sin40°

Given , BC = 8 cm

So, 2r sin40° = 8
84 4
r= 2 × sin 40 ° = 0.6428 = 6.22

ie, radius of circle = 6.22 cm

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi