In a right-angled triangle ABC with a right angle C, the outer angle at the vertex A is 120, AC + AB = 18cm. Find AC and AB.

Outside angle at vertex A – angle BAD = 120 degrees. The angles BAD and BAC (angle A) are adjacent, together they make up an unfolded angle, which is 180 degrees. Then:
angle BAD + angle BAC = 180 degrees;
120 + angle BAC = 180;
angle BAC = 180 – 120;
angle BAC = 60 degrees.
In triangle ABC, angle C = 90 degrees (by condition), angle A = angle BAC = 60 degrees. Then, by the theorem on the sum of the angles of a triangle, the angle B is equal to:
angle A + angle B + angle C = 180 degrees;
60 + angle B + 90 = 180;
angle B = 180 – 150;
angle B = 30 degrees.
The AC leg lies opposite an angle of 30 degrees, therefore AC is equal to half the hypotenuse:
AC = AB / 2.
Let’s compose a system of equations:
AC = AB / 2;
AC + AB = 18.
Substitute the AC value from the first equation into the second:
AB / 2 + AB = 18;
(AB + 2AB) / 2 = 18;
3AB / 2 = 18;
3AB = 36;
AB = 36/3;
AB = 12 cm.
We substitute the obtained value AB into the first equation of the system of equations:
AC = AB / 2;
AC = 12/2;
AC = 6 cm.
Answer: AC = 6 cm, AB = 12 cm.