Let Ankur, Syed and David standing on the points P,Q and R.

Let `PQ=QR=PR=x`

Therefore, `DeltaPQR` is an equilateral triangle. Drawn altitudes PC, OD and RN form vertices to the sides of a triangle and intersect these altitude at the centre of a circle M.

As PQR is an equilateral, therefore these altitude bisects their sides.

In `DeltaPQC`

`PQ^2=PC^2+QC^2` (by pythagorus theorem)

`x^2=PC^2+((x)/(2))^2`

`rArr PC^2=x^(2)-(x^2)/(4)=(3x^2)/(4)" "(because QC=(1)/(2)QR=(x)/2)`

`rArrPC=sqrt(3x)/(2)`

Now, `MC=PC-PM=sqrt(3x)/(2)-20" "(becausePM="radius"=20 m)`

In `DeltaQCM`,

`QM^2=QC^2+MC^2`

`(20)^2=(x/2)^2+(sqrt(3x)/(2)-20)^2" "(because QM ="radius"=20 m)`

`rArr400=(x^2)/(4)+-20sqrt(3x)+400`

`rArr0=x^2-20sqrt(3x)`

`rArrx=20sqrt(3)m`

HEnce, `PQ=QR =PS= 20(sqrt3)m`.