Question

In an equilateral triangle ABC of side a units, AD is the perpendicular drawn onto side BC. Find $ADcot\theta$ in terms of a.

Answer 1

In $△$ABC,

$\angle A=\angle B=\angle C={60}^{\circ }$
(since $△ABC$ is equilateral)

Since altitude of an equilateral triangle bisects the angle of vertex.
$\therefore \angle BAD={30}^{\circ }$

Since altitude of an equilateral triangle bisects the side.
Therefore, BD = DC = $\frac{a}{2}$.

From Pythagoras theorem,
$A{D}^2+B{D}^2=A{B}^2$
$\Rightarrow AD=\sqrt{a^2-(\frac{a}{2}{)}^2}$
$\Rightarrow AD=\frac{\sqrt{3}a}{2}$
Now, $ADcot\theta =ADcot{30}^{\circ }=AD\times \sqrt{3}$
$\Rightarrow ADcot\theta =\frac{\sqrt{3}a}{2}\times \sqrt{3}$
$\Rightarrow ADcot\theta =\frac{3a}{2}$