Question 921064
F = Foot of the hill, T = Top of the hill, M = point 500 meters up the hill
(The observer walks from F to M.)
Horizontal and vertical imaginary lines are blue.
Lines of sight to the top are green.
Here are my diagrams:
{{{drawing(300,300,-180,720,-120,780,
line(-180,0,0,0),locate(-10,-1,F),
blue(rectangle(508,250,538,280)),
red(arc(538,641,400,400,90,130)),locate(420,535,red(40^o)),
red(arc(538,641,260,260,90,105)),locate(510,590,red(15^o)),
blue(rectangle(508,0,538,30)),blue(line(538,0,538,641)),
blue(line(0,0,720,0)),line(0,0,433,250),
green(line(0,0,538,641)),blue(line(433,250,720,250)),
green(line(538,641,433,250)),locate(216,125,500),
locate(540,680,T),locate(435,245,M),
locate(60,70,30^o),red(arc(0,0,280,280,-30,0)),
red(arc(0,0,800,800,-50,0)),locate(400,150,50^o),
locate(560,380,75^o),red(arc(433,250,300,300,-75,0))
)}}} --> {{{drawing(300,300,-180,720,-120,780,
locate(-10,-1,F),
line(0,0,433,250),
green(line(0,0,538,641)),
green(line(538,641,433,250)),locate(216,125,500),
locate(540,680,T),locate(435,245,M),
locate(100,150,20^o),red(arc(0,0,260,260,-50,-30)),
locate(340,320,135^o),red(arc(433,250,200,200,150,285)),
locate(430,540,25^o),red(arc(538,641,260,260,105,130))
)}}} --> {{{drawing(300,300,-180,720,-120,780,
line(-180,0,0,0),locate(-10,-1,F),
blue(line(0,0,720,0)),blue(rectangle(508,0,538,30)),
green(line(0,0,538,641)),blue(line(538,0,538,641)),
locate(540,680,T),locate(545,350,blue(h)),
red(arc(0,0,260,260,-50,0)),locate(50,75,50^o)
)}}}
The first diagram allows us to figure out the angles for the second diagram.
The second diagram and the law of sines allow us to find the length of FT.
{{{FT/sin(135^o)=500/sin(25^o)}}}
{{{FT/0.7071=500/0.4226}}}
{{{FT=0.7071*500/0.4226}}}
{{{FT=836.6}}}
The third diagram and the length of FT allow us to find height {{{blue(h)}}} .
{{{h=FT*sin(50^o)}}}
{{{h=836.6*0.7660}}}
{{{highlight(h=641)}}} (rounded to 3 significant figures).
The height of the hill (rounded to 3 significant figures) is {{{highlight(641)}}} meters.