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in Trigonometry by (36.2k points)
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From the top of a hill h metres high, the angles of depressions of the top and the bottom of a pillar are α and β respectively. The height (in metres) of the pillar is

(a) \(\frac{h(tanβ-tanα)}{tanβ}\)

(b) \(\frac{h(tanα-tanβ)}{tanα}\)

(c) \(\frac{h(tanβ-tanα)}{tanβ}\)

(d) \(\frac{h(tanβ-tanα)}{tanα}\)

1 Answer

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Best answer

(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 = \(\frac{h(tan\,β-tan\,α)}{tan\,β}\)

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