Problems from The sexagesimal system and its operations

Define two numbers $$a$$ and $$b$$ in sexagesimal system.

a) Compute the sum of both $$a+b$$

b) Compute the subtractions $$a-b$$ and $$b-a$$

c) Multiply $$a\cdot7$$

d) Divide $$\dfrac{b}{6}$$

See development and solution

Development:

$$a=28^\circ \ 36' \ 54''$$ and $$b=75^\circ \ 43' \ 12''$$

a) Step 1:

$$\begin{eqnarray} & & \ \ 28^\circ \ 36' \ 54'' \\\\ &+ & \underline{\ \ 75^\circ \ 43' \ 12''} \\\\ & & 103^\circ \ 79' \ 66'' \end{eqnarray}$$

Step 2:

$$\dfrac{66}{60}=1+\dfrac{6}{60}$$

We obtain,

$$103^\circ \ 80' \ 6''$$

Step 3:

Same procedure for the minutes,

$$\dfrac{80}{60}=1+\dfrac{20}{60}$$

And we obtain,

$$a+b=104^\circ \ 20' \ 6''$$

$$$\\\\$$$

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

$$\begin{eqnarray} & & 75^\circ \ 43' \ \fbox{12}'' \\\\ &- & \underline{28^\circ \ 36' \ \fbox{54}''} \end{eqnarray}$$

Step 1:

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

$$\begin{eqnarray} & & 75^\circ \ 42' \ \fbox{72}'' \\\\ &- & \underline{28^\circ \ 36' \ \fbox{54}''} \\\\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ 18'' \end{eqnarray}$$

Step 2:

Minutes and hours are subtracted

$$\begin{eqnarray} & & 75^\circ \ 42' \ 72'' \\\\ &-& \underline{28^\circ \ 36' \ 54''} \\\\ & & 47^\circ \ \ 6' \ 18'' \end{eqnarray}$$

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

$$-47^\circ \ \ 6' \ 18'' $$

$$$\\\\$$$

c) Step 1:

$$\begin{eqnarray} & & 28^\circ \ \ \ 36' \ \ \ 54'' \\\\ & \times & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \\\\ & & \overline{140^\circ \ 180' \ 270''} \end{eqnarray}$$

Step 2:

More than $$60$$ seconds are obtained,

$$\dfrac{270}{60}=4+\dfrac{30}{60}$$

And the product is

$$140^\circ \ 184' \ 30''$$

Step 3:

The same procedure for the minutes

$$\dfrac{184}{60}=3+\dfrac{4}{60}$$

Finally,

$$a\times5=143^\circ \ 4' \ 30''$$

$$$\\\\$$$

d) Step 1:

We start by dividing the hours (or degrees):

$$\dfrac{75}{6}=12+\dfrac{3}{6}$$

$$12$$ will be the final hours, and $$3\times60$$ will be added to the minutes.

Step 2:

The same with the minutes $$180' + 43' = 223'$$

$$\dfrac{223}{6}=37+\dfrac{1}{6}$$

$$37$$ will be the final minutes and $$1\times60$$ will be added to the seconds.

Step 3:

The same with the seconds $$60'' + 12'' =72''$$

$$\dfrac{72}{6}=12''$$

And so,

$$\dfrac{b}{6}=12^\circ \ 37' \ 12''$$

Solution:

$$a=28^\circ \ 36' \ 54''$$ and $$b=75^\circ \ 43' \ 12''$$

a) $$a+b=104^\circ \ 20' \ 6''$$

b) $$b-a=47^\circ \ \ 6' \ 18''$$, $$a-b=-47^\circ \ \ 6' \ 18''$$

c) $$a\times5=143^\circ \ 4' \ 30''$$

d) $$\dfrac{b}{6}=12^\circ \ 37' \ 12''$$

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