1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: General Trig Problem

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data
    Cell phone tower #1 has a broadcast radius of 50 miles. Tower #2 has a broadcast radius of 60 miles and is located at a bearing of N 64 W of tower #1. The towers are 100 miles apart

    the questions:
    A.How much area has coverage by both towers? (dont need the N 64 W)

    B.A certain teacher who shall remain nameless, wants to build a house slightly out of range of either tower, in a general southwesterly direction from Tower #1. If the teacher will be lobbying for a new tower to be added so s/he will have coverage, where should Tower #3 be located? Note Tower 3 will have a broadcast radius of only 30 miles and should be spaced evenly as possible from the two other towers. Give the location as both bearing and distance from each of the other towers

    hint: for part b) remember form geometry how the find the points that are equidistant from two given points? Look it up.

    Heres a pic of what the picture looks like on the paper.
    untitled5.jpg


    3. The attempt at a solution

    For part a do you use the sector area formula then subtract some stuff?
    Then for part b i honestly have no idea what to do. There are a few questions after it but I will be able to solve them after I learn how to do these ones.

    Any help is welcomed
    Thanks,
    Xhanger
     
  2. jcsd
  3. May 2, 2010 #2
    For part a) you add the area of circle 1 and the area of circle 2, then subtract the segments. There might be an easier way to do it, but I'd have to spend more time researching that...

    For part b) , you might recall that any point on the perpendicular bisector of the segment through the centers of the circles will be equidistant. So we want a point on the perp. bisector!
    It has to be at least 60 miles from tower 2 (and therefore from tower 1 also) since we require it to be out of range.

    Hope that helps!
     
  4. May 3, 2010 #3
    Wrong circles, any circles of equal radius will do; but the broadcast area circles are not and won't. The question is whether any location outside the overlap of coverage is not beyond the 30 miles. Perhaps the solution is the line of all equidistant points. Probably not.
     
  5. May 3, 2010 #4
    "Any circles" --??? I'm afraid I don't understand what you are talking about. There are three circles with distinct radii.

    "probably not" --??? Then what do you propose?
     
  6. May 4, 2010 #5
    There appears to be something critical missing from your problem that would tie the location of tower #3 down to one point, since it asks for range and bearing to this point.

    Also, as stated, the house is outside #1 and #2; but there is no such restriction on the location of #3. My "probably not" referred to the possibility that there would be some location on the line of equidistance ("spaced evenly as possible from the two towers") between the other two where #3's radius did not extend beyond the coverage of the other two; however, there isn't. So as I now see it #3 could be anywhere on that line.

    Perhaps, you've left something out?
     

    Attached Files:

  7. May 4, 2010 #6
    The way I read it (which could be wrong..), the circle of radius 30, centered at point #3, is to have NO overlap with either of the other circles.
    With this understanding, #3 cannot be "anywhere" on the line. If we go down that line, say 500 miles, we can meet the conditions that the tower be equally spaced from #1 and #2. But to find the closest... there is exactly one point. No more information is needed.

    p.s. the "house" has no significance in this problem, except to give validity (or some sense of realism) to the ACTUAL problem of locating tower #3
     
  8. May 4, 2010 #7
    I guess we assume the house is to the southwest because of the original diagram.

    I have drawn out what you said, and it looks fairly straightforward.
     

    Attached Files:

  9. May 4, 2010 #8
    That's awesome (!), and exactly what I had in mind. What software do you use?
     
  10. May 4, 2010 #9
    I just use Turbocad. It doesn't have anything to do with the math.

    If #3 isn't equidistant from the other two, it's a little messy, by the way.
     

    Attached Files:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook