Question

Find the value of $\sin {60}^o$ geometrically.

Answer 1

$\Rightarrow$ Draw an equilateral $△ABC$

$\Rightarrow$ In $△ADB,\angle D={90}^o$

$\Rightarrow$ Let $a$ be the side of equilateral triangle.

$\therefore$ $AB=BC=AC=a$

$\Rightarrow$ $A{B}^2=A{D}^2+B{D}^2$

$\Rightarrow$ $a^2=A{D}^2+(\frac{a}{2}{)}^2$

$\Rightarrow$ $A{D}^2=a^2-(\frac{a}{2}{)}^2$

$\Rightarrow$ $A{D}^2=a^2-\frac{a^2}{4}$

$\Rightarrow$ $A{D}^2=\frac{3a^2}{4}$

$\therefore$ $AD=\frac{\sqrt{3}a}{2}$

$\Rightarrow$ $\sin {60}^o=\frac{Opp.side}{Hypotenuse}=\frac{AD}{BD}=\frac{\frac{\sqrt{3}a}{2}}{a}$

$\therefore$ $\sin {60}^o=\frac{\sqrt{3}}{2}$