Find the acute corners of a right-angled triangle if its hypotenuse is 12 and the area is 18.

Let it be a triangle ABC, with sides a, b, and c = 12. As you know, the area of any triangle is equal to:

s = a * c * sin (<ABC) = sin (<B).

Let’s write down what the sine of angle B is equal to:

sin B = a / c = (a / 12), then the area of the triangle ABC is equal to:

s = 1/2 * c * f * sin B = 1/2 * a * 12 * (a / 12) = a ^ 2/2 = 18, whence

a ^ 2 = 18 * 2 = 36, a = √36 = 6. (take only the positive root. Now you can find sin B = a / c = 6/12 = 0.5.

Determine the angle B by the value of its sine:

<B = arc sin (0.5) = 30 °, then <A = 90 ° – 30 ° = 60 °.