A triangle has two corners with angles of `( pi ) / 3` and `( pi )/ 6`. If one side of the triangle has a length of `12`, what is the largest possible area of the triangle?

First, we know that the angels of a triangle has to add up to `pi`. Therefore, we know that the leftover angle is `pi-(pi/3+pi/6)=pi/2` which is a right angle. Therefore, we know that the triangle is a right triangle. We also know that this is the special "30-60-90" triangle. Therefore, we can figure out the sides while assuming 12 being one of each: the shortest, the middle, the longest.

Remember that in a "30-60-90" triangle, the shortest side is `a`, the second longest (middle) side is `asqrt3` and the longest side is `2a`.
Also, you multiply the two legs and then divide it by two to find the area of a right triangle.

When 12 is the hypotenuse, we know that the shortest side is 6 and the middle side is `6sqrt3` with the area of `18sqrt3`

When 12 is the shortest side, we don't really care about the hypotenuse. The middle side is `12sqrt3` with the area of `72sqrt3`.

When 12 is the middle side, we again don't care about the hypotenuse. The shortest side would be `12/sqrt3` which really is `4sqrt3` with the area of `24sqrt3`. Out of these three, we see that `72sqrt3` is the largest possible area.