Question 870539
You can find the tangent of {{{30^o}}} using the special 30-60-90 right triangle, which is half of an equilateral triangle.
Consider this triangle
{{{drawing(300,300,-1,0.2,-0.55,0.55,
green(triangle(-0.866,0,0,-0.5,0,0.5)),
triangle(-0.866,0,0,0.5,0,0),
rectangle(0,0,-0.05,0.05),
locate(-0.03,-0.2,1),locate(-0.433,-0.25,2),
locate(-0.03,0.3,1),locate(-0.433,0.25,2),
locate(-0.4,0.05,x),locate(-0.75,0.09,red(30^o)),
red(arc(-0.866,0,0.5,0.5,-30,0))
)}}} {{{x=sqrt(2^2-1^2)=sqrt(4-1)=sqrt(3)}}} and {{{tan(30^o)=1/sqrt(3)}}}
There will always be a square root in {{{tan(30^o)}}} , but math teachers do not like square roots in denominators.
To make it a more elegant expression we do this
{{{tan(30^o)=1/sqrt(3)=(1/sqrt(3))(sqrt(3)/sqrt(3))=1*sqrt(3)/(sqrt(3)*sqrt(3))=highlight(sqrt(3)/3)}}}
One way to get rid of a square root in a denominator
is to multiply times that square root
both of the expressions above and below the fraction line
(numerator and denominator).