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 $\alpha$ and $\beta$ respectively. Prove that the height of the house is $h( 1+tan\alpha cot\beta)$ metres.
Given: 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 $\alpha$ and $\beta$ respectively.
To do: To prove that the height of the house is $h( 1+tan\alpha cot\beta)$ metres.
Solution:
Let the height of other house be '$H$'.
Now, In $\vartriangle DEC$,
$\Rightarrow tan\alpha =\frac{DC}{EC}$
$=\frac{H-h}{EC}$
$\Rightarrow EC=\frac{H-h}{tan\alpha}\ .....\ ( i)$
In $\vartriangle EBA$,
$\Rightarrow tan\beta =\frac{EA}{AB}$
$=\frac{h}{EC}\ .......\ [\therefore AB=EC]$
$\Rightarrow EC=\frac{h}{tan\beta }\ ....... ( ii)$
From $( i)$ & $( ii)$, we get
$\Rightarrow \frac{h}{tan\beta}=\frac{H−h}{tan\alpha}$
$\Rightarrow htan\alpha =Htan\beta −htan\beta $
$\Rightarrow H tan\beta =h(tan\alpha +tan\beta )$
$\Rightarrow H=h( \frac{tan\alpha}{tan\beta}+\frac{tan\beta}{tan\beta})$
$\Rightarrow H=h(1+tan\alpha cot\beta )$
Hence, the answer is proved.
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