Thread: Finding the length of side in a 3d diagram with a few data only

Hello guys,
Here is the question:

"The following three-dimensional diagram shows the four point A, B, C and D
A, B and C are in the same horizontal plane and AD is vertical.
Angle ABC is 45°
BC = 50m
Angle ABD = 30°
and the last on is that Angle ACD is 20°

Using the cosine rule in the trianlge ABC, or otherwise, find AD!"

So the point D is directly above A.
So you can imagine this as a pyramid with different long sides.
So one triangle with D above A as it says.

And now this is as far as I got:
It says that AD is vertical so that we can say that the angles CAD and BAD must be 90 °

Now we have the two triangles on the sides with only 2/3(we don´t need the third angle i think).

To make all algebra easier I named BA as c
and named CA as b
and AD is X what we are looking for

because we have now to rectangular triangles we can use tan twice.

tan 30 = x/c
and tan 20 = x/b

or in the end: c tan 30 = b tan 20

i looked back to the question and saw "use the cosine rule".
The only combination which would make sense is this:

b²=a²+c²-2ac cos (beta)
or
b²=50²+c²-100c cos 45°

We also could change: c tan 30 = b tan 20 to
b=(c tan 30)/tan20 and square it

b²= (c² x (tan 30)²)/(tan 20)²

now we could "eliminate b²) couldn´t we to find c so therefore we have:

(c² x (tan 30)²)/(tan 20)² = 50²+c²-100c cos 45°

I´ve tried several ways but I cant get to c in this equation because I always end up with a c² and a c but cant devide the c.
I hope that is right what I´ve done and that someone can help me please.

Thank you