Please find the attached diagram for a better understanding of the solution provided here.
In the attached diagram, OP is the height of the tree.
Rest of the diagram should be sufficiently self explanatory.
In [tex] \Delta OPH [/tex],
[tex] tan(30^0)=\frac{OP}{x} [/tex]
[tex] \therefore \frac{OP}{tan(30^0)}=x [/tex].........................Equation #1
Again, in [tex] \Delta OPL [/tex],
[tex] tan(20^0)=\frac{OP}{PH+HL}=\frac{OP}{x+50} [/tex]
[tex] \therefore \frac{OP}{tan(20^0)}-50=x [/tex] .........................Equation #2
Thus, equating the equations Equation #1 and Equation #2, we have:
[tex] \frac{OP}{tan(30^0)}=\frac{OP}{tan(20^0)}-50 [/tex]
Evaluating we get:
[tex] 1.732(OP)=2.747(OP)-50 [/tex]
[tex] 1.015(OP)=50 [/tex]
[tex] \therefore OP=\frac{50}{1.015}\approx49.26 ft [/tex]
Thus the height of the tree is approximately 49.26 feet