Question

Find the value of $\tan {30}^o$ geometrically.

Answer 1

Consider an equilateral triangle $\Delta ABC$ with sides $AB=BC=CA=2a$

$\angle A=\angle B=\angle C={60}^{\circ }$

Draw $AO\perp BC$

In $\Delta ABD$ and $\Delta ACD$

AD=AD (common)

$\angle AOB=\angle ADC=\left({90}^{\circ }\right)$

$AB=AC(\Delta ABC$ is equilateral)

$\therefore \Delta ABC\cong \Delta ACD$

$BD=DC(C.P.C.T)$

$\therefore BD=\frac{2a}{2}=a$ and $\angle BAD=\frac{60}{2}={30}^{\circ }$

$AD=\sqrt{2a^2-a^2}=\sqrt{3}a$ (Pythagoras theorem)

So, $\tan 30=\frac{perpendicular}{base}=\frac{a}{\sqrt{3}a}$

$\tan 30=\frac{1}{\sqrt{3}}$