From a window (h m high above the ground) of a house in a street, the angles of elevations and depression of the top and front of another house on the opposite side of the street are $θ$ and $ϕ$ respectively, show that the height of the opposite house is $h (1 + t a n θ . c o t ϕ) m$

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Solution

Let the height of the opposite house be H m.
In right $Δ A D E$
$t a n θ = \frac{D E}{A E} \Rightarrow t a n θ = \frac{H - h}{A E} \Rightarrow A E = \frac{H - h}{t a n θ} \dots \dots (1)$
In right $Δ A C E$
$t a n ϕ = \frac{C E}{A E} \Rightarrow t a n ϕ = \frac{h}{A E} \Rightarrow A E = \frac{h}{t a n ϕ} \dots \dots (2)$
From (1) and (2)
$\frac{H - h}{t a n θ} = \frac{h}{t a n ϕ} \Rightarrow H c o t θ - h c o t θ = h c o t ϕ \Rightarrow H = \frac{h (c o t θ + c o t ϕ)}{c o t θ} \Rightarrow H = h (c o t θ + c o t ϕ) t a n θ \Rightarrow H = h (1 + t a n θ c o t ϕ) (∵ t a n θ c o t θ = 1)$
Thus, the height of the opposite house is $h (1 + t a n θ c o t ϕ) m$

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