Question

The angle of depression of the top and bottom of 8 metre call building from the top of multistoried building are 30 and 45 respectively. Find the height of multistoried building and the distance between two buildings. $(\sqrt{3}=1.73)$

Answer 1

From the figure we can see that rectangle $BCDE$ has sides $BC=8$and $CD=BE=d$ (which is the distance between the two buildings).

We can see that, $\angle ADC=\angle DAF={45}^{\circ }$ as $AF$ and $CD$ are parallel and $AD$ is the transverse line ,and if you recall, we know that $co-interior$ $angles$ $are$ $equal$.

Similarly, $\angle AEB=\angle EAF=30$.

In triangle $ADC$

$\Rightarrow$$tan\left(\angle D\right)=\frac{AC}{CD}$

$\Rightarrow$$tan\left(45\right)=\frac{AC}{CD}=1$

$\therefore AC=CD$

Which implies, $H=d$. ....(eq 1)

Now ,

$\Rightarrow$$AB+BC=AC$

$\Rightarrow$$AB=AC-BC=H-8$

$\therefore AB=H-8$ .....(eq 2)

In triangle $AEB$,

$\Rightarrow$$tan\left(\angle E\right)=\frac{AB}{BE}$

$\Rightarrow$$tan\left(30\right)=\frac{AB}{BE}=\frac{1}{\sqrt{3}}$

$\Rightarrow$$\frac{H-8}{d}=\frac{1}{\sqrt{3}}$.....(from eq2)

$\Rightarrow$$(H-8)\left(\sqrt{3}\right)=d=H$ ....(from eq 1)

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

$\therefore H=\frac{8\sqrt{3}}{\sqrt{3}-1}=\frac{8\times 1.73}{1.73-1}=18.9589$

Final Answers,

$H=d=18.9589m$.