In an isosceles triangle ABC, the base AB is 12 and the angle at the vertex C is 120 degrees. Find the height AH.

Since the ABC triangle is obtuse, the height drawn from the top of the acute angle intersects the lateral side on its continuation.

Let’s construct the height of the СD, which is also the median and the bisector of the triangle ABC.

Then the angle AСD = AСВ / 2 = 120/2 = 600, BP = ВD = AB / 2 = 12/2 = 6 cm.

In the right-angled triangle of the AСD, we determine the length of the hypotenuse of the AС.

Sin60 = AD / AC.

AC = AD / Sin60 = 6 / (√3 / 2) = 12 / √3 = 4 * √3 cm.

The external angle ACН is adjacent to the ACB angle, then the ACН angle = (180 – 120) = 600.

Then, in the right-angled triangle ACН, we determine the length of the leg AH.

SinАСН = АН / АС.

AH = AC * Sin60 = 4 * √3 * √3 / 2 = 6 cm.

Answer: The length of the height AH is 6 cm.