I have trouble with this exercise...

[tzeen4]

Hello, everyone. I have this exercise in my book its the last on the trigonometry chapter so it's the hardest. I've spent 2 days on it already and I can't solve it.

So here it is:
At each corner of the square ABCD with side a=14 meters is standing a flag pole. The flag pole at corner A measures 7 meter, at B - 10, at C - 15. In the square's interior there is a point P which is located at equal distance from every flag-pole top.
a// find the location of this point P, therefore calculate its shortest distance from P to AD and from P to AB.
b// given the distace from P to flag pole D is the same distance as the one to A, B and C, find the flag pole height at corner D.

that is my drawing, havent noted the right angles there.
Thanks in advance.

 

[tzeen4]

Its not in the centre, because AA' BB' and CC' have different values and P is at equal distance to all of them, if it was it woudve been a piece of cake.
p.s. I have no idea, because I dont know how long to draw D

 

[Ishuda]

P = (x, y, 0)
A = (0,0,0)
B = (0,14,0)
C = (14,14,0)
D = (14,0,0)
A' = (0,0,7)
B' = (0,14,10)
C = (14,14,0)
D' = (14,0,u)

Write down the three equations fro P to A', P to B', and P to C'. Since they are the same, create two other equation by subtracting, for example, the "A'" equation from the "B'" equation and the "B'" equation from the "C'" equation. This will give you a linear system in x and y. Solve that and continue.

 

[HallsofIvy]

BUT this is what part b) states:
"b// given the distance from P to flag pole D is the same distance as the one to A, B and C,
find the flag pole height at corner D.
That CLEARLY says AP=BP=CP=DP ; so P MUST be at center of square.
Has nothing to do with AA', BB' and CC'.

NO. The problem says that P is equidistant from the flagpole tops. That is A'P= B'P= C'P= D'P, not AP, BP, CP, and DP.