Answer :
[tex]Look\ at\ the\ picture.\\\\tan30^o=\frac{\frac{a}{2}}{\frac{a\sqrt3}{2}}=\frac{a}{2}:\frac{a\sqrt3}{2}=\frac{a}{2}\cdot\frac{2}{a\sqrt3}=\frac{1}{\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}=\frac{\sqrt3}{3}[/tex]
Here is a mantra that I learned in Geometry class in high school. We were
all very excited, as angles had been invented only a few years earlier:
In a 30-60 right triangle ...
the side opposite the 30 is half the hypotenuse
the side opposite the 60 is half the hypotenuse times the square root of three.
So the tangent of the 30 would be (opposite) / (adjacent) = (1/2h) / (1/2h√3) .
That's (1)/(√3) or (√3)/(3) .
all very excited, as angles had been invented only a few years earlier:
In a 30-60 right triangle ...
the side opposite the 30 is half the hypotenuse
the side opposite the 60 is half the hypotenuse times the square root of three.
So the tangent of the 30 would be (opposite) / (adjacent) = (1/2h) / (1/2h√3) .
That's (1)/(√3) or (√3)/(3) .
