Find the acute corners of a right-angled triangle with a hypotenuse of 32cm, knowing that its area is 128cm2.

Let’s denote the legs of this right-angled triangle through a and b.

Since the area of ​​this triangle is 128 cm ^ 2, and the hypotenuse is 32 cm, the following relations take place:

a * b / 2 = 128;

a ^ 2 + b ^ 2 = 32 ^ 2.

We solve the resulting system of equations.

Multiplying both sides of the first equation by 4 and adding the resulting ratio with the second equation, we get:

a ^ 2 + b ^ 2 + 4 * (a * b / 2) = 32 ^ 2 + 4 * 128;

a ^ 2 + b ^ 2 + 2a * b = 1024 + 512;

(a + b) ^ 2 = 1536;

a + b = 16√6.

Multiplying both sides of the first equation by 4 and subtracting the resulting ratio from the second equation, we get:

a ^ 2 + b ^ 2 – 4 * (a * b / 2) = 32 ^ 2 – 4 * 128;

a ^ 2 + b ^ 2 – 2a * b = 1024 – 512;

(a – b) ^ 2 = 512;

a – b = 16√2.

Adding the ratios a + b = 16√6 and a – b = 16√2, we get:

a + b + a – b = 16√6 + 16√2;

2a = 16√6 + 16√2;

a = (16√6 + 16√2) / 2 = 8√6 + 8√2.

Find b:

b = 16√6 – a = 16√6 – (8√6 + 8√2) = 8√6 – 8√2.

We find the sine of the angle α, which lies opposite the leg a:

sin (α) = (8√6 + 8√2) / 32 = (√6 + √2) / 4.

We find the sine of the angle β, which lies opposite leg b:

sin (β) = (8√6 – 8√2) / 32 = (√6 – √2) / 4.

Therefore, α = arcsin ((√6 + √2) / 4) and β = arcsin ((√6 – √2) / 4).

Answer: arcsin ((√6 + √2) / 4) and arcsin ((√6 – √2) / 4).