Find the angles of an isosceles triangle if one of them is 27 more than the other.

1. Vertices of the triangle – A, B, C. ∠A is less than ∠B by 27 °.

2. ∠А is less than ∠В by 27 ° according to the problem statement, that is, ∠В = ∠А + 27 °.

3. ∠А = ∠С as angles at the base of an isosceles triangle.

3. The total value of all angles of the triangle, according to its properties, is 180 °:

∠А + ∠В + ∠С = 180 °. We replace in this expression ∠B by (A + 27 °), ∠C by ∠A:

∠А + А + 27 ° + ∠А = 180 °.

3∠А = 180 ° – 27 ° = 153 °

∠А = 51 °.

∠В = 51 ° + 27 ° = 78 °.

∠С = 51 °

Answer: ∠А = 51 °, ∠В = 78 °, ∠С = 51 °.