I do not understand where Po shen loh got the 3√3. How did he get that?

Module 2 Week 1 Day 3 Challenge Part 3
I do not understand where Po shen loh got the 3√3 for the bottom length of the triangle. How did he get that?

@mirthfulostrich Thank you for asking! This is a very special right triangle called a \(30^{\circ}60^{\circ}90^{\circ}\) right triangle, since it has angles of \(30^{\circ}, 60^{\circ}, \text{ and } 90^{\circ}.\) It is one of the most popular triangles.
After finishing Module 2, you will have probably seen this triangle \(50\) times! If you participate in any math competition at the middleschool or highschool level, you will probably see this triangle pop up twice on any exam! Prof. Loh has probably done \(2,000\) math problems in his lifetime so far which involve this special \(30^{\circ}60^{\circ}90^{\circ}\) right triangle! Consequently, he has memorized the ratios of its lengths, and in not too long, you will, too!
This triangle first made its appearance in the Day 1 Challenge Part 2 lesson, when we were finding the area of a regular hexagon.
Someone has already asked about how to find the side lengths of a \(30^{\circ}60^{\circ}90^{\circ}\) right triangle, so please visit this post to learn about how to derive the ratios of the sides of the triangle!The basic idea is this:
\( \textcolor{red}{\text{ Take half of an equilateral triangle. This gives us two of the sides (the short leg and the hypotenuse).}}\)
\( \textcolor{red}{\text{ Then, use the Pythagorean Formula to find the long leg. }} \) 
@debbie he's done 1,000 exams? That's a sheesh moment