Question 927333
Here is a sketch with the necessary imaginary horizontal and vertical lines drawn in green:
{{{drawing(400,400,-70,10,-10,70,
green(triangle(-200,0,0,0,0,200)),
green(triangle(-300,0,100,0,100,300)),
green(rectangle(-2,0,0,2)),
line(-60.59,0,10,17.86),line(0,15.33,0,58),
red(line(-60.59,0,0,58)),
arc(-60.59,0,15,15,-43.6,0),locate(-54.5,8,43.6^o),
arc(-60.59,0,30,30,-14.2,0),locate(-45.59,5,14.2^o),
locate(-61,0,P),locate(0,0,U),
locate(0.5,15.3,B),locate(0.5,60,T),
locate(0.5,37,h)
)}}} We need to calculate {{{h}}} , the height of tower {{{BT}}} .
We can use law of sines applied to triangle BPT:
{{{h/sin(BPT)=62.5m/sin(PTB)}}}
We just need to calculate the measure of those angles
{{{BPT=43.6^o-14.2^o=29.4^o}}}
and PTB is part of right triangle PTU, so
{{{PTB=90^o-43.6^o=46.4^o}}}
So, {{{h/sin(29.4^o)=62.5m/sin(46.4^o)}}}
Using approximate values,
{{{h/0.490904=62.5m/0.724172}}}
{{{h=0.490904*62.5m/0.724172}}}
{{{highlight(h=42.4m)}}} (rounded to one decimal place).