# In an isosceles triangle ABC between the height AH of the triangle and its lateral side AB is 10

**In an isosceles triangle ABC between the height AH of the triangle and its lateral side AB is 10 degrees. Find the BCD Outside Angle Degree.**

The height of an isosceles triangle divides it into two equal triangles. This means that the angles at the vertex A are equal:

∠ВAN = ∠СAН = 10 °.

The ACH triangle is rectangular because the height is perpendicular to its base. Hence ∠AНC = 90 °.

in order to calculate the degree measure ∠AСН you need:

∠АСН = 180 – (∠АНС + ∠НАС); since the sum of the angles of the triangle is 180 °.

∠АСН = 180 – (90 + 10) = 180 – 100 = 80 °.

The degree measure of the external angle is 180 °, which means that the external angle at the vertex C is equal to:

∠BCA = 180 ° – 80 ° = 100 °.

Answer: the degree measure of the external angle at the vertex C is equal to 100 °.