Question

A pole stands vertically inside a triangle park ABC. If the angle of elevation of the top of the pole from each corner of the park is the same, then in triangle ABC, the foot of the pole is at the

(a) centroid

(b) in centre

(c) cricumcentre

(d) orthocentre

Answer 1

(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.