In a right-angled triangle, one of the legs is 5 times larger than the other, the perimeter of the triangle is 64 cm

In a right-angled triangle, one of the legs is 5 times larger than the other, the perimeter of the triangle is 64 cm, the hypotenuse is 40 cm, find the acute corners of the triangle.

Let the length of the smaller leg be X cm, AB = X cm, then, by condition, the length of the leg BC = 5 * X cm.

The perimeter of the triangle ABC is equal to: Ravs = AC + AB + BC.

64 = 40 + X + 5 * X.

6 * X = 64 – 40 = 24 cm.

X = AB = 24/6 = 4 cm.

Then BC = 5 * AB = 5 * 4 = 20 cm.

Let’s define the acute angles of the triangle ABC.

SinBAC = BC / AC = 20/40 = 1/2.

Angle BAC = arcsinBAC = arcsin (1/2) = 30.

Then the angle BAC = 90 – 30 = 60.

Answer: The acute angles of the triangle are 30 and 60.