Subject: Optional Mathematics

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

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.

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

- Sexagesimal system (degree system)
- Centesimal system (grade system)
- Circular system (radian 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 (")

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 (").

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° =π^{c}

**Relation between Sexagesimal and Centisimal System**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)°**Relation between Radian System and Sexagesimal System**Relation between Radian Systema and Sexagesimal System

We know that,

180° =π^{c}

or, x° = (\(\frac{π}{180}\)^{c}

Similarly,

1c = (\(\frac{180}{π}\))°

So, x^{c}= (\(\frac{180}{π}\)×x)°**Relation between Radian System and Centesimal System**Relation between Radian System and Centesimal System

We know,

200g= π^{c}

or, xg = (\(\frac{π×x}{200}\))^{c}

Again

π^{c }= 200g

or, 1^{c}= (\(\frac{200}{π}\))g

So,x^{c }= (\(\frac{200×x}{π}\))g

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) = \(\frac{360°}{n}\)

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.

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, \(\frac{l}{r}\) = \(\frac{θ^c}{1^c}\)

∴θ^{c} = \(\frac{l}{r}\)

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:

- θ
^{c}= \(\frac{l}{r}\) - l = rθ
- r = \(\frac{l}{θ}\)

When the central angleθ^{c} 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.

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°.

→ 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

→ Radian system (circular system)

- π
^{c}= 180° = 200^{g}

- 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.

Convert into seconds:

27^{o} 38' 42"

Soln:

Here, 27^{o} = 27× 60' = 1620'

27^{o}38' = (1620 + 38)' = 1658'

Then, 1658' = (1658 × 60)" = 99480"

∴ 27^{o}38'42" = (99480 + 42)" = 99522". Ans.

Convert into seconds:

60^{o}

Soln:

Here, 60^{o} = 60× 60' = 600'

3600' = (3600× 60") = 216000"

∴ 60^{o} = 216000" Ans.

Convert into degrees:

20^{o} 10' 12"

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} = 0.17^{o}

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

Convert in degrees:

20^{o} 10' 10"

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} = 0.1694^{o}

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

Convert into degrees:

57^{o} 13' 43"

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} = 0.2286^{o}

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

Convert into degrees:

60^{o} 15' 30"

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} = 0.2583^{o}

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

Convert into grades:

6^{g} 14' 25"

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} = 0.1425^{g}

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

Convert into grades:

80^{g} 2' 85"

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} = 0.0285^{g}

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

Convert into grades:

75^{g} 5' 90"

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} = 0.059^{g}

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

Convert into grades:

65^{g} 70"

Soln:

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

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

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

Convert into sexagesimal system:

25^{g} 26' 23"

Soln:

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

Changing into grade

25.2623^{g} =(25.2623× \(\frac{9}{10}\))^{o}

= 22.73607^{o} = 22^{o} 44' 9.85". Ans.

Convert into sexagesimal:

30^{g} 29' 19.75"

Soln:

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

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

= 27.262777^{o} = 27^{o} 15' 46" Ans.

Convert into centesimal system:

26^{o }46'

Soln:

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

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

= 29^{g} 74' 6.6" Ans.

In a right angle triangle if one of acute angle is \(\frac{3}{10}\) of right angle. What is the measurement of other angle in grade?

Soln:

Here, given

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

If the third angle = x^{o}

We know sum of angles of the triangle = 180^{o}

or, x^{o} + 27^{o} + 90^{o} = 180^{o}

or, x^{o} = 180^{o} - 117^{o} = 63^{o}

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

63^{o} = ( 63 ×\(\frac{10}{9}\))^{g} = 70^{grd}

∴ Third angle of the triangle = 70^{g}. Ans.

One of the angles of triangle is 72^{o} and the ratio of the other two angles is 1:3 find the angles in grade.

Soln:

Here given,

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

Ratio of the remaining two angles is 1: 3

Let, these two angles be k^{g} and 3k^{g.}.

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

80 + k + 3k = 200

or, 4k = 200 - 80

or, 4k = 120

∴ k = \(\frac{120}{4}\) = 30

∴ First angle = 80^{g}

Second angle = k = 30^{g}

and Third angle = 3k = 3× 30 = 90^{g}

∴ Three angles = 30^{g} 80^{g}90^{g}. Ans.

To prove: Radian is a constant angle:

**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.

Then, by definition of radian,∠AOP = 1 radian (1^{c}).

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 = 180^{o}.

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, \(\frac{1^c}{180^o}\) =\(\frac{r}{πr}\)

∴ 1^{c} = \(\frac{180^o}{Π}\)

i.e.π^{c} = 180^{o}

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

The exterior angle of a regular polygon is equal to \(\frac{1}{4}\) of its interior angle. Find the no. of sides of the polygon.

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

Find the angle made by the minute hand and hour hand of a clock at half past three.

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°.

A central angle of a circle of radius 20 cm intercepts an arc of 8 cm. Find the angle in degree.

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°

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