The hypotenuse of a right triangle is 4 and the perimeter is 10. Find the sum of the sines of the angles of this triangle.

Let’s denote the legs of this right-angled triangle through x and y.

According to the condition of the problem, the hypotenuse of a given right-angled triangle is 4, therefore, using the Pythagorean theorem, we can write the following relation:

x² + y² = 4².

It is also known that the perimeter of a given right-angled triangle is 10, therefore, we can write the following relationship:

x + y + 4 = 10.

Simplifying the resulting ratio, we get:

x + y = 10 – 4;

x + y = 6.

We solve the resulting system of equations:

x² + y² = 4²;

x + y = 6.

Substituting into the first equation the value x = 6 – y from the second equation, we get:

(6 – y) ² + y² = 4²;

36 – 12y + y² + y² = 16;

2y² – 12y + 36 – 16 = 0;

2y² – 12y + 20 = 0;

y² – 6y + 10 = 0;

y = -3 ± √ (3² – 10) = -3 ± √ (9 – 10) = -3 ± √ (-1).

Since the discriminant of this quadratic equation is negative, this quadratic equation has no roots, therefore, a right-angled triangle with a hypotenuse equal to 4 and a perimeter equal to 10 does not exist.