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 °.

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