The height of an isosceles triangle, drawn to its lateral side, forms an angle of 20 ° with the other lateral side.

The height of an isosceles triangle, drawn to its lateral side, forms an angle of 20 ° with the other lateral side. Find an angle at the base of an isosceles triangle.

Let the triangle ABC be given, from the vertex A the height AH is drawn to the side of the BC. Consider the triangle AВН, in which the following angles are known: <ВAН = 20 °, <ВНA = 90 °, we find <AВН = <B = 180 ° – <ВAН – <ВНA = 180 ° – 20 ° – 90 ° = 70 °.

After finding the angle at the apex, <B, you can find two equal angles at the base:

<BAC = <BCA = (180 ° – <ABC) / 2 = (180 ° – 70 °) / 2 = 110 ° / 2 = 55 °.

Answer: The angles at the base of the triangle are 55 °.