You know, I hope, that an equilateral triangle, with all side of the same length, also has all angles of the same size. Since, in degrees, the angles add to 180, in an equilateral triangle, the angles are all 180/3= 60 degrees.
Further, drawing a line from one vertex perpendicular to the opposite side divides the equilateral triangle into two right triangles in which one acute angle is 60 degrees and the other is 60/2= 30 degrees. That is, into two 30-60-90 right triangles.
That perpendicular also bisects the opposite side so if each side of the equilateral triangle has length "s", each 30-60-90 right triangle has hypotenuse of length "s" and one leg, opposite the 30 degree angle, of length "s/2". You can use the Pythagorean theorem to show that the other leg, opposite the 60 degree angle, has length \(\displaystyle \sqrt{s^2- (s/2)^2}= \sqrt{s^2- s^2/4}= \sqrt{3s^2/4}= \frac{\sqrt{3}}{2}s\).
Here, you are given that the side opposite the 30 degree angle has length \(\displaystyle \frac{\sqrt{6}}{12}\) so the length of the hypotenuse is twice that, \(\displaystyle \frac{\sqrt{6}}{6}\), and the other leg has length \(\displaystyle \frac{\sqrt{3}}{2}\frac{\sqrt{6}}{6}=\)\(\displaystyle \frac{\sqrt{18}}{12}= \)\(\displaystyle \frac{\sqrt{9(2)}}{12}= \frac{3\sqrt{2}}{12}= \frac{\sqrt{2}}{3}\). Yes, that is exactly what you have! Good work!
