so, here it is
take a good look at the drawing
do you see how you can calculate distances from the viewer to every tree?
AA'/OA' = tan 18˚ => OA' = 138.46 m
BB'/OB' = tan 30˚ => OB' = 77,94 m
now, we apply pitagora to find the distance between trees
(viewer is behind this fence south of tree A wich means angle OA'B'=90˚)
A'B' sqr = OA' sqr + OB' sqr = 19171.17 + 6075 = 159 sqr
=> A'B' = 159 m
pfff... dunno, looks pretty fit for training to me
