In an isosceles triangle ABC with base AC equal to 12 dm and angle B equal to 120 °.

In an isosceles triangle ABC with base AC equal to 12 dm and angle B equal to 120 °. Find the distance from point A to line BC.

Let’s build the height BP of the triangle ABC.

The height of an isosceles triangle, drawn to the base, is also its median and bisector. Then the angle ABP = 120/2 = 60, AP = AC / 2 = 6 dm.

In a right-angled triangle ABP, Sin60 = AP / AB.

AB = AP / Sin60 = 6 / (√3 / 2) = 12 / √3 = 4 * √3 dm.

The angle ABН is adjacent to the angle ABC, the sum of which is 180, then the angle ABН = 180 – 120 = 60.

In a right-angled triangle ABН, Sin60 = AH / AB.

AH = AB * Sin60 = 4 * √3 * √3 / 2 = 12/2 = 6 dm.

Answer: From point A to straight BC 6 cm.