Draw the coordinate system.

Construct the unit circle with your compass.

Use a triangle to draw a $60\mathrm{°}60°$ angle with respect to the $xx$-axis. Mark the intersection of the resulting straight line with the unit circle.

Draw the perpendicular to the $xx$-axis that passes through the marked point.

Now you can measure the length of the sine with your ruler.

$\sin (60\mathrm{°})\approx 0.87\backslash sin(60°)\backslash approx0.87$

In a right triangle with a $60\mathrm{°}60°$ angle, the third interior angle has $30\mathrm{°}30°$. In the unit circle the hypotenuse has length $11$.

Extend the triangle to an equilateral triangle by mirroring the given triangle at the edge with $\sin (60\mathrm{°})\backslash sin(60°)$. The triangle must be equilateral because it has three $60\mathrm{°}60°$ angles.

In an equilateral triangle all sides have the same length, i.e. length $11$.

$\cos (60\mathrm{°})\backslash cos(60°)$ is exactly half the side length of the triangle, i.e. $\frac{1}{2}\backslash frac\{1\}\{2\}$.

Now use the Pythagorean theorem in the right triangle to calculate $\sin (60\mathrm{°})\backslash sin(60°)$.

Plug in the value for the cosine. ${(\cos (60\mathrm{°}))}^2+{(\sin (60\mathrm{°}))}^2=1^2\backslash left(\backslash cos(60°)\backslash right)^2+\backslash left(\backslash sin(60°)\backslash right)^2=1^2$

${\left(\frac{1}{2}\right)}^2+{(\sin (60\mathrm{°}))}^2=1\backslash left(\backslash frac\{1\}\{2\}\backslash right)^2+\backslash left(\backslash sin(60°)\backslash right)^2=1$

$\frac{1}{4}+{(\sin (60\mathrm{°}))}^2=1\backslash frac\{1\}\{4\}+\backslash left(\backslash sin(60°)\backslash right)^2=1$ $\mid -\frac{1}{4}\backslash mid-\backslash frac\{1\}\{4\}$

${(\sin (60\mathrm{°}))}^2=\frac{3}{4}\backslash left(\backslash sin(60°)\backslash right)^2=\backslash frac\{3\}\{4\}$ $\mid \sqrt{}\backslash mid\backslash sqrt\{\; \}$

$\sin (60\mathrm{°})=\pm \sqrt{\frac{3}{4}}\backslash sin(60°)=\backslash pm\backslash sqrt\{\backslash frac\{3\}\{4\}\}$ since $\sin (60\mathrm{°})\backslash sin(60°)$ was applied above the $xx$-axis, only the positive result comes into question.

$\sin (60\mathrm{°})=\sqrt{\frac{3}{4}}\backslash sin(60°)=\backslash sqrt\{\backslash frac\{3\}\{4\}\}$

$\sin (60\mathrm{°})=\frac{1}{2}\sqrt{3}\backslash sin(60°)=\backslash frac\{1\}\{2\}\backslash sqrt\{3\}$