A circle with a radius of 1 cm is inscribed in a right-angled triangle, the angle ABC is 60. Find the area of the triangle.

From the point O, the center of the circle, we will construct the radii OK, OH and OM to the points of tangency.

The СKOН quadrangle is a square, then CH = R = 1 cm.

Segments BН and BM are tangent to the circle drawn from one point, then ОВ is the bisector of the angle ABC, which means the angle ОBН = 60/2 = 30.

In a right-angled triangle ОBН, tg30 = ОН / BН.

BH = OH / tg30 = 1 / (1 / √3) = √3 cm.

Then ВС = СН + ВН = 1 + √3 cm.

In a right-angled triangle ABC tgB = tg60 = AC / BC.

AC = BC * tg60 = (1 + √3) * √3 = 3 + √3 cm.

Then Savs = AC * BC / 2 = (1 + √3) * (3 + √3) / 2 = (6 + 4 * √3) / 2 = 3 + 2 * √3 cm2.

Answer: The area of the triangle is 3 + 2 * √3 cm2.