Question

answer this question!!!✿●‿●✿​

Answer 1

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Answer 2

let,the poles be AB awnd CD

now,AB=a,

DC=b

separation(PC)=p

now

from the figure

EF is the required height

here

since EF is the height ,,so it will perpendicular to BC

also

AB is perpendicular to BC

so,BY corresponding angle property we can say

EF//AB

so,in triangle ABC,EF is a line which is parallel to one if it's side

by mid-point theorem

EF=AB/2=2/a

similarly

EF//CD

so

EF=CD/2=2/b

adding both

2EF=2(a+b)/ab

EF=(a+b)/ab