Question

Find value of $\csc {30}^o$. Geometrically.

Answer 1

$\therefore \angle A=\angle B=\angle C=60$

Draw $AD\perp BC$

In $△ABD$ and $△ACD$

$AD=AD$ [common]

$\angle ADB=\angle ADC$

$AB=AC$ [$△$ ABC is an equilateral triangle]

$\therefore △ABD\equiv △ACD$

$BD=DC$ [cpct]

$\angle BAD=\angle CAD$ [cpct]

$\therefore BD=\frac{2a}{a}=aand\angle BAD=\frac{60}{2}=30$

In right angled $△ABD$

$\sin 30=\frac{BD}{AB}$

$\sin 30=\frac{a}{2a}=\frac{1}{2}$

And we

$\csc \theta =\frac{1}{\sin \theta }$

So,

$\csc 30=\frac{1}{\sin 30}=\frac{1}{\frac{1}{2}}=2$