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Due south of a tower which is leaning towards north, there are

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Due south of a tower which is leaning towards north, there are two stations at distance $x, y, x, y$ respectively from its foot. If $α, β$ respectively be the angles of elevation of the top of the tower at these stations, show that the inclination of the tower to the horizon is
$cot^{- 1} \frac{y cot α - x cot β}{y - x}$

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Due south of a tower which is leaning towards north, there are two stations at distance

x, y, x, y

respectively from its foot. If

α, β

respectively be the angles of elevation of the top of the tower at these stations, show that the inclination of the tower to the horizon is

cot^{- 1} \frac{y cot α - x cot β}{y - x}

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Text solutionVerified

A Q = x, BQ = y

and

PQ

is a tower leaning towards North at an angle

θ

. Apply

m - n

theorem on

Δ PBQ

(y - x + x) cot α = x cot β (y - x) cot (π - θ)

y cot α - x cot β = (y - x) cot θ

∴ θ =

etc.

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Question Text	Due south of a tower which is leaning towards north, there are two stations at distance $x, y, x, y$ respectively from its foot. If $α, β$ respectively be the angles of elevation of the top of the tower at these stations, show that the inclination of the tower to the horizon is $cot^{- 1} \frac{y cot α - x cot β}{y - x}$
Answer Type	Text solution:1
Upvotes	150