Is There a Mistake in Solving this Right Angle Triangle Problem?

[chemistry1]

The number of degrees in one acute angle of a right-angled triangle is equal to the number of grades in the other; express both the angles in degrees.

So I have found the following answers :

810/17=47,05... degrees and 810/17=47,05... grades which gives 42,35... degrees

Now, the real answer is the following :

900/19=47,3... degrees and 900/19=47,3... grades which gives 42,63... degrees

The only problem with my answer is the following :

810/17degrees=900/17grades

so : 900/17 grades+ 810/17 grades = 100,5.. grades (but for the rest, everything is fine, I get get 90 degrees perfectly and respect all conditions.)

Would you count this as an error ?

Here's what I did : (Help me see the error)

x degrees= (x+x/9) grades

x grades= (x-x/9) degrees

so..

x degrees+(x-x/9)degrees=90 degrees

17x/9 degrees=90 degrees

17x=810

x=810/17 degrees

By the formulaiton of the problem, we also have 810/17 grades

Conversion

810/17 degrees=(810/17+(810/17)/9) grades=900/17 grades

and

810/17 grades=(810/17-(810/17)/9) degrees= 720/17 degrees

Can somebody tell me where I went wrong ? (By the way, I see how to obtain the "real" answer, but I don't see why mine would be wrong ...)

Thank you !

 

[scurty]

Can you explain what the term "grades" means? I've never heard of that term and a Google search isn't showing any useful links.

 

[chemistry1]

Try this : http://en.wikipedia.org/wiki/Gradian

 

[scurty]

Here's what I did : (Help me see the error)

x degrees= (x+x/9) grades

x grades= (x-x/9) degrees

Thanks for the link! Took me a little bit, but if $1^{\circ} = \frac{10}{9}^g$, then $1^g = \frac{9}{10}^{\circ}$.

$1^{\circ} = (1 + \frac{1}{9})^g \implies x^{\circ} = (x + \frac{x}{9})^g$
$1^g = (1-\frac{1}{10})^{\circ} \implies x^g = (x-\frac{x}{10})^{\circ}$

Basically, the fraction is supposed to have a denominator of 10, not 9.

So you solve the equation $x + x - \frac{x}{10} = 90$

 

[chemistry1]

Thanks for the link! Took me a little bit, but if $1^{\circ} = \frac{10}{9}^g$, then $1^g = \frac{9}{10}^{\circ}$.

$1^{\circ} = (1 + \frac{1}{9})^g \implies x^{\circ} = (x + \frac{x}{9})^g$
$1^g = (1-\frac{1}{10})^{\circ} \implies x^g = (x-\frac{x}{10})^{\circ}$

Basically, the fraction is supposed to have a denominator of 10, not 9.

So you solve the equation $x + x - \frac{x}{10} = 90$

Oh,wow, I just noticed my error!(With the 10) Thank you ! Thats why I was getting a wrong answer !

 

[chemistry1]

Yes, it's working ! Those are the times when I feel really ashamed of myself for doing stupid errors like this one -_____- Thank you agian!

 

[scurty]

Yes, it's working ! Those are the times when I feel really ashamed of myself for doing stupid errors like this one -_____- Thank you agian!

We've all been there, I assure you!

 

[HallsofIvy]

I am used to that being called "grads", not "grades". (And "gradian" is too easily confused with "radian".) There are 100 grads in a right angle so that it measures "percentage slope".