If a,b,c,d are lengths of sides of cyclic quadrilateral then cosine of angle B is given by \[cos(B)=\frac{a^2+b^2-c^2-d^2}{2(ab+cd)}\]
Find the angle B of the cyclic quadrilateral given the four sides as 6cm ,5cm, 4cm, 3cm .
\[cos^{-1}(\frac{6^2+5^2-4^2-3^2}{2(30+12)})=64.623\] degrees
