The angles of depression of the top and bottom of \( 8 \mathrm{~m} \) tall building from the top of a multistoried building are \( 30^{\circ} \) and \( 45^{\circ} \) respectively. Find the height of the multistoried building and the distance between the two buildings.
Given:
The angles of depression of the top and bottom of \( 8 \mathrm{~m} \) tall building from the top of a multistoried building are \( 30^{\circ} \) and \( 45^{\circ} \) respectively.
To do:
We have to find the height of the multistoried building and the distance between the two buildings.
Solution:
Let $AB$ be the tall building and $CD$ be the height of the multistoried building.
Let point $D$ be the point of observation.
From the figure,
$\mathrm{AB}=8 \mathrm{~m}, \angle \mathrm{DAE}=30^{\circ}, \angle \mathrm{DBC}=45^{\circ}$
Let the height of the multistoried building be $\mathrm{CD}=h \mathrm{~m}$ and the distance between the two buildings be $\mathrm{AE}=\mathrm{BC}=x \mathrm{~m}$.
This implies,
$\mathrm{DE}=h-8 \mathrm{~m}$
We know that,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { DC }}{BC}$
$\Rightarrow \tan 45^{\circ}=\frac{h}{x}$
$\Rightarrow 1(x)=h$
$\Rightarrow h=x \mathrm{~m}$..........(i)
Similarly,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { DE }}{AE}$
$\Rightarrow \tan 30^{\circ}=\frac{h-8}{x}$
$\Rightarrow \frac{1}{\sqrt3}=\frac{h-8}{h}$ [From (i)]
$\Rightarrow h=(h-8)\sqrt3 \mathrm{~m}$
$\Rightarrow h=h\sqrt3-8\sqrt3 \mathrm{~m}$
$\Rightarrow h(\sqrt3-1)=8\sqrt3 \mathrm{~m}$
$\Rightarrow h=\frac{8\sqrt3}{\sqrt3-1} \mathrm{~m}$
$\Rightarrow h=\frac{8\sqrt3}{\sqrt3-1}\times\frac{\sqrt3+1}{\sqrt3+1} \mathrm{~m}$
$\Rightarrow h=\frac{8\sqrt3(\sqrt3+1)}{(\sqrt3)^2-1^2} \mathrm{~m}$
$\Rightarrow h=\frac{8(3+\sqrt3)}{3-1} \mathrm{~m}$
$\Rightarrow h=4(3+\sqrt3) \mathrm{~m}$
$\Rightarrow x=4(3+\sqrt3) \mathrm{~m}$
Therefore, the height of the multistoried building is $4(3+\sqrt3) \mathrm{~m}$ and the distance between the two buildings is $4(3+\sqrt3) \mathrm{~m}$.
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