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Calculating Heights and Distances

Two stations ...

Question

Two stations due South of a leaning tower which leans towards the North, are at distances $a$ and $b$ from its foot. If $α$ and $β$ are the elevations of the top of the tower from these stations, then prove that its inclination $θ$ to the horizontal is given by $cot θ = \frac{b cot α - a cot β}{b - a}$

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Solution

height of the tower $D E = h$
Distance between first station to foot of tower $A D = a + x$
Distance between second station to foot of tower $B D = b + x$
Distance between $C a n d D = x$
Give α, β are the angle of elevation two stations to top of the tower
$t h a t i s ∠ D A E = α, ∠ D B E = β, ∠ D C E = θ$
$I n Δ A D E$
$C o t θ = \frac{x}{h} - - - - - - - - - - - - - - - - \to (1)$
$I n Δ B D E$
$C o t β = \frac{b + x}{h}$
$(b + x) = h C o t β$ (multiply a on both sides )
$(a b + a x) = h a C o t β - - - - - - - - - - - - - - - - \to (2)$
$I n Δ C D E$
$C o t α = \frac{a + x}{h}$
$(a + x) = h C o t α$ (multiply b on both sides )
$(a b + b x) = h b C o t α - - - - - - - - - - - - - - - - \to (3)$
$s u b s t r a c t (3) - (2)$
$(b - a) x = h (b C o t α - a C o t β)$
$\frac{x}{h} = \frac{b C o t α - a C o t β}{b - a}$
$C o t θ = \frac{b C o t α - a C o t β}{b - a} .$

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