# A quadrangle with sides of 8 and 15 cm is inscribed in a circle, and the angle between them

**A quadrangle with sides of 8 and 15 cm is inscribed in a circle, and the angle between them is 60 degrees, find the other two sides if one side is 1 cm larger than the other.**

Let’s construct a diagonal BD.

In triangle ABD, by the cosine theorem, we define the length of the side BD.

BD ^ 2 = AB ^ 2 + AD ^ 2 – 2 * AB * AD * Cos60 = 225 + 64 – 2 * 15 * 8 * 1/2 = 169.

ВD = 13 cm.

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

Let the side length ВС = X cm, then, by condition, СD = (X + 1) cm.

In a triangle ВСD by the cosine theorem:

ВD ^ 2 = ВС ^ 2 + СD ^ 2 – 2 * ВС * СD * Cos120.

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

169 = X ^ 2 + X ^ 2 + 2 * X + 1 + X ^ 2 + X.

3 * X ^ 2 + 3 * X – 168 = 0.

Let’s solve the quadratic equation.

X = BC = 8 cm.

Then CD = 8 + 1 = 9 cm.

Answer: The lengths of the other two sides are 8 cm and 9 cm.