Find the outer corners of an isosceles triangle if one of the inner corners is 114.

1. Let us first find the values ​​of the angles in a given triangle. Let it be a triangle ABC.

We accept that ∠В = 114 °. Then ∠А = ∠С = (180 ° – 114 °) / 2 = 66/2 = 33 °.

2. An angle adjacent to any inside corner of the triangle will be outside. That is, two outer corners can be drawn at each vertex. These angles will be vertical and therefore equal. In addition, the outer corner will be the complement of the inner corner to a 180 ° flattened corner.

Taking this into account, the external angles at the vertex C:

180 ° – 114 ° = 66 °;

The outer angles at the vertices A and C will be the same and equal:

180 ° – 33 ° = 147 °.