Trigonometry - Problems in 3 Dimensions

Question

So my question is that I've been given the following question, but some of it doesn't make sense ...

A person stands on a 15m diving board. On his left is his coach, who looks up at him from the ground at a 28 degree angle, while his parents are on his right looking up from the ground at a 25 degree angle. If the angle of the platform between his coach and his parents is 95 degrees, how far apart are they?

My question basically is that is the 95 made with the ground? Also, how is it relevant in finding the distance? I thought you could just solve how far the parents are from the board using tangent (15/tan25, 15/tan28), and do the same for the coach and then add the distance. But this seems deceptively simple.

Answer 1

The horizontal distance $c$ of the coach is $c = 15 / tan(25) = 32.168$

The horizontal distance $p$ of the parents is $p = 15 / tan(28) = 28.211$

The angle (in the xy plane) between coach and parents is $95$.

The square of the distance $d$ between coach and parents is given by the cosine rule

$d^2 = c^2 + p^2 − 2*c*p*cos(95)$

So

$d^2 = 32.168^2 + 28.211^2 - 2 * 32.168 * 8.211 * cos(95)$

$d^2 = 1876.68$

Which gives $d = 43.32$

Answer 2

Looking down on a plan view, you have a triangle with the three vertices as the coach, the diver and the parents. The angle at the diver is $95$ deg, the ground length c of the side from diver to coach is $\frac{15}{\tan(28)}$ and the ground length p of the side from diver to parents is $\frac{15}{\tan(25)}$. The ground length x of the side from coach to parents is therefore $x = \sqrt{(c^2 + p^2 - 2\cdot p\cdot c\cdot cos(95))}$.