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Posted by on Thursday, October 11, 2012 at 4:24am.

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 - , Thursday, October 11, 2012 at 5:27am

    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 - , Thursday, October 11, 2012 at 8:30am

    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 - , Thursday, October 11, 2012 at 11:10am

    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 - , Thursday, October 11, 2012 at 2:13pm

    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 - , Thursday, October 11, 2012 at 3:03pm

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

  • trigonometry - , Friday, October 12, 2012 at 4:41am

    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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