Problems from The sexagesimal system and its operations

Development:

$a={28}^{\circ }{36}^{\prime }{54}^{″}$ and $b={75}^{\circ }{43}^{\prime }{12}^{″}$

a) Step 1:

$\begin{array}{rcl}& & {28}^{\circ }{36}^{\prime }{54}^{″}\\ \\ & +& \underset{―}{{75}^{\circ }{43}^{\prime }{12}^{″}}\\ \\ & & {103}^{\circ }{79}^{\prime }{66}^{″}\end{array}$

Step 2:

${\displaystyle \frac{66}{60}}=1+{\displaystyle \frac{6}{60}}$

We obtain,

${103}^{\circ }{80}^{\prime }{6}^{″}$

Step 3:

Same procedure for the minutes,

${\displaystyle \frac{80}{60}}=1+{\displaystyle \frac{20}{60}}$

And we obtain,

$a+b={104}^{\circ }{20}^{\prime }{6}^{″}$

$$$$

b) $b-a$ is reduced first since $a$ is less than $b$,

$\begin{array}{rcl}& & {75}^{\circ }{43}^{\prime }{\boxed{\text{12}}}^{″}\\ \\ & -& \underset{―}{{28}^{\circ }{36}^{\prime }{\boxed{\text{54}}}^{″}}\end{array}$

Step 1:

We convert a minute into $60$ seconds to obtain a positive number of seconds after having subtracted.

$\begin{array}{rcl}& & {75}^{\circ }{42}^{\prime }{\boxed{\text{72}}}^{″}\\ \\ & -& \underset{―}{{28}^{\circ }{36}^{\prime }{\boxed{\text{54}}}^{″}}\\ \\ & & {18}^{″}\end{array}$

Step 2:

Minutes and hours are subtracted

$\begin{array}{rcl}& & {75}^{\circ }{42}^{\prime }{72}^{″}\\ \\ & -& \underset{―}{{28}^{\circ }{36}^{\prime }{54}^{″}}\\ \\ & & {47}^{\circ }{6}^{\prime }{18}^{″}\end{array}$

The subtraction $a-b$ will give a result of:

$-{47}^{\circ }{6}^{\prime }{18}^{″}$

$$$$

c) Step 1:

$\begin{array}{rcl}& & {28}^{\circ }{36}^{\prime }{54}^{″}\\ \\ & \times & 5\\ \\ & & \overline{{140}^{\circ }{180}^{\prime }{270}^{″}}\end{array}$

Step 2:

More than $60$ seconds are obtained,

${\displaystyle \frac{270}{60}}=4+{\displaystyle \frac{30}{60}}$

And the product is

${140}^{\circ }{184}^{\prime }{30}^{″}$

Step 3:

The same procedure for the minutes

${\displaystyle \frac{184}{60}}=3+{\displaystyle \frac{4}{60}}$

Finally,

$a\times 5={143}^{\circ }{4}^{\prime }{30}^{″}$

$$$$

d) Step 1:

We start by dividing the hours (or degrees):

${\displaystyle \frac{75}{6}}=12+{\displaystyle \frac{3}{6}}$

$12$ will be the final hours, and $3\times 60$ will be added to the minutes.

Step 2:

The same with the minutes ${180}^{\prime }+{43}^{\prime }={223}^{\prime }$

${\displaystyle \frac{223}{6}}=37+{\displaystyle \frac{1}{6}}$

$37$ will be the final minutes and $1\times 60$ will be added to the seconds.

Step 3:

The same with the seconds ${60}^{″}+{12}^{″}={72}^{″}$

${\displaystyle \frac{72}{6}}={12}^{″}$

And so,

${\displaystyle \frac{b}{6}}={12}^{\circ }{37}^{\prime }{12}^{″}$

Solution:

$a={28}^{\circ }{36}^{\prime }{54}^{″}$ and $b={75}^{\circ }{43}^{\prime }{12}^{″}$

a) $a+b={104}^{\circ }{20}^{\prime }{6}^{″}$

b) $b-a={47}^{\circ }{6}^{\prime }{18}^{″}$, $a-b=-{47}^{\circ }{6}^{\prime }{18}^{″}$

c) $a\times 5={143}^{\circ }{4}^{\prime }{30}^{″}$

d) ${\displaystyle \frac{b}{6}}={12}^{\circ }{37}^{\prime }{12}^{″}$

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