# In a triangle ABC AC = BC. Angle B is 70 degrees. Find the outer corner of the ACD.

Since by the condition in ΔABC the side AC is equal to the side BC, then ΔABC is isosceles with base AB.

According to the rule, in an isosceles triangle, the angles at the base are equal.

That is, the values of the angles A and B are equal.

1.) Now let’s write for angles A and B.

A = B = 70 °.

2.) Before finding the ACD angle, calculate the C angle.

Let us write for the angles ΔABC.

A + B + C = 180 °.

2A + C = 180 °.

C = 180 ° – 2A.

C = 180 ° – 2 × 70 °.

C = 180 ° – 140 °.

C = 40 °.

3.) Now we write the expression for the unfolded corner of the BCD.

In this case, we recall the rule according to which the angle BCD is 180 ° and what was previously calculated – the angle C = ACB = 40 °.

BCD = ACB + ACD.

40 ° + ACD = 180 °.

ACD = 180 ° – 40 °.

ACD = 140 °.

Answer: 140 °.