# trigonometry

posted by on .

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 - ,

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) / (1-6.115/x)

130/x = (105 + .05824 x) / (x - 6.115)

130 (x-6.115) = 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 - ,

Unless I am totally mis-reading 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 - ,

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 - ,

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 - ,

Dang - you are correct, Damon. Sorry, Reiny.

• trigonometry - ,

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".