Find the angles at the base AC of an isosceles triangle ABC. if its outer angle at the vertex B is equal to 112.

В △ ABC AB = BC – sides, AC – base. Outer apex angle В ∠CBK = 112 °.
The outside angle theorem for a triangle states that an outside angle adjacent to one of the angles of a triangle is equal to the sum of the other two angles of the triangle that it is not adjacent to. Thus:
∠CBK = ∠A + ∠C.
Since ∠A and ∠C are the angles at the base of an isosceles triangle, they are equal. We denote ∠A and ∠C as x, then:
x + x = 112 °;
2 * x = 112 °;
x = 112 ° / 2;
x = 56 °.
Thus, ∠A = ∠C = x = 56 °.
Answer: ∠A = ∠C = 56 °.


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