Help with understanding a trigo problem

Two buildings face each other on opposite sides of a city square. The angle of depression from the top of the taller building to the base of the shorter is 38°. The angle of depression from the top of the shorter building to the base of the taller is 23°. If one building is 30m higher than the other, how wide is the square?

I sketched it and since the taller building is 30m higher than the shorter one, the taller building is h+30 and the short one is h, therefore h=h and I labelled the base as x.So the angle depression of the taller one is 38° from the base of the shorter one so I know that the taller building is tan38°=h+30/x and the shorter one is tan23°=h/x. Please help me, I can't solve neither the base or the height because they are both unknown.

You're on the right track. We just need to solve for h and substitute. Since tan(23)=h/x, we can multiply both sides by x and we get: xtan(23)=h

Now, we can plug that into the h in tan(38)=h+30/x, so:
tan(38)=xtan(23)+30/x

We can multiply both sides by x, and we get:

xtan(38)=xtan(23)+30

Now, take that xtan(23) to the left side:
xtan(38)-xtan(23)=30

Factor out an x:
x[tan(38)-tan(23)]=30

Divide both sides by [tan(38)-tan(23)]:
x=30/[tan(38)-tan(23)]

x is approx 84.1m

tan 38= (h+30)/x or x= (h+30)/tan38

tan 23= h/x
or
x= h/tan23

let
(h+30)/tan38= h/tan23
solve for h