Is it possible to get the distance between two people when they both don't make a straight line? The problem is as follows:

Richard is resting below the shadow of a tree situated at $20$ meters
  in the direction $N53^{\circ}E$ from his home and his friend Mark is
  situated at $5\,m$ from the direction $S37^{\circ}W$ of Richard's
  house. Mark moves formard with a constant speed of $5\frac{m}{min}$
  in the direction $N37^{\circ}W$. Find the distance between them after
  $4$ minutes have elapsed.

The alternatives in my book are as follows:
$\begin{array}{ll}
1.&3\sqrt{41}\,m\\
2.&41\,m\\
3.&\sqrt{689}\,m\\
4.&31\,m\\
\end{array}$
I tried all sorts of arrangements to get this correctly but I'm still confused if it can actually be solved. Can someone help me with the bearing angles?.
So far what I was able to spot is summarized in my drawing from below. But there's a triangle which I cannot relate to obtain the requested length.

So can someone guide me in the right path for this problem?. I'd appreciate that an answer must contain a drawing to see exactly how can I make a side for that triangle or a way to find that distance.
 A: If a triangle has sides of length $3,4,5$ (or anything in proportion to $3,4,5,$ for example $12,16,20$), it is a right triangle with angles of approximately $37$ degrees and $53$ degrees at the other two vertices.
So I assume when I come across an exercise like this that we are supposed to assume that any right triangle with an angle of $37$ or $53$ degrees is supposed to have sides in the ratio $3,4,5.$ Therefore if the hypotenuse is $20$ then the other two sides are $12$ and $16.$
Another thing about a problem like this is that it works well in Cartesian coordinates. You can put the origin at any convenient point and work out the coordinates of other points.
Using both of these ideas, I took your diagram, gave the coordinates $(0,0)$ to the tree, and used the line between each pair of objects as the hypotenuse of a right triangle. The other two sides of the right triangle run either east-west or north-south.

So the red triangle has hypotenuse $20$; it has a $53$ degree angle so its sides are in proportion $3,4,5$; that means one side is $\frac35$ times as long as the hypotenuse and the other is $\frac45$ times as long.
Moreover the longer side is the one opposite the $53$ degree angle.
So you can mark the length of each side of the triangle and from that you can determine the coordinates $(x_1,y_1)$ at Richard's house.
Then take similar steps to find the sides of the dark blue triangle and the coordinates $(x_2,y_2)$ at Mark's starting spot.
Then do the green triangle to find the coordinates $(x_3,y_3)$ where Mark is after $4$ minutes.
Then $(x_3,y_3)$ and $(0,0)$ are the coordinates at each end of the hypotenuse of the light blue triangle; this is the length you want to compute.
So we just use the formula for distance between points with Cartesian coordinates, which is really nothing more than applying the Pythagorean Theorem to the light blue triangle:
$$ \sqrt{(x_3 - 0)^2 - (y_3 - 0)^2} .$$
You get a bit lucky with this particular exercise, however, because the green triangle is actually much larger than you have drawn it and the proportions are a little different. So when you actually work out what $x_3$ and $y_3$ should be, the answer is even simpler than it looks in the diagram.
A: I'm sure there's a more elegant approach to this... I notice that the two 20 m distances are perpendicular to each other, for example, but here's one way to get the answer:
From Mark's spot after 20 minutes to Mark's original position,
horizontal displacement $= 20 \sin 37^\circ = 20 \cos 53^\circ$
vertical displacement $= -20 \cos 37^\circ = -20 \sin 53^\circ$
From Mark's original position to Richard's house,
horizontal displacement $= 5 \sin 37^\circ = 5 \cos 53^\circ$
vertical displacement $= 5 \cos 37^\circ = 5 \sin 53^\circ$
From Richard's house to the tree,
horizontal displacement $= 20 \sin 53^\circ$
vertical displacement $= 20 \cos 53^\circ$
Overall horizontal displacement is $ 25 \cos 53^\circ + 20 \sin 53^\circ$
Overall vertical displacement is $ -15 \sin 53^\circ + 20 \cos 53^\circ$
Distance between $=\sqrt {\left( 25 \cos 53^\circ + 20 \sin 53^\circ \right)^2+\left( -15 \sin 53^\circ + 20 \cos 53^\circ \right)^2} \approx 31$
