# Circles of radii 3 and 9 with centers O1 and O2, respectively, touch at point A. The straight line passing through point A

**Circles of radii 3 and 9 with centers O1 and O2, respectively, touch at point A. The straight line passing through point A intersects the smaller circle at point B, and the larger one at point C. Find the area of the triangle BCO2 if the angle ABO1 = 30 °.**

The desired triangle ВСО2 consists of two triangles, AO2C, ABO2.

Triangle AO1B is isosceles, since O1A = O1B = 3 cm, angle O1AB = 30.

The angle CAO2 = 30 as vertical, then the angle AO2C = (180 – 30 – 30) = 120.

Let us determine the area of the triangle AO2C.

Sao2s = О2А * О2С * Sin120 / 2 = 9 * 9 * √3 / 2/2 = 81 * √3 / 4 cm2.

From the triangle AO1B, by the cosine theorem, we determine the length AB.

AB ^ 2 = O1A ^ 2 + O1B ^ 2 – 2 * O1A * O1B * Cos150 = 9 + 9 – 2 * 9 * (-1/2) = 18 + 9 = 27 cm.

AB = 3 * √3 cm.

Determine the area of the triangle ABO2.

Savo2 = O2A * AB * Sin150 / 2 = 9 * 3 * √3 / 2/2 = 81/2 = 27 * √3 / 4 cm2.

Then Svco2 = Sao2c + Sawo2 = (81 * √3 / 4) + 27 * √3 / 4 = 108 * √3 / 4 = 27 * √3 cm2.

Answer: The area of the BCO2 triangle is 27 * √3 cm2.