A window of a house is $h$ metre above the ground. From the window, the angles of elevation and depression of the top and bottom of another house situated on the opposite side of the lane are found to be $α$ and $β$ respectively. Prove that the height of the house is $h (1 + tan α tan β) m e t r e s$ .

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Solution

Let the height of other house be $^{'} H^{'} .$
Now, In $△ D E C,$
$\Rightarrow tan α = \frac{D C}{E C} = \frac{H - h}{E C}$
$\Rightarrow E C = \frac{H - h}{tan α} ⟶ (1)$
In $△ E B A,$
$\Rightarrow tan β = \frac{E A}{A B} = \frac{h}{E C}$ $[∴ A B = E C]$
$\Rightarrow E C = \frac{h}{tan β} ⟶ (2)$
From $(1) & (2),$ we get
$\Rightarrow \frac{h}{tan β} = \frac{H - h}{tan α}$
$\Rightarrow h tan α = H tan β - h tan β$
$\Rightarrow H tan β = h (tan α + tan β)$
$\Rightarrow H = h (1 + tan α cot β)$
Hence, the answer is proved.

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Q. A window of a house is h metre above the ground . From the window , the angles of elevation and depression of the top and bottom of another house situated on the opposite side of the lane are found to be

α

and

β

respectively. Prove that the height of the house is

h (1 + \tan α \tan β)

metres.

From a window (h metres high above the ground) of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are $θ$ and $ϕ$ respectively. Find the height of the opposite house.

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A window of a house is h metre above the ground. From the window, the angles of elevation and depression of the top and bottom of another house situated on the opposite side of the lane are found to be α and β respectively. Prove that the height of the house is h(1+tanα tanβ) metres.

Let the height of other house be ′H′.Now, In △DEC,⇒tanα=DCEC=H−hEC⇒EC=H−htanα⟶(1)In △EBA,⇒tanβ=EAAB=hEC [∴AB=EC]⇒EC=htanβ⟶(2)From (1)&(2), we get ⇒htanβ=H−htanα⇒htanα=Htanβ−htanβ⇒Htanβ=h(tanα+tanβ)⇒H=h(1+tanαcotβ)Hence, the answer is proved.