# How to get this formula using trip functions

• February 26th 2009, 01:48 PM
eric0784
How to get this formula using trip functions
Hi, i got this homework problem i just can't figure out. I'm was given the formula S= sqrt((LD^2)/(L+D)). I must find out how that formula was derived using the basic trig functions (sin,cos,tan, Pythagorean theory, etc...). We where given the diagram below to try and figure this problem out.

http://i39.tinypic.com/312d2l2.jpg

any help you guys can give me will be a great help.
• February 26th 2009, 02:02 PM
arpitagarwal82
Consider the fig below
in triangle ABC

$/sin x = (D/2) / (L + D/2)$ ------eq1

In triangle
CDE
$\tan x = (S/2)/L$ ------eq2

Now use $cosec^2 x = 1 + \cot^2 x$

Replace the value of cosec x from eq1 and cot x from eq2 to get the required value of S
• February 26th 2009, 04:03 PM
eric0784
thanks very much, that was a great help. the only thing i can't figure out now is how to get rid of the + 1 as i factor to get that formula I need. Is there something I am missing as I factor or does the + 1 go away with Cot^2x when i sub the equation for tanx in for it?
• February 26th 2009, 04:13 PM
arpitagarwal82
Quote:

Originally Posted by eric0784
thanks very much, that was a great help. the only thing i can't figure out now is how to get rid of the + 1 as i factor to get that formula I need. Is there something I am missing as I factor or does the + 1 go away with Cot^2x when i sub the equation for tanx in for it?

Continued from above solution
from eq1
$\sin x = D/(2L + D)$
so $\csc x = (2L + D)/D$
from eq2
$\tan x = S/2L$
so $\cot x = 2L/s$

So from $\csc^2 x = 1 + \cot^2 x$

${(2L + D)/D}^2 = 1 + 4L^2/S^2$

$=> (4L^2 + 4LD + D^2)/D^2 -1 = 4L^2/S^2$

=> $(4L^2 + 4LD)/D^2 = 4L^2/S^2$

=> $S^2 = 4L^2D^2/ (4L^2 + 4LD)$
$=> S^2 = LD^2/(L+D))$
• February 26th 2009, 05:04 PM
eric0784
arpitagarwal, i really do appreciate all your help with this. but i still don't understand how between these two steps:

http://www.mathhelpforum.com/math-he...6829cb5c-1.gif

=> http://www.mathhelpforum.com/math-he...7950a6cb-1.gif

how do you get rid of the D^2 inside the first set of the brackets and the -1 but besides that i was able to follow ya?
• February 26th 2009, 05:47 PM
arpitagarwal82
Quote:

Originally Posted by eric0784
arpitagarwal, i really do appreciate all your help with this. but i still don't understand how between these two steps:

http://www.mathhelpforum.com/math-he...6829cb5c-1.gif

=> http://www.mathhelpforum.com/math-he...7950a6cb-1.gif

how do you get rid of the D^2 inside the first set of the brackets and the -1 but besides that i was able to follow ya?

see.. how do you calculate $a/b -1$
its $(a-b)/b$ (taking LCM of denominator

same thing is done here

http://www.mathhelpforum.com/math-he...6829cb5c-1.gif

is same as
$((4L^2 + 4LD + D^2) - D^2 )/D^2$

= http://www.mathhelpforum.com/math-he...7950a6cb-1.gif
• February 26th 2009, 06:41 PM
eric0784
ah seems so obvious now lol. thank you very much