Question

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

Answer 1

In $△$ABC, $\angle A=\angle B=\angle C={60}^{\circ }$, since it is equilateral.
Consider $△$ABC,
$\angle$ADB = ${90}^{\circ }$
$\angle$ABC = ${60}^{\circ }$
$⟹\angle BAD={180}^{\circ }-{60}^{\circ }-{90}^{\circ }={30}^{\circ }$
Similarly, $\angle CAD={30}^{\circ }$
Now, $△ABD\cong △ACD$
According to SAS congruency,
Therefore, BD = DC = $\frac{a}{2}$.
From Pythagoras' theorem,
$A{D}^2+B{D}^2=A{B}^2$
$⟹AD=\sqrt{a^2-(\frac{a}{2}{)}^2}$
$⟹AD=\frac{\sqrt{3}a}{2}$
$AD\times cot\theta =AD\times cot{30}^{\circ }=AD\times \frac{AD}{BD}=\frac{\sqrt{3}a}{2}\times \frac{\frac{\sqrt{3}a}{2}}{\frac{a}{2}}$
$⟹AD\times cot\theta =\frac{3a}{2}$