First, let's draw an equilateral triangle where the sides are length n and the angles are pi/3 .
Second, draw a line that bisects the angle. Since the other two sides are the same length, it also bisects the opposite side and forms a right angle. This forms the angles (pi/3)/2 = pi/6
Now, let's look at this third triangle. Let's call the length of the remaining side " a " .
We can find " a " using the Pythagorean theorem.
a2 + (n/2)2 = n2
a = \(\sqrt{n^2-(\frac{n}{2})^2}=\sqrt{\frac{4n^2-n^2}{4}}=\frac{\sqrt3n}{2}\)
tan( pi/6 ) = opposite / adjacent = \(\frac{n}{2}\,/\,\frac{\sqrt3n}{2}=\frac{n}{2}\,*\,\frac2{\sqrt3n}=\frac1{\sqrt3}=\frac{\sqrt3}{3}\)
No calculator needed !
First, let's draw an equilateral triangle where the sides are length n and the angles are pi/3 .
Second, draw a line that bisects the angle. Since the other two sides are the same length, it also bisects the opposite side and forms a right angle. This forms the angles (pi/3)/2 = pi/6
Now, let's look at this third triangle. Let's call the length of the remaining side " a " .
We can find " a " using the Pythagorean theorem.
a2 + (n/2)2 = n2
a = \(\sqrt{n^2-(\frac{n}{2})^2}=\sqrt{\frac{4n^2-n^2}{4}}=\frac{\sqrt3n}{2}\)
tan( pi/6 ) = opposite / adjacent = \(\frac{n}{2}\,/\,\frac{\sqrt3n}{2}=\frac{n}{2}\,*\,\frac2{\sqrt3n}=\frac1{\sqrt3}=\frac{\sqrt3}{3}\)
No calculator needed !