(c) cricumcentre
Let PQ be the pole standing vertically in the centre of the triangular park ABC. Then PQ being the pole standing vertically is perp. to all sides of the park, i.e. QB, QC and BC.
∴ ∠PQB = 90°,∠PQA = 90°,∠PQC = 90°.
Also, given,
n ∠PBQ = ∠PAQ = ∠PCQ = θ (say)
In rt. Δ PAQ,
tan θ = \(\frac{PQ}{AQ}\)
⇒ AQ = PQ cot θ ...(i)
In rt. Δ PBQ,
tan θ = \(\frac{PQ}{BQ}\)
⇒ BQ = PQ cot θ ...(ii)
In rt. Δ PCQ,
tan θ = \(\frac{PQ}{CQ}\)
⇒ CQ = PQ cot θ ...(iii)
From (i), (ii), and (iii) AQ = BQ = CQ
⇒ Q is equidistant from the vertices A, B and C
⇒ Q is the circumcentre of ΔABC.