Question

From a window 15 m high above the ground, in a street, the angles of elevation and depression of the top and foot of another house on the opposite side of the street are 30° and 45° respectively. Show that the height of the opposite house is 23.66 m.

Answer 1

Let $AB$ be the house and $CD$ be the window.
We have:
$CD=15$ m such that $BE=15$ m, ∠$ADE={30}^o$ and ∠$DBC={45}^o$.
Let $AB=h$ m such that $AE=(h-15)$ m and let $BC=x$ m such that $DE=x$ m.

In the right ∆$DBC$, we have:
$\frac{DC}{BC}=\tan {45}^o=1$

⇒ $\frac{15}{x}=1$
⇒ $x=15$ m

Now, in the right ∆$ADE$, we have:
$\frac{AE}{DE}=\tan {30}^o=\frac{1}{\sqrt{3}}$
⇒ $\frac{h-15}{x}=\frac{1}{\sqrt{3}}$
On putting $x=15$ in the above equation, we get:
$\frac{h-15}{15}=\frac{1}{\sqrt{3}}$
⇒ $\sqrt{3}(h-15)=15$
⇒ $\sqrt{3}h-15\sqrt{3}=15$
⇒ $\sqrt{3}h=40.98$
⇒ $h=\frac{40.98}{\sqrt{3}}=\frac{40.98}{1.732}=23.66$ m
Hence proved.