### 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

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