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Heights of two vertical poles = 36 m & 28m
From the shorter pole construct a line parallel to ground upto longer pole.
Now, p will be = 8 m 
and h given is 17 m
From pythagoras th. 
⇒b^2=(17m)^2 - (8m)^2
⇒b^2=289m^2 - 64m^2
⇒b^2=225 m^2
⇒b= root under (225 m^2)
⇒b=15 m
Hence, distance between two poles is 15 m. (Ans)
1 4 1
Let AB be the first pole and CD be the second
AB=36m and CD=28m
let E be the point where a line parallel to the ground meets AB
we have to find CE
AC= hypotenuse of the right ΔACE=17m
and AE= 36-28=8m
in ΔACE, By Pythagoras Theorem,
 AC^{2}= AE^{2} + CE^{2}
 CE^{2}= 17^{2}  - 8^{2}
 CE^{2}= 289-64=225
as therefore they are separated by a distance of 225m
1 4 1
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