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