# A quadrilateral is inscribed in the circle, the two sides of which are 16 cm and 30 cm

**A quadrilateral is inscribed in the circle, the two sides of which are 16 cm and 30 cm, and the angle between them is 60 °. Find the other two sides of this quadrilateral if their difference is 2 cm.**

Let’s draw the diagonal BD of the quadrangle.

By the cosine theorem, we determine the length of the side BD from the triangle ВСD.

ВD ^ 2 = ВС ^ 2 + СD ^ 2 – 2 * ВС * СD * Cos60 = 256 + 900 – 2 * 16 * 30 * 1/2 = 1156 – 480 = 676.

ВD = 26 cm.

Since ABCD is inscribed in a circle, the sum of its opposite angles is 180, then the angle BAD = 180 – 60 = 120.

Let the side length AD = X cm, then, by condition, AB = (X + 2) cm.

Then from the triangle ABD, by the cosine theorem, we determine the length AD.

BD ^ 2 = AB ^ 2 + AD ^ 2 – 2 * AB * AD * Cos120.

676 = (X + 2) ^ 2 + X ^ 2 – 2 * (X + 2) * X * (-1/2).

676 = X ^ 2 + 4 * X + 4 + X ^ 2 + X ^ 2 + 2 * X.

3 * X ^ 2 + 6 * X – 672 = 0.

X ^ 2 + 2 * X – 224 = 0.

Let’s solve the quadratic equation.

X1 = -16. (Does not fit, since it is negative).

X2 = AD = 14 cm.

Then AB = 14 + 2 = 16 cm.

Answer: The lengths of the sides are 14 cm and 16 cm.