Sexagisimal and centisimal system

[Drain Brain]

please give me some tips on to start solving the ff:

1. Prove that the number of Sexagisimal minutes in any angle is to the number of Centesimal minutes in the same angle as 27:50.

2. Divide $44^{\circ}\,8^{'} $into two parts such that the number of Seagesimal seconds in one part may be equal to the number of Centesimal seconds in the other part.

 

[Evgeny.Makarov]

1. Prove that the number of Sexagisimal minutes in any angle is to the number of Centesimal minutes in the same angle as 27:50.

Unfortunately, this problem is ambiguous. This statement may mean that a certain angle is expressed as

$x^\circ\,y'\,z''$ using degrees and sexagesimal minutes and seconds

and as

$u^g\,v'\,w''$ using grades and centesimal minutes and seconds,

then $y'/v'=27/50$. But this is not what it means. It means that if an angle is expressed entirely in sexagesimal seconds (e.g., 1234 seconds) and entirely in centesimal seconds, the ratio of those numbers is $27:50$.

Fortunately, the second meaning is easier to prove. For example, a centimeter is 1/100 of a meter, so a certain length expressed in centimeters is 100 times (which is the inverse of 1/100) the same length expressed in meters. Similarly, it is sufficient to show that the ratio of sexagesimal second to centesimal second is $50:27$. This can be done by expressing both quantities as fractions of the right angle.

 

[Drain Brain]

I'm still confused about it. :(

Why did you change it to the ratio of Sexagesimal second to Centesimal second?

 

[Evgeny.Makarov]

Why did you change it to the ratio of Sexagesimal second to Centesimal second?

I gave an example of centimeters and meters.

For example, a centimeter is 1/100 of a meter, so a certain length expressed in centimeters is 100 times (which is the inverse of 1/100) the same length expressed in meters.

If one unit of measure is $n$ times larger than another unit of measure, then the number of times the first unit fits in a certain magnitude is $n$ times fewer than the number of times the second unit fits into the same magnitude. Since the first unit is larger, it fits fewer times. For example, a yard equals three feet, so if a field is 100 yards long, then it is 300 feet long. Larger units — fewer units and vice versa. More precisely, the ratio of the number of yards to the number of feet that fit into a certain length is inverse to the ratio of a yard to a foot. Thus, 100 (yards) / 300 (feet) = 1 / (yard / foot). If a unit increases $n$ times (e.g., 3 times increase from a foot to a yard), the measure of a magnitude decreases $n$ times (was: 300 feet, now: 100 yards). That's why
\begin{align*}
&\frac{\text{the number of times a sexagesimal minute fits in an angle}}
{\text{the number of times a centesimal minute fits in an angle}}\\
&\quad=\frac{\text{centesimal minute}}{\text{sexagesimal minute}}\\
&\quad=\frac{27}{50}\qquad\text{(you need to show this).}
\end{align*}

For the second problem, let's denote $n$ centesimal seconds by $n^{\prime\prime c}$ and $n$ sexagesimal seconds by $n^{\prime\prime s}$, and similarly for minutes. First you need to find the ratio of a centesimal second to a sexagesimal second. Suppose this ratio is $r$:
\[
\frac{1^{\prime\prime c}}{1^{\prime\prime s}}=r.
\]
Then $1^{\prime\prime c}=r\cdot 1^{\prime\prime s}$. According to the assumption, $n$ sexagesimal seconds plus $n$ centesimal seconds equal $44^\circ\,8^{\prime s}$. Convert the latter angle to sexagesimal seconds; suppose the result is $m^{\prime\prime s}$. Then
\[
n^{\prime\prime s}+n^{\prime\prime c}=
n^{\prime\prime s}+r\cdot n^{\prime\prime s}=
(1+r)n^{\prime\prime s}=m^{\prime\prime s}.
\]
Solving this equation for $n$ gives you the number of sexagesimal seconds in one part and the number of centesimal seconds in the other part.

 

[Drain Brain]

I gave an example of centimeters and meters.

$\begin{align*}
&\frac{\text{the number of times a sexagesimal minute fits in an angle}}
{\text{the number of times a centesimal minute fits in an angle}}\\
&\quad=\frac{\text{centesimal minute}}{\text{sexagesimal minute}}\\
&\quad=\frac{27}{50}\qquad\text{(you need to show this).}
\end{align*}$

do I need to use actual number here to prove that that is the case? How do I show it? please bear with me.

 

[Evgeny.Makarov]

do I need to use actual number here to prove that that is the case?

I am not sure I understand the question. Which number: $27/50$? If you don't use this number, then what are you proving?

How do I show it?

As I wrote in post #2,

This can be done by expressing both quantities as fractions of the right angle.

How many degrees are there in the right angle? How many sexagesimal minutes are in one degree? Therefore, how many minutes are there in the right angle? Then answer similar questions about grades and centesimal minutes.

 

[Drain Brain]

$90^{\circ}\cdot \frac{60^{'}}{1^{\circ}}=5400^{'}$

and

$100^{g}\cdot \frac{100^{'}}{1^g}=10,000^{'}$

then,

$\frac{5400}{10,000}=\frac{27}{50}$

Am I correct?can you provide another way of doing it. thanks!

 

[Evgeny.Makarov]

Yes, this is correct. It's important to understand, though, whether this is the ratio of a centesimal minute to a sexagesimal one or vice versa. Intuitively, a centesimal minute is smaller, so it's the former. More precisely, a centesimal minute is $1/10\,000$th of the right angle and a sexagesimal minute is $1/5\,400$th of the right angle, so the ratio of centesimal to sexagesimal is
\[
\frac{1/10\,000}{1/5\,400}=\frac{5\,400}{10\,000}=\frac{27}{50}.
\]