Radians

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1 Summary
1.1 What are Radians?
1.2 Example #1
1.3 Length of Arc
1.4 Area of a sector
1.5 Area of a segment
1.6 Example #2
1.6.1 Reference

Summary

  • 1 Revolution = 360°
  • \pi radian = 180°
  • 1 radian = \frac { 180}{ \pi } = 57.3°
  • 1° =  \frac { \pi }{ 180 } radian  = 0.175 radian
  • Length of arc  =\quad r\theta
  • Area of a Sector   =\quad \frac { 1 }{ 2 } { r }^{ 2 }\theta
  • Area of a segment   =\quad \frac { 1 }{ 2 } { r }^{ 2 }\theta \quad -\quad \frac { 1 }{ 2 } { r }^{ 2 }sin\theta

The most common system of measuring the angles is that of degrees. One complete revolution is divided into 360 equal parts and each part is called one degree (1°). Furthermore, Half revolution is equivalent to \frac { 360 }{ 2 } \quad =\quad 180°.

Also for smaller angles, 1° is divided into 60 parts and each part is called 1 minute which is further divided into 60 parts and each part is now called 1 second.

1 Revolution = 360°
1 Degree = 60 minutes
1 minute = 60 seconds

What are Radians?

Radians are a very important way of measuring angles. A radian is defined as the angle subtended at the centre of a circle by an arc whose length is equal to that of the radius of the circle.

The diagram ‘Fig 1’ shows a circle with centre 0 and radius r. The length of the arc AB is also r. In this position, the angle made by the arc AB at the centre is defined as 1 radian.

If AB = r, then  A\hat { o } B\quad =\quad 1  radian where  A\hat { o } B  is the angle subtended at the center.

If AB = 2r then  A\hat { o } B\quad =\quad 2  radian

If  AB\quad =\quad 2\pi r  which is the circumference then  A\hat { o } B\quad  is  2\pi   radian. Shown in Fig 2.

As  2\pi r  is the circumference , then  OA makes a complete revolution 2\pi \quad =\quad 360°

\pi radian = 180°
1 radian = \frac {180 }{ \pi } = 57.3°
1° =  \frac { \pi }{ 180 } radian = 0.175 radian

Example #1

To convert an angle from degrees to radians we multiply the angle by \frac { \pi }{ 180 }  radian e.g:

Q. Convert 150° angle into radian.

Solution:

150°=  150\quad \times \quad \frac { \pi }{ 180 }  radian

=\quad \frac { 5\pi }{ 6 } \quad \quad \quad rad

Length of Arc

Let s be length of an arc AB which subtends an angle ???? radian at the centre O of a circle of radius r units. Shown in Fig 3.

From the definition of radian, we calculate that:

\frac { arc\quad length\quad of\quad AB }{ circumference\quad of\quad the\quad circle } \quad=

\quad \frac { Angle\quad subtended\quad by\quad AB\quad at\quad O }{ Angle\quad subtended\quad by\quad the\quad circle\quad at\quad 0 }

\frac { s }{ 2\pi r } \quad =\quad \frac { \theta }{ 2\pi }

s\quad =\quad r\theta    where  \theta  is in radians

Area of a sector

To find the area of the sector which contains angle ???? radian at the centre of the circle

as shown in Fig 4, we consider the sector as a fraction of the circle hence:

\frac { area\quad of\quad the\quad sector\quad AOB }{ area\quad of\quad the\quad circle } \quad =

\quad \frac { Angle\quad subtended\quad by\quad AB\quad at\quad O }{ Angle\quad subtended\quad by\quad circle\quad at\quad O }

\frac { A }{ \pi { r }^{ 2 } } \quad =\quad \frac { \theta }{ 2\pi }

A\quad =\quad \frac { \pi { r }^{ 2 }\theta }{ 2\pi }

A\quad =\quad \frac { 1 }{ 2 } { r }^{ 2 }\theta

Area of a segment

The area of the minor segment as shown in Fig 5

is obtained by the arc AB of the centre O is given by:

Area\quad of\quad the\quad Segment\quad =\quad Area\quad of\quad the\quad sector\quad AOB\quad -\quad Area\quad of\quad the\quad triangle\quad AO

=\quad \frac { 1 }{ 2 } { r }^{ 2 }\theta \quad -\quad \frac { 1 }{ 2 } { r }^{ 2 }sin\theta       where  \theta  is in radians

Example #2

Q. The diagram (Fig 6) shows the sector OPQ of a circle with centre O and radius r cm.

The angle POQ is  \theta  radians and the perimeter of the sector is 20 cm.

i) Show that   \theta \quad =\quad \frac { 20 }{ r } \quad -\quad 2

ii) Find the area of the sector in terms of r.

Solution:

i) Perimeter = 20

OP + OQ + PQ =  20

r\quad +\quad r\quad +\quad r\theta \quad =\quad 20

2r\quad +\quad r\theta \quad =\quad 20

\theta \quad =\quad \frac { 20 }{ r } \quad -\quad \frac { 2r }{ r }

Hence shown:

\theta \quad =\quad \frac { 20 }{ r } \quad -\quad 2

ii) Area of the sector OPQ\quad =\quad \frac { 1 }{ 2 } { r }^{ 2 }\theta

Substituting  \theta   from part (i) above

Area\quad =\quad \frac { 1 }{ 2 } { r }^{ 2 }\frac { 20 }{ r } \quad -\quad 2

Reference
  1. Edexcel AS and A level Modular Mathematics C2