How to calculate this indeterminate trigonometric problem? I have made all the relationships that have occurred to me, but I feel that I am missing something, or the problem is very simple and I went off on a tangent.
The drawing is what emerges from a hill, I managed to calculate some things but I did not get to the specifics of calculating $d$. The problem is to find $d$ in the problem below.

I managed to calculate some variables that I put myself.
$$y=30\sqrt{3}$$
$$x=\frac{30}{\tan 25} -30\sqrt{3}$$
$$d_1 =30\sqrt{\frac{1}{\tan^2 25} +1}$$
I got these results by applying ratios of $\tan 25$ and $\tan 30$
Can someone help me? Thanks in advance
Edit
Here is the original problem:


Step 1: Find an expression for $d$ in terms of $H$
We need H in order to solve for d. Therefore, we can first consider the largest triangle with hypotenuse $d+d_1$.
We see that $\sin(25)$ = $H$ / $(d+d_1)$ for the larger triangle.
Step 2: Find another expression for $d$ in terms of $H$
And for the smaller triangle with hypotenuse equal to $d$, we can again perform some simple trigonometry to see that $\sin(25)$ = $(H-30)$ / $d$. Note that this is equivalent to $H$ = $d\sin(25)$ $+$ $30$
Step 3: Solve the equations simultaneously to find $d_1$
Therefore, we can now solve for $d$ by substitution.
This gives us $\sin(25)$ = $(d\sin(25)$ $+$ $30)$ / $(d+d_1)$
We rearrange for $d_1$ to give us:
$d_1$ = $30/\sin(25)$
Step 4: Final Observations
There are infinitely many possible values of $d$ using the information that you have provided. The length $Z$ is arbitrary and you could change it to whatever you wanted and this would produce a new value for $d$.
You can see this for yourself if you let $Z=1$, $Z=2$, $Z=3$, etc. The diagram will still make sense in all of these situations, however, the value of $d$ will be different for all of them.
Therefore, we can spot immediately that we need more information so that we can find $Z$. Otherwise, what I have done so far is the best that we can hope to achieve for this question.
Perhaps if you posted the full question, that would make it easier?
