Convert Angles from Sexagecimal to Centesimal System

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To convert angles from the sexagecimal system to the centesimal system, first convert minutes and seconds into a decimal fraction of a degree. Then, convert the total degree measurement into grads, where one grad equals a right angle. The conversion process involves adjusting the decimal places to represent minutes and seconds in grad measure. A potential typo was noted in the conversion example, where the last part should indicate seconds instead of minutes. The use of grads is less common today, primarily having historical significance in certain regions.
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How do you convert angles from the sexagecimal system to centesimal one ?

For example

63 degrees 14 minutes 51 seconds reduced to centesimal ??

Here's how it's done but I don't understand the steps...

http://postimg.org/image/4tat51y63/

Thank you
 
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chemistry1 said:
How do you convert angles from the sexagecimal system to centesimal one ?

For example

63 degrees 14 minutes 51 seconds reduced to centesimal ??

Here's how it's done but I don't understand the steps...

http://postimg.org/image/4tat51y63/

Thank you

What they're doing is converting the minutes and seconds to a decimal fraction of a degree, and then converting the 63.xxyy° to grads, where 1 grad = a right angle.

After the conversion to grads + fractional grads, they converted the first two decimal places to whatever is the equivalent of minutes in grad measure, and they converted the 3rd and 4th decimal places to whatever is the equivalent of seconds in grad measure.

It looks like they have a typo in the very last part, where they have 70g 275` = 70g 27.5` = 70g 27` 50`. That part at the end (50`) should be 50``, I believe.

Angle measure in degrees (°) and radians is very common, but angle measure in grads is rarely used, as far as I know. I believe it was used in Russia and/or the Soviet Union.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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