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

September 30, 2021 | education

| **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.

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