In an acute-angled triangle ABC AB = BC, the angle A is 50 degrees, CD is the height drawn from the vertex C

In an acute-angled triangle ABC AB = BC, the angle A is 50 degrees, CD is the height drawn from the vertex C. Find the degree measure of the angle DCB

By definition, an isosceles triangle and two sides must be equal and the angles at the base are also equal.

By the condition of the problem AB = BC, hence the conclusion that the triangle ABC is isosceles.

hence the angle A = C, or BAC = BCA = 50;

the sum of the angles of a triangle is 180. (one)

You can write the equality BAC + BCA + ABC = 180;

we find:

ABC = 180 O – BAC – BCA = 180 – 50 – 50 = 80;

CD is the height drawn from the vertex C. Therefore:

right angles and angle BDC = CDA = 90;

Let’s move on to the rectangular BDC triangle. remember (1):

BDC + CBD + DCB = 180;

DCB = 180 – BDC – CBD = 180 – 90 – 80 = 10;