From a window $h$ m high above the ground in a street, the angle of elevation and depression of the top and foot of the other house on the opposite side of the street are $θ$ and $ϕ$ respectively. Show that the height of the opposite house is $h$ (1+ tan $θ$ cot $ϕ$ ).

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Solution

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Let the height of the opposite house be H m

In right $Δ A D E$ ,

$t a n θ = \frac{D E}{A E}$

$\Rightarrow t a n θ = \frac{H - h}{A E}$

$\Rightarrow A E = \frac{H - h}{t a n θ} - - - - - - - (1)$

In right $Δ A C E$
$t a n ϕ = \frac{h}{A E}$

$\Rightarrow A E = \frac{h}{t a n ϕ} - - - - - - - - (2)$

From (1) and (2),
$\frac{H - h}{t a n θ} = \frac{h}{t a n ϕ}$

$\Rightarrow H c o t θ - h c o t θ = h c o t ϕ$

$\Rightarrow H = \frac{h (c o t θ + c o t ϕ)}{c o t θ}$

$\Rightarrow H = h (c o t θ + c o t ϕ) (t a n θ)$

$\Rightarrow H = h (1 + t a n θ c o t ϕ)$

Thus the height of the opposite house is $h (1 + t a n θ c o t ϕ) m$

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