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.

$$1h\to 60\text{min}\to 60\cdot 60=3.600\text{s}$$ $$1^{\circ }\to {60}^{\prime }\to 60\cdot 60={3.600}^{″}$$

Operations with sexagesimal numbers

Sum

• Step 1

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

$\begin{array}{rcl}& & {38}^{\circ }{24}^{\prime }{55}^{″}\\ \\ & +& \underset{―}{{40}^{\circ }{49}^{\prime }{17}^{″}}\\ \\ & & {78}^{\circ }{73}^{\prime }{72}^{″}\end{array}$

• 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.

${\displaystyle \frac{72}{60}}=1+{\displaystyle \frac{12}{60}}$

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

${78}^{\circ }(73+1{)}^{\prime }{12}^{″}={78}^{\circ }{74}^{\prime }{12}^{″}$

• Step 3

Repeat the same procedure for the minutes:

${\displaystyle \frac{74}{60}}=1+{\displaystyle \frac{14}{60}}$

Then,

${78}^{\circ }{74}^{\prime }{12}^{″}={79}^{\circ }{14}^{\prime }{12}^{″}$

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.

$\begin{array}{rcl}& & {52}^{\circ }{23}^{\prime }{18}^{″}\\ \\ & -& \underset{―}{{43}^{\circ }{49}^{\prime }{25}^{″}}\end{array}$

If the subtraction of the seconds results in less than zero, add ${60}^{″}$ to the seconds and $1^{\prime }$ is reduced to minutes in the number on top,

${52}^{\circ }{23}^{\prime }{18}^{″}={52}^{\circ }{22}^{\prime }{78}^{″}$

$\begin{array}{rcl}& & {52}^{\circ }{22}^{\prime }{78}^{″}\\ \\ & -& \underset{―}{{43}^{\circ }{49}^{\prime }{25}^{″}}\\ \\ & & {23}^{″}\end{array}$

• Step 2

Repeat the same procedure with the minutes.

$\begin{array}{rcl}& & {51}^{\circ }{82}^{\prime }{78}^{″}\\ \\ & -& \underset{―}{{43}^{\circ }{49}^{\prime }{25}^{″}}\\ \\ & & 8^{\circ }{33}^{\prime }{23}^{″}\end{array}$

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: $\begin{array}{rcl}& & {51}^{\circ }{82}^{\prime }{78}^{″}\\ \\ & \times & 5\\ \\ & & \overline{{255}^{\circ }{410}^{\prime }{390}^{″}}\end{array}$

• 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

${\displaystyle \frac{390}{60}}=6+{\displaystyle \frac{30}{60}}$

${255}^{\circ }{410}^{\prime }{390}^{″}={255}^{\circ }{416}^{\prime }{30}^{″}$

• Step 3

Repeat the same step for the minutes,

${\displaystyle \frac{416}{60}}=\boxed{\text{6}}+{\displaystyle \frac{\boxed{\text{56}}}{60}}$

$(255+\boxed{\text{6}}{)}^{\circ }{\boxed{\text{56}}}^{\prime }{30}^{″}={261}^{\circ }{56}^{\prime }{30}^{″}$

Division by a number

We want to divide ${37}^{\circ }{48}^{\prime }{25}^{″}$ by $5$

• Step 1

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

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

• Step 2

The same procedure with the minutes,

${48}^{\prime }+{120}^{\prime }={168}^{\prime }$

${\displaystyle \frac{168}{5}}=33+{\displaystyle \frac{3}{5}}$

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

• Step 3

Finally, the same procedure with the seconds,

${25}^{″}+{180}^{″}={205}^{″}$

${\displaystyle \frac{205}{5}}=41$

Then, the final result is:

$7^{\circ }{33}^{\prime }{41}^{″}$

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