trigonometry
posted by Cathy .
a flagpole 25 feet high stands on the top of a tower, which is 105 feet high. at what distance from the base of the tower will the flagpole subtend an angle of 3 degrees, 20 minutes?

trigonometry 
Damon
angle a is from tan a = 105/x
angle b = 3.33333 so tan b = 0.05824
tan (a+b) = 130/x
tan (a+b) = (tan a + tan b)/(1  tan a tan b)
130/x = (105/x + .05824) / (16.115/x)
130/x = (105 + .05824 x) / (x  6.115)
130 (x6.115) = 105 x + .05824 x^2
130 x  795 = 105 x + .0524 x^2
.0524 x^2  25 x + 795 = 0
x^2  477 x + 15172 = 0
x = [ 477 +/ sqrt (227529 60688)]/2
x = [ 477 +/ 408 ] / 2
x = 442 or 34.5 
trigonometry 
Reiny
Unless I am totally misreading the question .....
height of tower + flag pole = 130
distance from base of tower  x
tan 3°20' = 130/x
x = 130/tan3°20' = appr 2232 ft 
trigonometry 
Steve
When you get an answer, be sure to do a sanity check. Considering what a small angle 3°20' is, you'd expect the distance to be very large. Even 442 would make for an angle much greater than 3°20'.
Without even doing the math, Reiny's answer makes much more sense that Damon's. Sorry, Man. :(
Just a comment. 
trigonometry 
Damon
Ah, it did not ask for the angle between the base of the building and the top of the pole. It asked for the tiny angle subtended by the 25 foot pole alone.
To get a rough approximation do
fan 3.33 = 25/x
x = 430 ft 
trigonometry 
Steve
Dang  you are correct, Damon. Sorry, Reiny.

trigonometry 
Cathy
Thanks all of you for the help...
Damon's process is correct. However, he just failed to write the equations properly for the last 6 lines.
"0.0524" should be "0.05824".
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