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Old 02-12-10, 08:01 PM   #1
Tom Stormcrowe
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Trig Function Question

It's been years since I've worked any trig, and I was working an algebraic equation determining a distance between 2 points when the vectors are on a 90 degree tangent to each other from a common origin point (2 people walking, one at 2 mph and one at 4 mph). If it were a trig question, I'd have a known hypotenuse of 3 miles, and unknown values for a and b, other than b=2a.


I just plain can't remember beyond this point.
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Old 02-12-10, 08:06 PM   #2
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so we know hyp., which is 3 and b=2a

http://faculty.uoit.ca/kay/G10_AppWebPage/

Pythagorean theorem says: c^2=(a^2)+(b^2)
since b=2A and c=3, you can substitute it into the equation and write it as
9=(a^2)+(2a^2)
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Old 02-12-10, 08:07 PM   #3
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AAAAH, Thanks! I knew I was on the right track.

So, then a=SQRT3 and b=2(SQRT3), then. I ignored the negative result since I was using it for distance in the real world, by the way.
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Old 02-12-10, 08:26 PM   #4
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Two parts couch, one part match. Sorry Tom, I killed my Trig brain cells years ago.

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Old 02-12-10, 08:35 PM   #5
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no worries, couch. Mine are very rusty.
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Old 02-12-10, 08:47 PM   #6
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Quote:
Originally Posted by Tom Stormcrowe View Post
AAAAH, Thanks! I knew I was on the right track.

So, then a=SQRT3 and b=2(SQRT3), then. I ignored the negative result since I was using it for distance in the real world, by the way.
Tom, you didn't quite get there:=

9=a^2+(2a)^2
9=a^2+4a^2
9=5a^2
9/5=a^2
SQRT(9/5)=a

Enjoy
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Old 02-12-10, 08:53 PM   #7
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Gah, OK, thanks.
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