One of the acute corners of a right-angled triangle is 2 times smaller than the other, and the difference between

One of the acute corners of a right-angled triangle is 2 times smaller than the other, and the difference between the hypotenuse and the smaller leg is 15 cm. Find the hypotenuse and the smaller leg?

1. Find the degree measures of the acute angles of a right-angled triangle (ABC) given by condition. By the theorem on the sum of the angles of a triangle:
angle A + angle B + angle C = 180 degrees.
Angle A = 90 degrees (right angle by condition). Angle B = x, angle C = 2x. Then:
90 + x + 2x = 180;
3x = 90;
x = 30 degrees.
The smaller the angle, the smaller the leg opposite to it, therefore the AC leg is smaller than the AB leg. In addition, the AC leg lies against an angle of 30 degrees, so it is equal to half of the BC hypotenuse. It turns out a system of equations:
BC – AC = 15 (by condition);
AC = BC / 2.
Let us solve the system of equations, for this we substitute the value of the AC from the second equation into the first equation:
BC – BC / 2 = 15;
(2BC – BC) / 2 = 15;
BC / 2 = 15;
BC = 30 cm.
Find the smaller AC leg:
AC = BC / 2 = 30/2 = 15 (cm).
Answer: AC = 15 cm, BC = 30 cm.