Pat needs to determine the height of a tree before cutting it down to be sure that it will not fall on a nearby fence. the angle of elevation of the tree from one position on a flat path from the tree is upper h equals 30 degrees commah=30°, and from a second position upper l equals 50 feetl=50 feet farther along this path it is upper b equals 20 degrees .b=20°. what is the height of the​ tree?

Respuesta :

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

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