Suppose that a quadrangle ABCD which is inscribed in a circle has the side lengths $$\mathrm { AB } = \sqrt { 2 } , \quad \mathrm { BC } = \mathrm { CD } = 2 , \quad \mathrm { DA } = \sqrt { 6 } .$$ (1) Let us set $\theta = \angle \mathrm { BAD }$. We have the two equalities $$\begin{aligned}
& \mathrm { BD } ^ { 2 } = \mathbf { A } - \mathbf { B } \sqrt { \mathbf { C } } \cos \theta , \\
& \mathrm { BD } ^ { 2 } = \mathbf { D } + \mathbf { E } \cos \theta .
\end{aligned}$$ Hence, $$\theta = \mathbf { F G } { } ^ { \circ } , \quad \mathrm { BD } = \mathbf { H } \text {. } \mathbf { I } \text {. }$$ (2) Furthermore, we have $$\angle \mathrm { BAC } = \mathbf { J K } ^ { \circ } , \quad \angle \mathrm { BCA } = \mathbf { L M } ^ { \circ } \text { and } \mathrm { AC } = \mathbf { N } + \sqrt { \mathbf { O } } \text {. }$$ We also have $$\sin \angle \mathrm { ADC } = \frac { \sqrt { \mathbf { P } } } { \mathbf{Q} } ( \sqrt { \mathbf { R } } + \mathbf { S } )$$ (3) Let us denote the point of intersection of the straight line AD and the straight line BC by E. We have $\mathrm { EB } = \mathbf { T } + \mathbf { U } \sqrt { \mathbf { V } }$.
Suppose that a quadrangle ABCD which is inscribed in a circle has the side lengths
$$\mathrm { AB } = \sqrt { 2 } , \quad \mathrm { BC } = \mathrm { CD } = 2 , \quad \mathrm { DA } = \sqrt { 6 } .$$
(1) Let us set $\theta = \angle \mathrm { BAD }$. We have the two equalities
$$\begin{aligned}
& \mathrm { BD } ^ { 2 } = \mathbf { A } - \mathbf { B } \sqrt { \mathbf { C } } \cos \theta , \\
& \mathrm { BD } ^ { 2 } = \mathbf { D } + \mathbf { E } \cos \theta .
\end{aligned}$$
Hence,
$$\theta = \mathbf { F G } { } ^ { \circ } , \quad \mathrm { BD } = \mathbf { H } \text {. } \mathbf { I } \text {. }$$
(2) Furthermore, we have
$$\angle \mathrm { BAC } = \mathbf { J K } ^ { \circ } , \quad \angle \mathrm { BCA } = \mathbf { L M } ^ { \circ } \text { and } \mathrm { AC } = \mathbf { N } + \sqrt { \mathbf { O } } \text {. }$$
We also have
$$\sin \angle \mathrm { ADC } = \frac { \sqrt { \mathbf { P } } } { \mathbf{Q} } ( \sqrt { \mathbf { R } } + \mathbf { S } )$$
(3) Let us denote the point of intersection of the straight line AD and the straight line BC by E. We have $\mathrm { EB } = \mathbf { T } + \mathbf { U } \sqrt { \mathbf { V } }$.