So with this triangle, it's considered a "special triangle". The 30-60-90 triangle states that if the short leg is x, the hypotenuse is 2x and the long leg is x√3.
Now since the 60° is across from 11, I can presume that 11 is the long leg, and to get the short leg HG, you would need to divide 11 by √3. However, we need to rationalize the denominator. To do that, multiply the fraction by √3/√3:
[tex]\frac{11}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{11\sqrt{3}}{3}[/tex]
Next, since we got the short leg, multiply that by 2 to get the hypotenuse:
[tex]\frac{11\sqrt{3}}{3}\times\frac{2}{1}=\frac{22\sqrt{3}}{3}[/tex]
In short, HG = [tex]\frac{11\sqrt{3}}{3}[/tex] and HI = [tex]\frac{22\sqrt{3}}{3}[/tex] . The correct option is D.
