(a) \(\frac{h(tan\,β-tan\,α)}{tan\,β}\)
Let PQ be the hill of height h metres.
Let AB be the pillar of height H metres.
Let AC || BQ meet PQ in C.
The angles of depression of the top and bottom of the pillar are given as α and β respectively.
∴ ∠PAC = α, ∠PBQ = β
Let BQ = AC = x metres.
In rt. Δ PCA,
tan α = \(\frac{PC}{AC}\) = \(\frac{h-H}{x}\)
⇒ x = \(\frac{h-H}{tan\,α}\) ...(i)
In rt. Δ PQB,
tan β = \(\frac{PQ}{QB}=\frac{h}{x}\)
⇒ x = \(\frac{h-H}{tan\,β}\) ...(ii)
∴ From (i) and (ii), we get
\(\frac{h-H}{tan\,α}\) = \(\frac{h-H}{tan\,β}\)
⇒ h tan α – H tan β = h tan α
⇒ h (tan β – tan α) = H tan β
⇒ H = \(\frac{h(tan\,β-tan\,α)}{tan\,β}\)