Question

The angles of depression of the top and bottom of $8$ m tall building from the top of a multi-storeyed building are ${30}^0$ and ${45}^0$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

Answer 1

Let $AB$ and $CD$ be the multi-storied building and the building respectively.

Let the height of the multi-storied building be $hm$ and $thedistancebetweenthetwobuildingbexm$

$AE=CD=8m$ [ Given ]

$BE=AB-AE=(h-8)m$ and

$AC=DE=xm$ [ Given ]

Now, in $△ACB$,

$\Rightarrow$ $\tan {45}^o=\frac{AB}{AC}$

$\Rightarrow$ $1=\frac{h}{x}$

$\therefore$ $x=h$ ---- ( 1 )

In $△BDE$,

$\Rightarrow$ $\tan {30}^o=\frac{BE}{ED}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{h-8}{x}$

$\therefore$ $x=\sqrt{3}(h-8)$ ------ ( 2 )

$\Rightarrow$ $h=\sqrt{3}h-8\sqrt{3}$

$\Rightarrow$ $\sqrt{3}h-h=8\sqrt{3}$

$\Rightarrow$ $h(\sqrt{3}-1)=8\sqrt{3}$

$\Rightarrow$ $h=\frac{8\sqrt{3}}{\sqrt{3}-1}$

$\Rightarrow$ $h=\frac{8\sqrt{3}}{\sqrt{3}+1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$\Rightarrow$ $h=\frac{8\sqrt{3}(\sqrt{3}+1)}{3-1}$

$\Rightarrow$ $h=\frac{8\sqrt{3}(\sqrt{3}+1)}{2}$

$\therefore$ $h=(12+4\sqrt{3})m$

$\therefore$ $x=(12+4\sqrt{3})m$ [ From ( 1 ) ]

$\therefore$ The height of the multi-storied building and the distance between the two buildings is $(12+4\sqrt{3})m$