Finding equidistant point using law of cosine/sine

Question

My work so far

I found the side lengths that I believe to be related to the problem but I do not know how to implement these to find the solution

Answer 1

Denote Dutch Vader’s spaceship by $B’’$ . Clearly , it is travelling along the line $A’E’’$ . Let the spaceship have flown $x$ miles , that is , $A’B’’=x$.

We have $A’E’= \frac{60000}{\tan{40}} , A’E’’=\frac{50000}{\tan{25}}$ .

In right $\triangle B’’E’B$ , $$B’’E’ = A’E’-x$$ In right $\triangle B’’E’’A$ , $$B’’E’’=A’E’’-x$$

We require the hypotenuses of the two right triangles to be equal . Therefore , by Pythagoras’ Theorem , we must have :-$$ \sqrt{\left(\frac{60000}{\tan{40}}-x\right)^2+60000^2}=\sqrt{\left(\frac{50000}{\tan 25}-x\right)^2+50000^2} $$ This equation yields $x \approx 73967.8$ miles .

We observe that this quantity is greater than $A’E’$ . This merely implies that Vader’s spaceship has flown past $E’$ .

The rest of the problem is trivial , and can be solved by appropriate substitutions .

Answer 2

Referring to the graph:

$\hspace{3cm}$

You need to find $DF$ (black or red). Make up the equations for black $DF$: $$\begin{cases}EF^2+BE^2=BF^2\\ (CE+EF)^2+AC^2=AF^2\end{cases} \Rightarrow \\\begin{cases}EF^2+60000^2=BF^2\\ CE^2+2CE\cdot EF+EF^2+50000^2=AF^2\end{cases} \Rightarrow \\ BF^2=AF^2 \Rightarrow \\ 60000^2=CE^2+2CE\cdot EF+50000^2 \Rightarrow \\ EF=\frac{60000^2-50000^2-CE^2}{2CE}=\frac{60000^2-50000^2-35720^2}{2\cdot 35720}\approx -2462,$$ which implies the point $F$ lies after the point $E$, i.e. the red one.

Hence: $$DF=DE+EF\approx 71505+2462=73967.$$