You are given an isosceles triangle ABC. BD – median, height. The DBC angle is 42 degrees.

You are given an isosceles triangle ABC. BD – median, height. The DBC angle is 42 degrees. Find the corners of the triangle.

Given: triangle ABC-isosceles; BD-median, height; angle DBC = 42 degrees.

Solution: DBC angle = 42 degrees (by condition);

angle ABC = angle ABD + angle DBC;

angle ABD = angle DBC (BD-bisector in an isosceles triangle);

angle ABC = 42 + 42 = 84 degrees;

angle BDC = 90 degrees (BD-height);

angle BCD = 180- (90 + 42) = 180-132 = 48 degrees;

angle BCD = angle BCA (same angle in different triangles);

angle BAC = angle BCA = 48 degrees (triangle ABC-isosceles);

Answer: 84; 48; 48.