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.