Help me bf2s, I can't remember Geometry for shit
Last edited by haffeysucks (2008-08-22 10:04:15)
"people in ny have a general idea of how to drive. one of the pedals goes forward the other one prevents you from dying"
pie r squared divided by the tanjent of angle xyz + 9 divide by 5
stick to englishSamTheMan:D wrote:
tanjent
"people in ny have a general idea of how to drive. one of the pedals goes forward the other one prevents you from dying"
tangent
lol'd
lol'd
c2 = a2 + b2 - 2abcos(theta)
edit: where c is the side opposite theta, and a and b are the other two sides
edit: where c is the side opposite theta, and a and b are the other two sides
Spoiler (highlight to read):
yes i know it's been done before
'dPoseidon wrote:
http://img.photobucket.com/albums/v702/ … /maths.jpg
Spoiler (highlight to read):
yes i know it's been done before
lol I already forgot the Cosine Law >.<
Coswhat?Ryan wrote:
lol I already forgot the Cosine Law >.<
(A*A) + (B*B) = (C*C)
Then find the square root of (C*C)
Can't be arsed to find the shortkeys, ill stick to a pencil
Then find the square root of (C*C)
Can't be arsed to find the shortkeys, ill stick to a pencil
Cosine Law!naightknifar wrote:
Coswhat?Ryan wrote:
lol I already forgot the Cosine Law >.<
That's Pythagoras's Theorem (sp?), not what he's looking forPhaytal wrote:
(A*A) + (B*B) = (C*C)
Then find the square root of (C*C)
He'll get the answer, but it won't be the right way to do it. Question asks to solve X using the Law of Cosines.Sup3r_Dr4gon wrote:
Draw a line from to apex to the base so you have two right angled triangles, then just work out the size for each then add them.
Last edited by Ryan (2008-08-22 10:37:46)
Draw a line from to apex to the base so you have two right angled triangles, then just work out the size for each then add them.
Should be somewhere at the bottom.
That only applies for right-angle triangles.Phaytal wrote:
(A*A) + (B*B) = (C*C)
Then find the square root of (C*C)
Law of Cosines (as taken from my Pre-Calculus notes)
c2 = a2 + b2 - 2ab(cos γ)
*Note: γ is angle directly opposite c side. The formula can be changed to suit the a or b sides as well. Basically, just use the above as an example, where you use c and gamma together, a and alpha, and b and beta. Other than that, you should be able to make it work.
c2 = a2 + b2 - 2ab(cos γ)
*Note: γ is angle directly opposite c side. The formula can be changed to suit the a or b sides as well. Basically, just use the above as an example, where you use c and gamma together, a and alpha, and b and beta. Other than that, you should be able to make it work.
thats the one, I am going to put all my math problems on BF2s from now onJenspm wrote:
http://upload.wikimedia.org/math/d/0/c/ … e75528.png
Uhmm Math!
الشعب يريد اسقاط النظام
...show me the schematic
...show me the schematic
You have a book right? Or better yet, you're connected to the internet right? Look up the law of cosines!
I envy all you clever people
YES!mtb0minime wrote:
You have a book right? Or better yet, you're connected to the internet right? Look up the law of cosines!
lol, not to be a dick but come on man... it says right on the problem use law of cosines. Its one equation that gives the answer. If you couldn't remember law of cosines look that up!
I know which questiosn knocked my gcse maths down thenliquidat0r wrote:
That only applies for right-angle triangles.Phaytal wrote:
(A*A) + (B*B) = (C*C)
Then find the square root of (C*C)
I always remembered it ass
O A O
S H C H T A
O = Opposite, A = Adjacent, S/C/T = The angle or whatever, so you fill in the 2 you have and follow the triangle rules
O A O
S H C H T A
O = Opposite, A = Adjacent, S/C/T = The angle or whatever, so you fill in the 2 you have and follow the triangle rules
Can't remember Maths wut
Good for you.... but it doesn't help at all in this scenarioPhaytal wrote:
I always remembered it ass
O A O
S H C H T A
O = Opposite, A = Adjacent, S/C/T = The angle or whatever, so you fill in the 2 you have and follow the triangle rules