In a right-angled triangle ABC, the height CD is drawn (angle C = 90) AC = 6, DB = 5. Find CB = x.

Let the sought side CB = X cm, and the segment AD = Y cm.

In a right-angled triangle ABC CosABC = CB / AB = X / (Y + 5).

In a right-angled triangle BCD CosCBD = BD / СB = 5 / X.

5 / X = X / (Y + 5).

X ^ 2 = 5 * (Y + 5). (one).

In a right-angled triangle ABC according to the Pythagorean theorem:

CB ^ 2 = AB ^ 2 – AC ^ 2.

X ^ 2 = (Y + 5) ^ 2 – 36. (2).

Equate equalities 1 and 2.

5 * (Y + 5) = (Y + 5) ^ 2 – 36.

Y ^ 2 + 10 * Y + 25 – 36 – 5 * Y – 25 = 0.

Y ^ 2 + 5 * Y – 36 = 0.

Let’s solve the quadratic equation.

Y = 4 cm.

Then X ^ 2 = 5 * (4 + 5) = 45.

X = CB = √45 = 3 * √5 cm.

Answer: The length of the CB side is 3 * √5 cm.