Question 1161720
{{{29^o}}}{{{"42'"=(29+42/60)^o=29.7^o}}} and {{{59^o}}}{{{"23'"=(59+23/60)^o=approximately59.38^o}}} (rounded)
 
{{{drawing(350,500,-4,31,-2,48,
green(rectangle(-5,0,400,31.2)),
triangle(0,0,27.9,0,27.9,47.1),
triangle(0,31.2,27.9,31.2,27.9,47.1),
rectangle(27.9,31.2,26.9,32.2),
rectangle(27.9,0,26.9,1),
rectangle(27.9,0,0,31.2),rectangle(0,0,-1,1),
locate(-1,0,A),locate(-1,31.2,B),
locate(28,48,C),locate(27.5,0,E),
locate(28,31.2,D),locate(0.2,19,31.2m),
locate(28,40,h),arc(0,0,20,20,-59.4,0),
locate(3,4,59.38^o),locate(4,34,29.7^o),
arc(0,31.2,20,20,-29.7,0),
locate(13.5,0,d),locate(18,31.2,green(horizontal)),
locate(16,2,green(horizontal))
)}}} Consider the right triangles ACE and BCD.
 
{{{CD=h}}} , {{{CE=h+31.2m}}} and {{{BC=d}}} , so {{{tan(59.38^o)=(h+31.2m)/d}}} and {{{tan(29.7^o)=h/d}}}
so
{{{tan(59.38^o)/tan(29.7^o)=(h+31.2m)/h}}} --> {{{1.6898/0.5704=1+31.2m/h}}} --> {{{2.96254=1+31.2m/h}}} --> {{{1.96254=31.2m/h}}} --> {{{h=31.2m/1.96254}}}
Rounding, we get {{{h=15.9m}}} and {{{h+31.2m=15.9m+31.2m=highlight(47.1m)}}}