A Geocaching race organizer is putting four caches along a route that starts and finishes at the same place.
While making his route, he recorded a few route details. Starting from point \(A\), the organizer travelled due west to point \(B\). From point \(B\), he travelled \(1500\) m southeast to point \(C\). While at \(C\), he noted that \(\angle BCA\) is \(45\degree\). From point \(C\), he travelled \(1000\) m in a direction \(30\degree\) north from east to point \(D\). Finally from point \(D\), he travelled due north to point \(E\) and noted that \(\angle DEC=30\degree\).
Determine the distance, rounded to the nearest metre, from the cache at point \(E\) to the race finish at point \(A\).
Since he travelled west from \(A\) to \(B\) then southeast from \(B\) to \(C\), we know that \(\angle ABC = 45\degree\). We are given that \(\angle BCA = 45\degree\). Thus, \(\angle BAC = 90\degree\), and so \(\triangle ABC\) is isosceles with \(AB=AC\) and \(BC = 1500\).
In \(\triangle ABC\), we can use the sine ratio to solve for length \(AC\): \[\begin{aligned} \sin(45\degree) &= \frac{AC}{BC}\\ \frac{1}{\sqrt{2}}&=\frac{AC}{1500} \\ AC &= \frac{1500}{\sqrt{2}} = 750\sqrt{2} \end{aligned}\]
Extend \(ED\) to \(F\) so that \(\angle CFD = 90\degree\). In \(\triangle CDE\), we are given \(\angle DEC = 30\degree\). Since \(\angle CDF = 60\degree\), we have \(\angle CDE = 180\degree - 60\degree = 120\degree\), and therefore \(\angle DCE = 180\degree - 120\degree - 30\degree = 30\degree\).
We can now determine \(CE\) using the sine law: \[\begin{aligned} \frac{CE}{\sin(120\degree)} &= \frac{1000}{\sin(30\degree)} \\ CE &= \frac{1000 \sin(120\degree)}{\sin(30\degree)} \\ &= \frac{1000\left(\frac{\sqrt{3}}{2}\right)}{\frac{1}{2}}\\ &=1000\sqrt{3} \end{aligned}\]
Now, \(\angle ACE = 90\degree - (30\degree + 30\degree) = 30\degree\).
Finally, knowing the lengths \(AC\) and \(CE\), along with \(\angle ACE\), we can apply the cosine law to \(\triangle ACE\), to determine \(AE\): \[\begin{aligned} AE^2 &= AC^2 + CE^2 - 2(AC)(CE) \cos(\angle ACE) \\ &= (750\sqrt{2})^2 + (1000\sqrt{3})^2 - 2(750\sqrt{2})(1000\sqrt{3}) \cos(30\degree) \\ &=1\,125\,000 + 3\,000\,000 - 2(750\sqrt{2})(1000\sqrt{3})\left(\frac{\sqrt{3}}{2}\right)\\ &= 4\,125\,000 -(750\sqrt{2})(1000)(3)\\ &= 4\,125\,000 -2\,250\,000\sqrt{2} \end{aligned}\] Therefore the distance, from the final cache location to the finish line, is \(\sqrt{4\,125\,000 -2\,250\,000\sqrt{2}}\approx 971\) m.