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Tom Stormcrowe 02-12-10 08:01 PM

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.

AEO 02-12-10 08:06 PM

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)

Tom Stormcrowe 02-12-10 08:07 PM

AAAAH, Thanks! I knew I was on the right track. :p

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.

couch_incident 02-12-10 08:26 PM

Two parts couch, one part match. Sorry Tom, I killed my Trig brain cells years ago.

Couch

Tom Stormcrowe 02-12-10 08:35 PM

no worries, couch. ;) Mine are very rusty. :p

graham hull 02-12-10 08:47 PM


Originally Posted by Tom Stormcrowe (Post 10397122)
AAAAH, Thanks! I knew I was on the right track. :p

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

Tom Stormcrowe 02-12-10 08:53 PM

Gah, OK, thanks. :p


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