Question

The angles of depression of the top and bottom of a 10 m high building from the top of a multi-storey building are 30° and 45° respectively. Find the height of the multi-storey building are 30° and 45° respectively. Find the height of the multi-storey building and the distance between the two buildings.

Answer 1

Let $AB$ be the multistorey building and $DE$ be the high building such that $DE=10$ m. Draw $CD$ ⊥ $AB$.
Thus, we have:
∠​$BDC={30}^o$ and ∠$BEA={45}^o$
Let $AB=h$ m such that $BC=(h-10)$ m and $AE=x$ m such that $CD=x$ m.

In the right ∆$BEA$, we have:
$\frac{AB}{AE}=\tan {45}^o=1$

⇒ $\frac{h}{x}=1$
⇒ $x=h$
Or,
$h=x$
Now, in the right ∆$BDC$, we have:
$\frac{BC}{CD}=\tan {30}^o=\frac{1}{\sqrt{3}}$

⇒ $\frac{(h-10)}{x}=\frac{1}{\sqrt{3}}$
⇒ $(h-10)\sqrt{3}=x$
By putting $x=h$ in the above equation, we get:
$(h-10)\sqrt{3}=h$
⇒ $h(\sqrt{3}-1)=10\sqrt{3}$
⇒ $h=\frac{10\sqrt{3}}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}=5\sqrt{3}(\sqrt{3}+1)=5(3+\sqrt{3})=23.66$ m
We have:
Height of the multistorey building = $AB=h=23.66$ m
Distance between the two buildings = $AE=x=h=23.66$ m