Question 617680
SHORT ANSWER:
The fastest way to write an answer seems to be invoking trigonometric ratios.
The long leg is opposite the {{{60^o}}} angle, so
{{{sin(60^o)=long_leg/hypotenuse}}} --> {{{hypotenuse=sin(60^o)*long_leg}}}
long_leg={{{6*sqrt(3)}}}
and {{{sin(60^o)=sqrt(3)/2}}}
Putting it al together
{{{hypotenuse=(sqrt(3)/2)(6*sqrt(3))=3*sqrt(3)*sqrt(3)=3*3=highlight(9)}}}
 
HOW WE KNOW THOSE TRIGONOMETRIC RATIOS:
A 30-60-90 triangle is half of an equilateral triangle:
{{{drawing(300,300,-2,10,-6,6,
triangle(0,5,0,-5,8.66,0),
rectangle(0,0,0.3,0.3),
red(line(0,0,8.66,0)),
locate(-.5,0.3,A), locate(-0.5,5.5,B),
locate(-.5,-5,D),locate(8.7,0.3,C),
locate(4.5,3,a),locate(0.1,2.8,c),locate(4,0.6,b)
)}}}
{{{AB=AD}}}, so {{{a=BC=AD=2*AB=2c}}} (The hypotenuse is twice as long as the short leg).
Using the Pythagorean theorem ({{{b^2+c^2=a^2}}}), we get
{{{b^2+c^2=(2c)^2}}} --> {{{b^2+c^2=4c^2}}} --> {{{b^2=4c^2-c^2}}} --> {{{b^2=3c^2}}} --> {{{b=sqrt(3c^2)}}} --> {{{b=sqrt(3)*c}}}
So the trigonometric ratios in a 30-60-90 triangle are
{{{sin(30^o)=cos(60^o)=c/a=c/(2c)=1/2}}}
{{{cos(30^o)=sin(60^o)=b/a=(sqrt(3)*c)/(2*c)=sqrt(3)/2}}}
{{{tan(30^o)=cot(60^o)=c/b=c/(sqrt(3)*c)=1/sqrt(3)=sqrt(3)/3}}}
{{{cot(30^o)= tan(60^o)=b/c=sqrt(3)*c/c=sqrt(3)}}}