$ABC$ is an equilateral triangle of side $2a$. Find each of its altitudes.

(N/A) Let $AD$ be the altitude in the given equilateral triangle,$\triangle ABC$.
We know that an altitude in an equilateral triangle bisects the opposite side.
$\therefore BD = DC = a$
In $\triangle ADB$,we have $\angle ADB = 90^{\circ}$.
Applying the Pythagoras theorem,we obtain:
$AD^2 + DB^2 = AB^2$
$\Rightarrow AD^2 + a^2 = (2a)^2$
$\Rightarrow AD^2 + a^2 = 4a^2$
$\Rightarrow AD^2 = 3a^2$
$\Rightarrow AD = a\sqrt{3}$
In an equilateral triangle,all the altitudes are equal in length.
Therefore,the length of each altitude is $a\sqrt{3}$.