Math Geometry

Try drawing a line from the intersection of d and c going straight down perpendicular to b. That should get you started.

if it is of any help, ( I promise I'll learn LaTex):

d = [ Area/c*sin(n) ] + [ c*cos(n)/2)
Area is the area of the trapezoid

d = [ P - c*( 1 + sin(n) + cos(n) ) ] /2

P is the perimeter of the trapezoid

>>8552008
Yes, take Calc I dummy.
It's a related rates problem.

If C stays the same and N increases, then D increase.

If C increases, and N stays the same, and the right side of D is connected to the end of C, then D increases.

But I'd assume C stays constant so that D is a straight line. So all you should have to know is a greater N means a greater D, and a lower N means a lower D.

>>8553133
TeX is easy. If you want to learn it, just rightclick someone else's TeX to get what they put inside the [math ] brackets, and practice it with the TeX compiler on here. (the \left and \right is optional, but it makes the () scale better, and \displaystyle makes it so fractions aren't tiny as hell)

[math]\displaystyle d=\frac {\text{Area}} {c\sin{ \left( n \right)}} + \frac {c\cos{\left(n\right)}} {2} [/math]

>>8553258
how does it work
[math\] y=x^2 [math\]

>>8553307
[math/] y=x^2 [math/]

>>8553594
[math] y=x^2 [/math]

>>8553307
>>8553594
[mäth] [/mäth] (substitute a in for ä, obviously)
Also, should've mentioned. If you right click the TeX that I put in my post, you see what exactly I put by going to "show math as" -> "TeX commands." If, for some reason, that doesn't work, here's what I put between those math tags:
\displaystyle d=\frac {\text{Area}} {c\sin{ \left( n \right)}} + \frac {c\cos{\left(n\right)}} {2}
You can test your TeX at https://www.codecogs.com/latex/eqneditor.php
Just note that you don't have to have the $ and ^ or whatever symbols at the beginning or end, it's just the [mäth] tags.

Well, [math]
\frac{d}{b}=\frac{a}{c}
\frac{ab}{2}=Area
\frac{b}{c}=\cos(n)
\frac{a}{c}=\sin(n)
\frac{a}{b}=\tan(n)
d = \frac{ab}{c}
\righarrow b=c\cos(n), a = c\sin(n), b\tan(n) = a,
c^2\cos(n)\sin(n)/2
d = \frac{Area}{a}+\frac{b}{2}
Area = ab-\frac{(b-d)a}{2}
d = \frac{a}{2}+b, \rightarrow d = 2b

[/math]

I want to know If I made a mistake, and if yes where ?

fuck, I forgot the \\
here you go :
[math]
d = \frac{Area}{a}+\frac{b}{2}\\
Area = ab-\frac{(b-d)a}{2} \\
d = \frac{a}{2}+b \\
\Rightarrow 2b=d
[/math]

forget, I missed with the notation of the side..

>>8552008
is there a reason that d is where it is?