## Measurement of Angle

Subject: Optional Mathematics

#### Overview

The following three systems are commonly used in the measurement of angles: a) Sexagesimal system (degree system): This system is also called British System. In this system, the unit of measurement is degree. So, this system is also known as degree system. b) Centesimal system (grade system): This system of measurement is also called the French system. In this system, the unit measurement is grade.So this system is also known as grade system

### Angle

Let OX be the original line. The original line is also called initial line. Let a revolving line OP start from OX and revolve to reach into the position OP. The line OP is also called the terminal line. These lines form an angle XOP at the point O. The point O is called the vertex and we write $\angle$XOP to mean angle XOP.

If the direction of revolving line OX is clockwise, $\angle$XOQ is called the negative angle. But, if the direction of revolving line OP is anticlockwise, then $\angle$XOP is called the positive angle.

### Measurement of Angles

The following three systems are commonly used in the measurement of angles:

1. Sexagesimal system (degree system)

#### Sexagesimal system

The system is also called British System. In this system, the unit of measurement is degree. So this system is also known as degree system. A right angle is divided into 90 equal parts and each part is called a degree. A degree is divided into 60 equal parts and each part is called a minute. A minute is also divided into 60 equal parts and each part is called a second. Therefore, we have

60 seconds = 1 minutes (60" =1')
60 minutes = 1 degree (60' = 1°)
90 degrees = 1 right angle
The degree, minute the second are denoted by (°), (') and (")

#### Centesimal System

This system of measurement is also called the French system. In this system, the unit of measurement is grade. So, this system is also known as grade system. A right angle is divided into 100 equal parts, each part is called a grade. A grade is divided into 100 equal part, each part is called a minute. A minute is also divided into 100 equal parts, each part is called a second. Therefore, we have,

100 seconds = 1 minute (100" = 1')
100 minutes = 1 grade (100' = 1g)
100 grades = 1 right angle
The grade, minute and second are respectively denoted by (g), (') and (").

#### Circular System

Draw a circle with centre O and radius OP. Take any point B on the circumference such that arc PQ is equal in length to the radius OP. Join QO. The angle POQ, so formed , is said to be 1 radian. A radian is denoted by (c). So, 1 radian is written as 1c. Here, ∠POQ = 1c and arc PQ = r.

An angle subtended by an arc equal in length to the radius, at the centre of a circle is defined as a radian.

Let OP = OQ = r. Produce PO to the point R on the circumference of the circle. Then PR is called the diameter of the circle.

Here, circumference of the circle = 2πr

∴ arc PQR =πr

We know angles subtended by different arcs at the centre of a circle are proportional to the corresponding arcs .

So, $\frac{∠POQ}{∠POR}$ = $\frac{arc PQ}{arc PR}$

or, $\frac{1°}{180°}$ = $\frac{r}{πr}$

∴ 180° =πc

1. Relation between Sexagesimal and Centisimal System
In sexagesimal system 1 right angle = 90°...............(i)
In centesimal system 1 right angle = 100g.............(ii)
So, 90° = 100g
or, 1° = ($\frac{100}{90}$)g
= ($\frac{10}{9}$)g
so, x° = ($\frac{10}{9}$×x)g
Again
100g = 90°
or, 1g = ($\frac{90}{100}$)°
or, 1g= ($\frac{9}{10}$)°
So, xg = ($\frac{9}{10}$×x)°

2. Relation between Radian System and Sexagesimal System
We know that,
180° =πc
or, x° = ($\frac{π}{180}$c
Similarly,
1c = ($\frac{180}{π}$)°
So, xc = ($\frac{180}{π}$×x)°

3. Relation between Radian System and Centesimal System
We know,
200g= πc
or, xg = ($\frac{π×x}{200}$)c
Again
π= 200g
or, 1c = ($\frac{200}{π}$)g
So,x= ($\frac{200×x}{π}$)g

#### Polygons

Polygon is a closed plane figure having three or more than three line segments. Triangles, quadrilateral, nonagon, octagon etc. are the examples of a polygon. A polygon having all sides equal in length is called a regular polygon. A regular polygon has the same measures of interior angles.
In regular polygons of sides n, each interior angle is θ = $\frac{(n-2) × 180°}{n}$
Similarly,
The exterior angle is a side of a regular polygon where an angle between any side of a shape and a line extended from the next side.
Exterior angle of polygon (Φ) = $\frac{360°}{n}$

#### Theorem on circular measure

The radian is a constant angle
Proof: Let O be the center of the circle and AB be the diameter. Let C be a point on the circumference such that OB = arc BC = r where r is the circle. Then, by definition,∠BOC = 1c.
We know, circumference of a circle = 2πr
∴ Length of arc BCA = $\frac{1}{2}$× 2πr =πr.
Now, angles subtended by different arcs at the center o0f a circle are proportional to the corresponding arcs.
Therefore we have
$\frac{∠BOC}{∠AOB}$ = $\frac{arc \: BC}{arc \: BCA}$
or, $\frac{1^c}{180^o}$ = $\frac{r}{πr}$
or, 1c = $\frac{180^o}{π}$ which is constant
Hence, the radian is a constant angle.

#### Theorem of arc length and radius with the angle at the center of the circle

The circular measure of an angle subtended by an arc of a circle at the center is expressed by the ratio of the arc to the radius of a circle.
i.e. angle subtended at the centre of a circle = $\frac{length \: of \: the \: arc \: of \: the \: circle}{radius \: of \: the \: circle}$
Proof: Let O be the center of a circle and r be its radius. Take any two points A and B on the circle such that arc AB = r.
Then, by the definition,∠AOB = 1c.
Let C be any other point on the circle such that arc ABC = \ and the angle subtended by arc ABC at the center of the circle be θc.
We know angles subtended by different arcs at the center of a circle are proportional to the corresponding arcs.
Therefore, we have
$\frac{Arc \: ABC}{Arc \: AB}$ = $\frac{∠AOB}{∠AOC}$
or, $\frac{l}{r}$ = $\frac{θ^c}{1^c}$
∴θc = $\frac{l}{r}$
Thus, angle at the center = $\frac{length \: of \: the \: arc \: of \: the \: circle}{radius \: of \: the \: circle}$
From this theorem, we have following three relations:

1. θc = $\frac{l}{r}$
2. l = rθ
3. r = $\frac{l}{θ}$

When the central angleθc becomes 2π, then the corresponding arc l becomes circumference of the circle. Then, by formula (ii) we have
l = rθ
or, circumference of circle = r× 2π = 2πr.

#### Clock

In a clock, the hands of the clock always rotate is the clock-wise direction. The minute hand, the hour hand and the second-hand together show us the time. All the hands of the clock are moving always but only the change in second hand can be observed as it makes a remarkable change within seconds. But all the hands of the clock are always in the state of change.

The second hand,

The second-hand makes 1 complete rotation in 1 minute.

i.e. In 1 minute = 60 sec, the second-hand turns 360.

$\therefore$ In 1 second, the second hand turns = $\frac{360°}{60}$ = 6°

The minute hand,

The minute hand makes 1 complete rotation in 1 hour.

i.e. In 1 minute = 60 min, the minute hand turns 360°.

$\therefore$ In 1 minute, the minute hand turns $\frac{360°}{60}$ = 6°

The hour hand,

The hour hand makes 1 complete rotation in 12 hours.

i.e. In 12 hours, the minute hand turns 360°.

$\therefore$ In 1 hour, the minute hand turns = $\frac{360°}{12}$ = 30°.

##### Things to remember

→  Sexagesimal system (British System)

• 1 right angle =90°
• 1° = 60 minutes
• 1 minutes = 60 second

→  Cetesimal system (French System)

• 1 right angle = 100 grade
• 1 grade = 100 minutes
• 1 minutes = 100 Seconds

• πc = 180° = 200g
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Measuring angles in degrees | Angles and intersecting lines | Geometry | Khan Academy

Soln:
Here, 27o = 27× 60' = 1620'

27o38' = (1620 + 38)' = 1658'

Then, 1658' = (1658 × 60)" = 99480"

∴ 27o38'42" = (99480 + 42)" = 99522". Ans.

Soln:

Here, 60o = 60× 60' = 600'

3600' = (3600× 60") = 216000"

∴ 60o = 216000" Ans.

Soln:

Here, 10" =($\frac{12}{60}$)

10' 12" =(10 +$\frac{12}{60}$)' = (10 + 0.2)' = 10.2'

10.2' =($\frac{10.2}{60}$)o = ($)\frac{102}{600}$o = 0.17o

∴ 20o 10' 12" = (20 + 0.17)o = 20.17o . Ans.

Soln:

Here, 10" = ($\frac{10}{60}$)' = ($\frac{1}{6}$)'

10' 10" = (10 + $\frac{1}{60}$) =($\frac{60 + 1}{6}$)' =$\frac{61'}{6}$

$\frac{61'}{6}$ =($\frac{61}{6}$× $\frac{1}{60}$)o =($\frac{61}{360}$)o = 0.1694o

∴ 20o 10' 10" = (20 + 0.1694)o = 20.1694o. Ans.

Soln:

Here, 43" =($\frac{43}{6}$)'

13'43" =(13 + $\frac{43}{60}$)' =($\frac{823}{60}$)'

($\frac{823}{60}$)' =($\frac{823}{60}$×$\frac{1}{60}$)o = 0.2286o

∴ 57o 13' 43" = ( 57 + 0.2286)o = 57.2286o Ans.

Soln:

Here, 30" =($\frac{30}{60}$)' =($\frac{1}{2}$)'

15'30" = (15 + $\frac{1}{2}$)' =($\frac{31}{2}$)'

($\frac{31}{2}$)' =($\frac{31}{2}$×$\frac{1}{60}$)o = 0.2583o

∴ 60o 15' 30" = ( 60 + 0.2583)o = 60.2583o Ans.

Soln:

Here, 25" = ($\frac{25}{100}$)' =($\frac{1}{4}$)'

14' 25" = ( 14 + $\frac{1}{4}$)' =($\frac{57}{4}$)'

($\frac{57}{4}$)' =($\frac{57}{4}$ × $\frac{1}{100}$)g =($\frac{57}{400}$)g = 0.1425g

∴ 6g 14' 25" = 6 + 0.1425g = 6.1425g Ans.

Soln:

Here, 85" = ($\frac{85}{100}$)' = ($\frac{17}{20}$)'

2'85" = ( 2 + $\frac{17}{20}$)' = ($\frac{57}{20}$)'

($\frac{57}{20}$)' =($\frac{57}{20}$×$\frac{1}{100}$)g =($\frac{57}{2000}$)g = 0.0285g

∴ 80g 2' 85" = ( 80 + 0.0285)g = 80.0285g. Ans.

Soln:

Here, 90" = ($\frac{90}{100}$)' = ($\frac{9}{10}$)'

5' 90" = ( 5 + $\frac{9}{10}$)' = ($\frac{59}{10}$)'

($\frac{59}{10}$)' =($\frac{59}{10}$×$\frac{1}{100}$)g = 0.059g

∴75g 5' 90" = (75 + 0.059)g = 75.059g. Ans.

Soln:

Here, 70" = ($\frac{70}{100}$)' =($\frac{7}{10}$)'

($\frac{7}{10}$)' =($\frac{7}{10}$×$\frac{1}{100}$)g = ($\frac{7}{1000}$)g = 0.007g

∴ 65g 70" = ( 65 + 0.007)g = 65.007g Ans.

Soln:

Here, 25g 26' 23" = 25.2623g [ 25 +$\frac{26}{100}$ + $\frac{23}{10000}$ = 25+0.26+0.0023]

25.2623g =(25.2623× $\frac{9}{10}$)o

= 22.73607o = 22o 44' 9.85". Ans.

Soln:

Here,30g 29' 19.75" = 30.291975g [ 30 + $\frac{29}{100}$ + $\frac{19.75}{10000}$ = 30 + 0.29 + 0.001975]

30.29195g = (30.291975× $\frac{9}{10}$)o

= 27.262777o = 27o 15' 46" Ans.

Soln:

Here,26o46' = (26 +$\frac{46}{60}$)o = ( 26 + 0.766)o = 26.7666o

26.7666o = ( 26.7666×$\frac{10}{9}$)g = 29.74066g

= 29g 74' 6.6" Ans.

Soln:

Here, given

One angle of a right angle triangle = $\frac{3}{10}$× 90 = 27o

If the third angle = xo

We know sum of angles of the triangle = 180o

or, xo + 27o + 90o = 180o

or, xo = 180o - 117o = 63o

but, 1o = ($\frac{10}{9}$)g so,

63o = ( 63 ×$\frac{10}{9}$)g = 70grd

∴ Third angle of the triangle = 70g. Ans.

Soln:

Here given,

One angle of triangle = 72o = ( 72× $\frac{10}{9}$)g = 80g

Ratio of the remaining two angles is 1: 3

Let, these two angles be kg and 3kg..

We know, the sum of three angle of triangle = 200g

80 + k + 3k = 200

or, 4k = 200 - 80

or, 4k = 120

∴ k = $\frac{120}{4}$ = 30

∴ First angle = 80g

Second angle = k = 30g

and Third angle = 3k = 3× 30 = 90g

∴ Three angles = 30g 80g90g. Ans.

Proof:

Let O be the centre and OA = r be the radius of a circle. Let AP be an arc which subtends an angle AOP on the centre O of the circle and the length of which is equal to the radius i.e. AP = r.

Now, producing AO up to B so that the arc length APB =$\frac{1}{2}$ of the circumference =$\frac{1}{2}$× 2Πr =πr

Where AOB = diameter = 2r.

and∠AOB = 180o.

Now, by the theories of geometry,

the angles at the centre of the circle are proportionalto the arcs on which they stand:

So, $\frac{∠AOP}{∠AOB}$ =$\frac{arc AP}{arc APB}$

or, $\frac{1^c}{180^o}$ =$\frac{r}{πr}$

∴ 1c = $\frac{180^o}{Π}$

i.e.πc = 180o

Since the value of 1c =$\frac{180^o}{Π}$ is independent of r, so the radian is a constant angl. Proved

Solution:

Let the no. of sides of the regular polygon be n.

Then, exterior angle = $\frac{}{}$ × interior angle.

i.e. $\frac{360}{n}$ = $\frac{1}{4}$ × $\frac{(n - 2) × 180°)}{n}$

or, $\frac{360}{n}$ = $\frac{45(n - 2)}{n}$

or, n - 2 = $\frac{360}{45}$

or, n - 2 = 8

$\therefore$ n = 10

Solution

When the clock strikes half past three.

The hour hand at 3 and minute hand at 6.

So, for real calculation,

The hour hand moves 30 minutes away from 3.

So, In 1 hour, the hour hand makes 30°.

$\therefore$ In $\frac{}{}$ hour, the hour hand makes 30 × $\frac{1°}{2}$ = 15°

Also, the angle between the hour and minute hand when the hand is at 3 and 6 is 90°.

So, the angle between the two hands at half past three 90° - 15° = 75°.

Solution:

length of a arc (l) = 8 cm

radius of circle (r) = 20 cm

Then, central angle $\theta$ = $\frac{l}{r}$ = $\frac{8}{20}$ = $\frac{2}{5}$ radian.

So, $\theta$ = $\frac{2}{5}$ = $\frac{180}{π}$ = 22.92°