Q2 As shown in the figure, a triangle $ABD$ is inscribed in a semi-circle with the diameter $BC$, where
$$\mathrm{AB}=3, \quad \mathrm{BD}=5, \quad \tan\angle\mathrm{ABD}=\frac{3}{4}.$$
We are to find the lengths of the three sides $BC$, $CD$ and $DA$ of the quadrangle $ABCD$ and the area $S$ of the quadrangle $ABCD$.
First, since $\cos\angle\mathrm{ABD}=\frac{\mathbf{L}}{\mathbf{M}}$, we have $\mathrm{DA}=\sqrt{\mathbf{NO}}$.
Also, since $\sin\angle\mathrm{ABD}=\frac{\mathbf{P}}{\mathbf{Q}}$, we have $\mathrm{BC}=\frac{\square\mathbf{R}\sqrt{\mathbf{ST}}}{\mathbf{U}}$ and thus $\mathrm{CD}=\frac{\mathbf{V}}{\mathbf{W}}$. From these we obtain
$$S = \frac{\mathbf{XXY}}{\mathbf{XY}}.$$

A quadrangle $ABCD$ which is inscribed in a circle $O$ satisfies
$$\mathrm { AB } = \mathrm { BC } = \sqrt { 2 } , \quad \mathrm { BD } = \frac { 3 \sqrt { 3 } } { 2 } , \quad \angle \mathrm { ABC } = 120 ^ { \circ } ,$$
where
$$AD > CD .$$
(1) Then $\mathrm { AC } = \sqrt { \mathbf { A } }$, and the radius of circle O is $\sqrt { \mathbf { B } }$.
(2) Set $x = \mathrm { AD }$. Since $\angle \mathrm { ADB } = \mathbf { CD }^\circ$, $x$ satisfies
$$4 x ^ { 2 } - \mathbf { E } \mathbf { F } x + \mathbf { G H } = 0 .$$
Also, set $y = \mathrm { CD }$. In the same way, it follows that $y$ satisfies
$$4 y ^ { 2 } - \mathbf { I J } y + \mathbf { K } \mathbf { L } = 0 .$$
Thus, noting (1), we obtain $AD$ and $CD$.

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 $\mathrm { EB } = \mathbf { T } + \mathbf { U } \sqrt { \mathbf { V } }$.

For a quadrilateral ABCD inscribed in a circle of radius 1, let $\mathrm { AB } : \mathrm { AD } = 1 : 2$ and $\angle \mathrm { BAD } = 120 ^ { \circ }$. Also, when the point of intersection of diagonals BD and AC is denoted by E, let $\mathrm { BE } : \mathrm { ED } = 3 : 4$.
We are to find the area of quadrilateral $\mathrm{ABCD}$.
In order to find the area of quadrilateral ABCD, we are to find the area of triangle ABD, denoted by $\triangle \mathrm { ABD }$, and the area of triangle BCD, denoted by $\triangle \mathrm { BCD }$.
First, let us find $\triangle \mathrm { ABD }$. Since
$$\mathrm { BD } = \sqrt { \mathbf { A } } , \quad \mathrm { AB } = \frac { \sqrt { \mathbf { BC } } } { \mathbf { D } } ,$$
we have
$$\triangle \mathrm { ABD } = \frac { \mathbf { E } \sqrt { \mathbf { F } } } { \mathbf { GH } } .$$
Next, let us find $\triangle \mathrm { BCD }$. Since
$$\triangle \mathrm { ABC } : \triangle \mathrm { ACD } = \mathbf{I} : \mathbf { J } ,$$
we see that $\mathrm { BC } : \mathrm { CD } = \mathbf { K } : \mathbf { L }$. (Give the answers using the simplest integer ratios.)
Hence we have $\mathrm { BC } = \frac { \mathbf{M} \sqrt { \mathbf { N } } } { \mathbf{O} }$ and
$$\triangle \mathrm { BCD } = \frac { \mathbf { PQ } \sqrt { \mathbf { R } } } { \mathbf { ST } }$$
Thus, from (1) and (2) we obtain the result that the area of quadrilateral ABCD is $\frac{\mathbf{U}\sqrt{\mathbf{V}}}{\mathbf{WX}}$.

In $\triangle A B C$, $\overline { A B } = \overline { B C } = 3$ and $\cos \angle A B C = - \frac { 1 } { 8 }$. On the circumcircle of $\triangle A B C$ there is a point $D$ satisfying $\overline { B D } = 4$ and $\overline { A D } \leq \overline { C D }$. Then $\overline { C D } = $ (17-1) $+$ $\sqrt{\text{(17-2)}}$. (Express as a simplified radical form.)