Question

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

• A. a
• B. $\frac{a}{2}$
• C. $\frac{\sqrt{3}a}{2}$
• D. $\frac{3a}{2}$

Answer 1

The correct option is D $\frac{3a}{2}$

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,
$A{D}^2+B{D}^2=A{B}^2$
$⟹AD=\sqrt{a^2-(\frac{a}{2}{)}^2}$
$⟹AD=\frac{\sqrt{3}a}{2}$
$ADcot\theta =ADcot{30}^{\circ }=AD\times \frac{AD}{BD}=\frac{\sqrt{3}a}{2}\times \frac{\frac{\sqrt{3}a}{2}}{\frac{a}{2}}$
$⟹ADcot\theta =\frac{3a}{2}$