In triangle ABC AB = BC, CD is height, angle BCD is 32. Find angle A.

1. In a right-angled triangle BCD with a right angle ∠BDC, the sum of the acute angles ∠B and ∠BCD is 90 °:

∠B + ∠BCD = 90 °, from here we get:
∠B = 90 ° – ∠BCD;
∠B = 90 ° – 32 ° = 58 °.
2. In an isosceles triangle ABC with sides AB and BC, the angles at the base AC are:

∠A = ∠С.

3. The sum of the angles of triangle ABC is 180 °:

∠A + ∠B + ∠С = 180 °.

Substitute instead of ∠С the angle вместоA equal to it:

2∠A + ∠B = 180 °;
2∠A = 180 ° – ∠B;
∠A = (180 ° – ∠B) / 2;
∠A = (180 ° – 58 °) / 2;
∠A = 122 ° / 2 = 61 °.
Answer: ∠A = 61 °.