Question 843524
Here is a sketch illustrating the problem:
{{{drawing(300,200,-5.5,0.5,-0.7,3.3,
line(-5.5,0,0.5,0),line(0,0,0,3),
blue(line(-5,0,0,3)),blue(line(-3,0,0,3)),
rectangle(-0.2,0.2,0,0),locate(-5.02,0.3,P),
blue(arc(-5,0,2,2,-31,0)),locate(-4.4,0.5,7^o),
blue(arc(-3,0,2,2,-45,0)),locate(-2.7,0.5,9.1^o),
locate(-4.22,-0.25,30m),locate(-1.55,-0.25,d),
arrow(-3,-0.2,-5,-0.2),arrow(-5,-0.2,-3,-0.2),
arrow(-1,-0.2,-3,-0.2),arrow(-2,-0.2,0,-0.2),
locate(-0.18,1.6,h)
)}}} To show those right triangles clearly, I had to draw the angles much wider than {{{7^o}}} and {{{9.1^o}}} .
The height of the tree is {{{h}}} .
After walking {{{30m}}} towards the tree, at a distance {{{d}}} from the base of the tree, the angle of elevation becomes {{{9.1^o}}} and
{{{tan(9.1^o)=h/d}}}<--->{{{d/h=1/tan(9.1^o)}}}
From point P, at a distance {{{d+30m}}}  from the base of the tree, the angle of elevation was {{{9.1^o}}} and
{{{tan(7^o)=h/(d+30m)}}}<--->{{{(d+30m)/h=1/tan(7^o)}}}
{{{(d+30m)/h-d/h=1/tan(7^o)-1/tan(9.1^o)}}}
{{{(d+30m-d)/h=1/tan(7^o)-1/tan(9.1^o)}}}
{{{30m/h=1/tan(7^o)-1/tan(9.1^o)}}}
From now on we calculate approximate values, rounding as needed.
{{{30m/h=1/0.122785-1/0.160174}}}
{{{30m/h=8.1443-1/6.2432}}}
{{{30m/h=1.9011}}}
{{{30m/1.9011=h}}}
{{{h=highlight(15.78m)}}}