Find the fourth side of a quadrilateral inscribed in a circle having one of its sides equal to 20 m as its diameter, and the other two sides

I am not sure at the moment if Chris and I have interpreted the question the same.

 (This diagram is drawn to scale)

Here is my interpretation.

$$\\

Let length BC = x

Now I will use the cosine rule:

$$\boxed{cosA=\frac{b^2+c^2-a^2}{2bc}}\\\\
so\\\\
cos\theta=\frac{8^2+16^2-x^2}{2\times8\times16}\quad and \quad cos\theta=\frac{\sqrt{336}^2+12^2-x^2}{2\times \sqrt{336}\times12}\\\\
Hence\\\\
\frac{8^2+16^2-x^2}{2\times8\times16}=\frac{\sqrt{336}^2+12^2-x^2}{2\times \sqrt{336}\times12}\\\\
\frac{320-x^2}{128}=\frac{480-x^2}{ 12\times 4\sqrt{21}}\\\\
\frac{320-x^2}{128}=\frac{480-x^2}{ 12\times 4\sqrt{21}}\\\\
\frac{320-x^2}{8}=\frac{480-x^2}{ 3\sqrt{21}}\\\\
3\sqrt{21}(320-x^2)=8(480-x^2)\\\\
3\sqrt{21}\times320-3\sqrt{21}\;x^2\;=\;3840-8x^2\\\\
960\sqrt{21}-3840\;=\;(-8+3\sqrt{21})\;x^2\\\\$$

$$\\x^2=\frac{960\sqrt{21}-3840}{-8+3\sqrt{21}}\\\\
x=+\sqrt{\frac{960\sqrt{21}-3840}{-8+3\sqrt{21}}}\\\\$$

$${\sqrt{{\frac{\left({\mathtt{960}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{21}}}}{\mathtt{\,-\,}}{\mathtt{3\,840}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{21}}}}\right)}}}} = {\mathtt{9.864\: \!242\: \!223\: \!858\: \!689}}$$

so 

The forth side is approx 9.86metres long

Since our answers are the same I guess we interpreted the question the same after all     

I'm assuming that you're saying that one side of the quadrilateral is the same as the circle's diameter, and we have two adjacent sides to this that are 8 and 12m, respectively. Then the circle's radius = 10 m

To find the remaining side, we need to first find the central angles intercepting the 8 and 12 m sides.....

We can use the Law of Cosines in both cases....for the 12m side, we have

12^2 = 10^2 + 10^2 - 2(10)(10)cosΘ   ......simplifying, we have

cosΘ= 7/25  = cos^-1(7/25) = about 73.739795291688°

For the 8m side, we have

8^2 = 10^2 + 10^2 - 2(10(10)cosΘ   ...simplifying again, we have

cosΘ = 17/25 = cos^-1(17/25) = about 47.156356956404°

So, since the diameter would span 180° of arc, the remaing side must span (180 - 73.739795291688 - 47.156356956404)° = 59.103847751908° of arc

So...the remaining side... (s).... is given by

s^2  = 10^2 + 10^2 - 2(10)(10)cos(59.103847751908)

s = √[10^2 + 10^2 - 2(10)(10)cos(59.103847751908)] = about 9.86 m

This makes sense....the greater side intercepts the greatest arc, the next greatest side intercepts the next greatest arc, etc.