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Answer :
Let Ankur, Syed and David standing on the point P, Q and R.
Let PQ = QR = PR = X
\(\therefore \) DPQR is an equilateral traingle.
Drawn altitudes PC, QD and RN from vertices to the sides of a traingle and intersect these altitudes at the centre of a circle M.
As PQR is an equilateral, therefore these altitudes bisects their sides.
In \(\triangle{PQC}\),
\({PQ}^2 = {PC}^2 + {QC}^2 \)
(By Pythagoras theorem)
\(\Rightarrow \) \({x}^2 = {PC}^2 + ( \frac{x}{2} )^2 \)
\(\Rightarrow \)\({PC}^2 = {x}^2 - (\frac{x}{2} )^2\)
\(\Rightarrow \) \({PC}^2 = {x}^2 - \frac{x^2}{4} = 3\frac{x^2}{4} \)
(Since, QC = \( \frac{1}{2} \) QR = \(\frac{x}{2} \) )
\(\therefore \) PC = \(\frac{\sqrt{3}x}{2} \)
Now, MC = PC - PM = \(\frac{\sqrt{3}x}{2} \) - 20
(Since, PM = radius)
Now, in \(\triangle{QCM}\),
\({QM}^2 = {QC}^2 + {MC}^2 \)
(By Pythagoras theorem)
\(\Rightarrow \) \( 20^2 = (\frac{x}{2} )^2 + (\frac{\sqrt{3}x}{2} - 20)^2 \)
(Since, QM = radius)
\(\Rightarrow \) \(400 = \frac{x^2}{4} +( \frac{\sqrt{3}x}{2})^2 - 20\sqrt{3}x + 400\)
\(\Rightarrow \) \(0 = {x}^2 - 20\sqrt{3}x\)
\(\Rightarrow \) \({x}^2 = 20\sqrt{3}x\)
\(\therefore \) \(x = 20\sqrt{3}\)
Hence, PQ = QR = PR = \(20\sqrt{3}\)m.