Explanation:

Start with an equilateral triangle of side 2. The interior angle at each vertex must be \(\displaystyle{\frac{{\pi}}{{{3}}}}\) since 6 such angles make up a complete \(\displaystyle{2}\pi\) circle.

Then bisect the triangle through a vertex and the middle of the opposite side, dividing it into two right angled triangles.

These will have sides of length 2,1 and \(\displaystyle\sqrt{{{2}^{{2}}-{1}^{{2}}}}=\sqrt{{{3}}}\). The interior angles of each right angled triangle are \(\displaystyle{\frac{{\pi}}{{{3}}}},\ {\frac{{\pi}}{{{6}}}}\) and \(\displaystyle{\frac{{\pi}}{{{2}}}}\) with the \(\displaystyle{\frac{{\pi}}{{{6}}}}\) coming from the fact that we have bicested one of the \(\displaystyle{\frac{{\pi}}{{{3}}}}\) angles.

Then:

\(\displaystyle{\sin{{\left({\frac{{\pi}}{{{6}}}}\right)}}}={\frac{{\text{opposite}}}{{\text{hypotenuse}}}}={\frac{{12}}{}}\)

Start with an equilateral triangle of side 2. The interior angle at each vertex must be \(\displaystyle{\frac{{\pi}}{{{3}}}}\) since 6 such angles make up a complete \(\displaystyle{2}\pi\) circle.

Then bisect the triangle through a vertex and the middle of the opposite side, dividing it into two right angled triangles.

These will have sides of length 2,1 and \(\displaystyle\sqrt{{{2}^{{2}}-{1}^{{2}}}}=\sqrt{{{3}}}\). The interior angles of each right angled triangle are \(\displaystyle{\frac{{\pi}}{{{3}}}},\ {\frac{{\pi}}{{{6}}}}\) and \(\displaystyle{\frac{{\pi}}{{{2}}}}\) with the \(\displaystyle{\frac{{\pi}}{{{6}}}}\) coming from the fact that we have bicested one of the \(\displaystyle{\frac{{\pi}}{{{3}}}}\) angles.

Then:

\(\displaystyle{\sin{{\left({\frac{{\pi}}{{{6}}}}\right)}}}={\frac{{\text{opposite}}}{{\text{hypotenuse}}}}={\frac{{12}}{}}\)