# The sexagesimal system and its operations

The sexagesimal system is a system of numeration in which every unit is divided into 60 smaller units. In other words, the base used is 60.

This system is used to measure time and angles.

$$1h \rightarrow 60 \ \mbox{min} \rightarrow 60\cdot 60=3.600 \ \mbox{s}$$$$$1^\circ \rightarrow 60' \rightarrow 60\cdot 60=3.600''$$$

## Operations with sexagesimal numbers

### Sum

• Step 1

Write the numbers to be added as follows, and add them column by column:

$$\begin{eqnarray} & & 38^\circ \ 24' \ 55'' \\\\ &+ & \underline{40^\circ \ 49' \ 17''} \\\\ & & 78^\circ \ 73' \ 72'' \end{eqnarray}$$

• Step 2

If the sum of seconds is bigger than $$60$$, we must divide the result by $$60$$; the remainder will be the seconds and the quotient will be added to the minutes.

$$\dfrac{72}{60}=1+\dfrac{12}{60}$$

Namely, the remainder is a $$12$$ and the quotient $$1$$. Then, the result is written as:

$$78^\circ \ (73+1)' \ 12'' = 78^\circ \ 74' \ 12''$$

• Step 3

Repeat the same procedure for the minutes:

$$\dfrac{74}{60}=1+\dfrac{14}{60}$$

Then,

$$78^\circ \ 74' \ 12'' = 79^\circ \ 14' \ 12''$$

### Subtraction

• Step 1

Write the numbers one above the other, the hours above the hours (or the degrees above the degrees), the minutes above the minutes etc.

$$\begin{eqnarray} & & 52^\circ \ 23' \ 18'' \\\\ & - & \underline{43^\circ \ 49' \ 25''} \end{eqnarray}$$

If the subtraction of the seconds results in less than zero, add $$60''$$ to the seconds and $$1'$$ is reduced to minutes in the number on top,

$$52^\circ \ 23' \ 18'' = 52^\circ \ 22' \ 78''$$

$$\begin{eqnarray} & & 52^\circ \ 22' \ 78'' \\\\ & - & \underline{43^\circ \ 49' \ 25''} \\\\ & &\ \ \ \ \ \ \ \ \ \ \ \ 23'' \end{eqnarray}$$

• Step 2

Repeat the same procedure with the minutes.

$$\begin{eqnarray} & & 51^\circ \ 82' \ 78'' \\\\ &- & \underline{43^\circ \ 49' \ 25''} \\\\ & & \ \ 8^\circ \ 33' \ 23'' \end{eqnarray}$$

Note: We must always substract the smallest number from the biggest. If we are working with angles, it is possible for us to calculate a negative angle (the subtraction is done with a value greater than zero and then the sign is changed).

If we are using temporary measurements, it does not make much sense to obtain negative times. Nevertheless, resolving a problem in which a time reference is defined $$t=0$$, it is possible to obtain a negative time for a previous moment.

### Multiplication by a number

• Step 1

Multiply seconds, minutes and hours (or degrees) by the number: $$\begin{eqnarray} & & 51^\circ \ \ \ 82' \ \ \ 78'' \\\\ & \times & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \\\\ & & \overline{255^\circ \ 410' \ 390''} \end{eqnarray}$$

• Step 2

If more than $$60$$ seconds are obtained, divide by $$60$$ and the remainder will be the seconds and the quotient will be added to the minutes

$$\dfrac{390}{60}=6+\dfrac{30}{60}$$

$$255^\circ \ 410' \ 390'' = 255^\circ \ 416' \ 30''$$

• Step 3

Repeat the same step for the minutes,

$$\dfrac{416}{60}=\fbox{6}+\dfrac{\fbox{56}}{60}$$

$$(255+\fbox{6})^\circ \ \fbox{56}' \ 30'' = 261^\circ \ 56' \ 30''$$

### Division by a number

We want to divide $$37^\circ \ 48' \ 25''$$ by $$5$$

• Step 1

Divide the hours (or degrees) by the number: $$\dfrac{37}{5}=7+\dfrac{2}{5}$$

The quotient, $$7$$, is the hours and the remainder, multiplied by $$60$$, $$(2\times60)$$, will be added to the minutes.

• Step 2

The same procedure with the minutes,

$$48'+120'=168'$$

$$\dfrac{168}{5}=33+\dfrac{3}{5}$$

$$33$$ will be the final minutes, and the remainder, multiplied by $$60$$ will be added $$(3\times60)$$ to the seconds.

• Step 3

Finally, the same procedure with the seconds,

$$25''+180''=205''$$

$$\dfrac{205}{5}=41$$

Then, the final result is:

$$7^\circ \ 33' \ 41''$$

Note: The last division might have a non empty remainder. In such a case, the seconds would be expressed with decimals.