In an isosceles triangle ABC with a base AC equal to 40 cm, the outer angle at apex B equal to 60

In an isosceles triangle ABC with a base AC equal to 40 cm, the outer angle at apex B equal to 60 degrees, find the distance from apex B to line AB.

The outer angle of the triangle at the vertex B is 60, then its adjacent inner angle is 180 – 60 = 120. ABC = 120.

Then the angle at the base will be (180 – 120) / 2 = 30.

Let us omit the height of the triangle from vertex B.

AD = AC / 2 = 40/2 = 20 cm.

Consider a right-angled triangle ABD, in which the angle D is straight, the angle A is 30 degrees, the leg AD = 20 cm.

Then the leg tg A = BD / AD.

ВD = tan A * АD = 20 / √3.

Answer: ВD = 20 / √3.