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Maths trigonometry help

  1. Jun 14, 2003 #1
    P is 2km from Q and R lies on the perpendicular bisector at the link between P and Q and distance from P to R is 3 km.

    1. Find Length for Plan 1
    2. " " " Plan 2
    3. " " " Plan 3
    4. " " " Plan 4

    5. Decide which plan would be used if minimizing the cost.


    are the diagrams.

    i solved question 1 + 2.

    just need help from 3 - 5.

    if some people do not wish to look at the picture i will give a visual description in words.

    diagram 3 = is in a Y shape. the distance in the middle from P to Q is x. and the stroke down which is R, opposite sides of R is x. so alltogether there are 3 unknowns.

    diagram 4 = is in a Y shape, P to Q is equal which has sides b. the distance in the middle from P to Q is 120. opposites of R is 120.

    i suggest you can please look at the web link

    http://www.jeack.com.au/~dngo/maths.bmp thanks.
  2. jcsd
  3. Jun 15, 2003 #2


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    Staff Emeritus
    Science Advisor

    I looked at the diagrams and they are very clear. However, I still don't know what the "cost" is because you said nothing about what these diagrams are or how money comes into it!

    Are these "pipes" or "roads" between areas? If so what is the cost of building km of it?

    If the diagrams are of some sort of connection to be built connecting P, Q, and R, and the cost "per km" is the same everywhere, then you are really asking for the distances.

    Picture 1 shows a connection directly from Q to P (distance 2 km) and then a connection from P to R (distance 3 km) so the total distance is 5 km.

    Picture 1 shows a connection from Q to R and then from P to R. We are told that the distance from P to R is 3 km. Since R is on the perpendicular bisector of the segment from P to Q, by symmetry, the distance from P to R is also 3 km. Total distance: 6 km.

    Picture 3 has links from P,Q, and R to a central point. We are told that the 3 distances must be the same (x). This may not be the simplest way to do it, but I set it up on a coordinate system. Take the x-axis to be the line through P and Q. Since P is on the perpendicular bisector of the segment through P and Q we can take P on the y-axis, P as (-1,0) and Q as (1,0). Then, if call P (0,p), the distance from P to Q is given by sqrt(1+ p^2)= 3. Squaring both sides, 1+p^2= 3 or p^2= 2 so p= sqrt(2). The "central point" we are looking for in (3) must also (again by symmetry) be on the y axis. Call it S= (0,s). Now we have: distance from P to S is sqrt(1+s^2) and that must be the same as the distance from R to S: sqrt(2)- s (Since they are both on the y-axis, we just subtract the y coordinates. Strictly speaking we should take absolute value but it is clear from the picture that s is less than sqrt(2).) The two distances are the same: sqrt(1+ s^2)= sqrt(2)-s. Squaring both sides
    1+ s^2= 2- 2sqrt(2)s+ s^2 or 2sqrt(2)s= 1 so s= 1/(2sqrt(2)). The total distance is 3s= 3/(2sqrt(2)).

    In (4) you have either misunderstood your own picture or miswritten your description. "120" is obviously not a "distance": it is the angle in degrees between each segment. (4) is the same as (3) except that instead of requiring each segment be of the same length we are requiring that each leg be the same angle from the others.
    Again, call the central point "S". Setting up a coordinate system as before we can see (again, from symmetry) that S is on the y-axis: S= (0,s). The angle between the segment from P to S and the segment from R to S is 120 degrees so the angle between the line from P to S and the horizontal is 120- 90= 30 degrees. Since PQ and that horizontal are parallel the angle between PQ and PR is also 30 degrees (alternating interior angles). The distance from P to the point bisecting PQ is 1 so s/1= tan 30= 1/sqrt(3). S is the point
    (0, 1/sqrt(3)). From that it is easy to calculate that PS= sqrt(1+ 1/3)= 2/sqrt(3). This is, by symmetry, the same as QS. Since R is still the point (0,sqrt(2)), the distance from R to S, RS, is
    sqrt(2)- 1/sqrt(3)= (sqrt(6)-1)/sqrt(3). The total distance, then, is
    2/sqrt(3)+ 2/sqrt(3)+ (sqrt(6)-1)/sqrt(3)= (3+ sqrt(6))/sqrt(3).

    (5) asks for the "plan" which will minimize cost. Assuming that the cost of each km is the same (If different areas require different costs, that would change everyhing!) then the "minimum cost" is the same as minimum distance. The distances are:
    (1) 5 km.
    (2) 6 km.
    (3) 3/(2sqrt(2)) km.
    (4) (3+ sqrt(6))/sqrt(3) km.

    I'll let you calculate those yourself and see which is smallest.
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