Question 842397
{{{drawing(600,300,-45,45,-7,38,
line(-45,0,45,0),locate(7,37.8,red(sunbeam)),
green(triangle(-41.687,0,41.687,0,0,5.119)),
rectangle(0,0,-1,1),
locate(-42,0,A),locate(-0.4,0,B),
locate(0.3,7.2,C),locate(0.3,32.6,D),
triangle(-41.687,0,0,32.569,0,0),
line(0,32.569,0,5.119),
red(line(-41.687,0,7,38.038)),
red(arc(-41.687,0,30,30,-38,0)),locate(-27.5,8,red(38^o)),
green(arc(-41.687,0,70,70,-7,0)),locate(-6.6,4,green(7^o)),
locate(-21,5,green(42ft)),locate(0.3,4,y),
locate(-21,-2,x),locate(0.3,20,h),
arrow(-19,-3,0,-3),arrow(-22,-3,-41.687,-3)
)}}} Segment CD represents the tree. The green triangle represents the mound.
{{{h}}}= height of the tree.
AC ={{{42ft}}}
ABC and ABD are right triangles.
{{{x=(42ft)*cos(7^o) =approximately}}}{{{41.687ft}}}
{{{y=(42ft)*sin(7^o) =approximately}}}{{{5.1185ft}}}
{{{y+h=x*tan(38^o) =approximately}}}{{{0.781286}}}
{{{h=x*tan(38^o)-y}}}
Substituting the approximate values above,
{{{h=(41.687ft)*0.781286-5.1185ft=approximately}}}{{{highlight(27.451ft)}}}
Working further with formulas instead,
{{{h=(42ft)*cos(7^o)*tan(38^o)-(42ft)*sin(7^o) =approximately}}}{{{highlight(27.451ft)}}}