Question 466173
We'll describe three cases:
1)The given side is the hypotenuse, equal a: Then the leg opposite to the 30 degree angle is a/2 and the other leg base on the Pythagorean theorem will be:

{{{sqrt(a^2-(a/2)^2)=sqrt(a^2-a^2/4)=sqrt(3a^2/4)=a*sqrt(3)/2}}}.

2)The given side is the side opposite to the 30 degree angle, equal b: Then the 
hypotenuse will be 2b, and the other leg base on the Pythagorean theorem will be

{{{sqrt((2b)^2-b^2)=sqrt(3b^2)=b*sqrt(3)}}}

3)The given side is the side opposite to the angle 60 degree, equal c: 
Assume that the angle opposite to the angle 30 degree is x, then the hypotenuse is 2x, and using the Pythagorean theorem we write:

{{{c^2=(2x)^2-x^2}}}<=>{{{c^2=3x^2}}}<=>{{{x=sqrt(c^2/3)=c*sqrt(3)/3}}}, and the hypotenuse will be: {{{2c*sqrt(3)/3}}}