(Simple) Trigonometry problem

[eternal plover]

it wouldnt be a function, it would be an expression

id have to think about this, ill let you know if i think of anything

[steady bluff]

Oh sry, I thought it would be an function since my goal is to plot it in excel. So when a variable changes, the output changes too

[steady bluff]

@eternal plover Sorry lets make it easier: I want to know the distance between A and B:

[eternal plover]

You have to find the x and y difference between the two points, then use the distance formula

To find the x difference, you just have to add/subtract stuff

for instance, here you would want to do (325-125), which equals something

And to find the y difference you would do the same for the given up-down lengths

Although we are missing a very crucial piece of information

We can't figure out the y difference

if we could, we would use the distance fomula: $\sqrt{y^{2}+x^{2}}$

[primal temple]

TheLord26

[steady bluff]

It should be possible since according to the CAD software its fully defined

I know how to caculate this distance, but not the other one

If it makes it easier, this angle is 32,62 degrees

[eternal plover]

i saw your other post where you said EBD and ABC are congruent, if thats the case then it is pretty simple

[steady bluff]

Yes they are

[eternal plover]

You can solve for side BD (and hence BC), then you can solve for EB using law of cosines (and hence find AB).

,,BD=\sqrt{125^{2}+310^{2}}

[primal temple]

TheLord26

[eternal plover]

,,\angle EBD = 180-32.62

[steady bluff]

But AB and CB are not equal

[primal temple]

TheLord26

[eternal plover]

obviously.

EB and AB are, aren't they?

We are going to use properties of triangles to solve for it.

[steady bluff]

yes

Btw, AD is not 180 degrees

[eternal plover]

dam, okay nvm then