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jan.harrison

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Two observers are 200 feet apart, in line with a tree containing a bird's nest. The angles of elevation to the bird's nest are 30 degrees and 60 degrees. How far is each observer from the base of the tree?
 
jan.harrison said:
Two observers are 200 feet apart, in line with a tree containing a bird's nest. The angles of elevation to the bird's nest are 30 degrees and 60 degrees. How far is each observer from the base of the tree?
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Draw sketch of the situation and use the definition of angle of elevation and tan(?).

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
 
How would tan help me if I don't know the height of the tree? The base of one of the triangles would be x + 200 but I can't figure out where to go from there.
 
I don't see how tangent is going to help me when I don't know the height of the tree. I do know that the base of the triangle is x + 200 but do not know where to go from here.
 
jan.harrison said:
I don't see how tangent is going to help me when I don't know the height of the tree. I do know that the base of the triangle is x + 200 but do not know where to go from here.
Click to expand...

Did you draw a sketch - or just staring at the screen?

tan(60) = h/x

tan(30) = h/(x+200)

Two equations - two unknowns. You surely know how to solve those!!!!
 
jan.harrison said:
Two observers are 200 feet apart, in line with a tree containing a bird's nest. The angles of elevation to the bird's nest are 30 degrees and 60 degrees. How far is each observer from the base of the tree?
Click to expand...

Alternatively, the figure:
A = the birds nest
B = the location of the farthest observor
C = the location of the second oobservor
D = the bottom of tree below the nest
/_ACD = 60º
/_ ABC = 30º
/_BAC = 30º
/_CAD = 60º
BC = 200
AD = h
Let AC = s
Then, 200/sin30 = s/sin30
Therefore, s = 200
CD = 200sin30 = 100
Therefore, the two observors are 300 and 100 feet from the tree base.
 
Subhotosh Khan said:
tan(60) = h/x......................................(1)

tan(30) = h/(x+200) ..............................(2)

(1)/(2)

(?3)/[1/(?3)] = (x+200)/x

3 = 1 + 200/x

x = 100
Click to expand...
 
THere was such a simplier way. I realized after looking at a sketch of the triangles that one was isosceles with the hypotenuse measuring 200. I was able to use cosine to work the problem. The comment concerning the two unknowns would not have worked for my students because we've not discussed that yet. Thanks for your replies!!!!!
 
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