In an equihedral triangle ABC, the base AC is the outer angle at the vertex B is 112. Find the angles of triangle ABC.

1) It is known that triangle ABC is isosceles.
AC is the base, then ∠ A = ∠ C.
2) If the outer angle at the vertex B is 112 °, then to find what the angle B is equal to, you need to subtract the outer angle B of the triangle ABC from 180 °.
∠ В = 180 ° – 112 ° = 80 ° – 12 ° = 70 ° – 2 ° = 68 °.
3) Find the sum of two angles, angle A and B.
∠ A + ∠ B = 180 ° – 68 ° = 120 ° – 8 ° = 112 °;
4) Since angle A is equal to angle B, then we divide 112 in half.
∠ A = ∠ B = 112 ° / 2 = 50 ° + 6 ° = 56 °.
Answer: ∠ A = ∠B = 56 ° and ∠ C = 68 °.