# Measurement of Angles

The measure of an angle is the amount of rotations required to get to the terminal side from the initial side. A common measure of an angle is derived by placing its vertex at the center of a circle of some fixed radius.

There are two commonly used measurements for angles: degrees and radians

Sexagesimal System (Degree, Minute and Second)

If the initial ray $\overrightarrow {OA}$ rotates in an anti-clockwise direction in such a way that it coincides with itself, the angle then formed is said to be of $360$ degrees ($360^\circ$).

One rotation (anti-clockwise) = $360^\circ$

$\frac{1}{2}$ rotation (anti-clockwise) = $180^\circ$ is called a straight angle
$\frac{1}{4}$ rotation (anti-clockwise) = $90^\circ$ is called a right angle.

1 degree ($1^\circ$) is divided into $60$ minutes ($60’$) and 1 minute ($1’$) is divided into $60$ seconds ($60”$). As this system of measurement of angles originates from English and because$90$, $60$ and multiplies of $6$ and $10$, it is known as the English System or Sexagesimal System.

Thus $1$ rotation (anti-clockwise)               =          $360^\circ$
One degree ($1^\circ$)                                =          $60’$
One minute ($1’$)                                        =          $60”$

Conversion from $D^\circ M’S”$ to a decimal form and vice versa.

(i)         $16^\circ \,30’$ =  $16.5^\circ$ ${\text{As}}\,\left( {30′ = \frac{{1^\circ }}{2} = 0.5^\circ } \right)$
(ii)        $42.25^\circ$ =   $45^\circ \,15’$ $\left( {0.25^\circ = \frac{{25^\circ }}{{100}} = \frac{{1^\circ }}{4} = 15′} \right)$

Example:

Convert $18^\circ \,\,6’\,\,21”$ to decimal form.

Solution:

$1′ = {\left( {\frac{1}{{60}}} \right)^\circ }$ and $1” = {\left( {\frac{1}{{60}}} \right)^\prime } = {\left( {\frac{1}{{60 \times 60}}} \right)^\circ }$

$\begin{gathered} \therefore 18^\circ \,\,6’\,\,21” = {\left[ {18 + 6\left( {\frac{1}{{60}}} \right) + 21\left( {\frac{1}{{60 \times 60}}} \right)} \right]^\circ } \\ \therefore 18^\circ \,\,6’\,\,21” = \left( {18 + 0.1 + 0.005833} \right)^\circ = 18.105833^\circ \\ \end{gathered}$
Example:

Convert $21.256^\circ$ to the $D^\circ M’S”$ form

Solution:

$\begin{gathered} 0.256^\circ = \left( {0.256} \right)\left( {1^\circ } \right) \\ 0.256^\circ = \left( {0.256} \right)\left( {60′} \right) = 15.36′ \\ \end{gathered}$

and

$\begin{gathered} 0.36′ = \left( {0.36} \right)\left( {1′} \right) \\ 0.36′ = \left( {0.36} \right)\left( {60”} \right) = 21.6” \\ \end{gathered}$

Therefore,

$\begin{gathered} 21.256^\circ = 21^\circ + 0.256^\circ \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 21^\circ + 15.36′ \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 21^\circ + 15′ + 0.36′ \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 21^\circ + 15′ + 21.6” = 21^\circ \,\,15’\,\,21”\,rounded{\text{ }}off{\text{ }}to{\text{ }}nearest{\text{ }}second \\ \end{gathered}$

Thus $\angle AOB = 1{\text{ radian}}$