Question

The angles of depressions of the top and bottom of $8m$ tall building from the top of a multistoried building are ${30}^o$ and ${45}^o$ respectively. Find the height of multistoried building and the distance between the two buildings.

Answer 1

Let AB be the multistoried building of height h and let the distance between two buildings be x meters

$\angle XAC=\angle ACB={45}^{\circ }$ [Alternate angles $\because$ AX || DE]

$\angle XAD=\angle ADE={30}^{\circ }$ [Alternate angles $\because$ AX || BC]

In $\Delta ADE$

$\tan {30}^{\circ }=\frac{AE}{ED}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h-8}{x}$ ($\because$ CB = DE = x)

$\Rightarrow x=\sqrt{3}(h-8)$ .......(i)

$\Delta ACB$

$\tan {45}^{\circ }=\frac{h}{x}$

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

$\Rightarrow$ $x=h$ .......(ii)

From (i) and (ii)

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

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

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

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

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

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

$\Rightarrow h=4\sqrt{3}(\sqrt{3}+1)$

$\Rightarrow h=4(3+\sqrt{3})meters$

Form (ii) $x=h$

So, $x=$ $4(3+\sqrt{3})meters$

Hence, height of multistoried bulding $=$ $4(3+\sqrt{3})m$

Distance between two building $=$ $4(3+\sqrt{3})m$