Question

From a window, 60 metres high above the ground, of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are 60° and 45° respectively. Show that the height of the opposite house is 60$(1+\sqrt{3})$ metres.

Answer 1

Let $AB$ be the house and $DE$ be the window such that $DE=60$ m. Draw $CE$ ⊥ $AB$.
Thus, we have:
∠$BEC={60}^o$ and ∠$EAD={45}^o$
Let $AB\hspace{0.17em}=h$ m and $CA=60$m such that $BC=(h-60)$ m and $AD=x$ m.

In the right ∆$ADE$, we have:
$\frac{DE}{AD}=\tan {45}^o=1$

⇒ $\frac{60}{x}=1$
⇒ $x=60$ m
Now, in the right ∆$BEC$, we have:
$\frac{BC}{EC}=\tan {60}^o=\sqrt{3}$

⇒ $\frac{(h-60)}{x}=\sqrt{3}$

On putting $x=60$ in the above equation, we get:
$\frac{(h-60)}{60}=\sqrt{3}$

⇒ $h-60=60\sqrt{3}$
⇒ $h=60\sqrt{3}+60=60(\sqrt{3}+1)$ m

Hence, the height of the opposite house is $60(\sqrt{3}+1)$ m.