Question

Find the value of $\mathrm{cosec}30°$ geometrically.

Answer 1

Finding the value of $\mathrm{cosec}30°$ geometrically

Let us consider an equilateral triangle $\mathrm{ABC}$, whose all sides are equal to $2\mathrm{a}$ respectively.

Let us draw a perpendicular $\mathrm{AD}$ that meets $\mathrm{BC}$ at $\mathrm{D}$.

In $∆\mathrm{ABD}\mathrm{and}\mathrm{ACD},$

$\angle \mathrm{ADB}=\angle \mathrm{ADC}=90°$

$\mathrm{AB}=\mathrm{AC}$ (equal sides of an equilateral triangle)

$\mathrm{AD}=\mathrm{AD}$ (common side)

Therefore, by RHS congruency,

$∆\mathrm{ABD}\cong \mathrm{ACD}$

Hence,

$$\begin{gathered} \mathrm{BD}=\mathrm{DC}(\mathrm{CPCT}\mathrm{rule}) \\ \mathrm{BD}+\mathrm{DC}=2\mathrm{a} \\ \mathrm{BD}=\mathrm{a} \end{gathered}$$

$\begin{array}{rcl}\mathrm{cosec}30°& =& \mathrm{AB}/\mathrm{BD}\\ & =& 2\mathrm{a}/\mathrm{a}\\ & =& 2\end{array}$

Hence the value of $\mathrm{cosec}30°$is $2$