Angled mountain

Consider the mountain as $AB$ , and the observer before and after ascending as $C$ and $D$ respectively. Let the perpendicular from $D$ to $BC$ and from $D$ to $AB$ be $E$ and $F$ respectively. Now, $\angle ACB=45$ . Thus, $\sin \angle ACB = \frac{AB}{BC} = 1$ . So, $AB=BC=x$ .

Now, consider $\triangle DEC$ . $\cos \angle DCE = \frac{CE}{CD} = \frac{\sqrt{3}}{2}$

Thus, $CE$ $=$ $500\sqrt{3}$ , and $EB=DF=x-500\sqrt{3}$

Now, in $\triangle ADF$ , $\sin \angle ADF = \frac{AF}{DF} = \frac{\sqrt{3}}{2}$

Thus, $AF = \sqrt{3}(x-500\sqrt{3})$ and thus, $x=\sqrt{3}x -1500 +500$

Solving, $x$ $=$ $\frac{1000}{\sqrt{3}-1}$

Rounding off, $x = \boxed{137}$ decameters.