Why does 90 degrees equal arccot(300/1000)?

[matthewmystar]

Hey everyone,

I have recently developed a strong interest in physics. To the point where I am changing my major from computer science to computational physics. Anyway, I am having a problem understanding a formula in a book I have on relativity. The problem is:
90$$\circ$$ = arcsec(300/1000). Now I know some Trig. but I haven't done it in awhile and don't remember arcsec at all. Can anyone explain to me why that problem equals 90 degrees.

Thanks,
Daniel

 

[Math Jeans]

secx is equal to 1/cosx. arcsecx would then be equal to the angle that causes sec to equal x.

However, because cosx is always <= 1, then 1/cosx cannot be less than 1. Therefore, if 300/1000=3/10, arcsec(300/1000) shouldn't exist.

 

[matthewmystar]

secx is equal to 1/cosx. arcsecx would then be equal to the angle that causes sec to equal x.

However, because cosx is always <= 1, then 1/cosx cannot be less than 1. Therefore, if 300/1000=3/10, arcsec(300/1000) shouldn't exist.

That is was the conclusion I was coming up with also. Maybe I am misunderstanding the problem.
This is the exact problem from the book:http://books.google.com/books?id=fz...q9xmXR&sig=ZM6PZ9v6XRaqWad6pn9UxlufThQ&hl=en"

 

[bassplayer142]

I think you are confusing trig arcsecant with arcseconds...
http://en.wikipedia.org/wiki/Arcsecond

 

[dst]

He is not.

The actual angle it says is 73.3 degrees. Which is correct - see here: http://www.analyzemath.com/Calculators_2/arccos_calculator.html (evaluates to 73.5 degrees for x=0.3)

 

[Phlogistonian]

The formula should say arccotangent, not arcsecant.