We have 3 triangles that are related in the image
We will use the sine relations to determine the value of x
For triangle ABD
[tex]\begin{gathered} \frac{24}{\sin d}=\frac{16}{\sin b}=\frac{2x+4+2x}{\sin a} \\ \\ \frac{24}{\sin d}=\frac{16}{\sin b}=\frac{4x+4}{\sin a} \\ \\ \end{gathered}[/tex]
For triangle ABC
[tex]\frac{24}{\sin c1}=\frac{E}{\sin b}=\frac{2x+4}{\sin a1}[/tex]
For triangle ACD
[tex]\frac{16}{\sin c2}=\frac{E}{\sin d}=\frac{2x}{\sin a2}[/tex]
Now solving the system of equations we have
[tex]\begin{gathered} 4x+4=20 \\ 4x=20-4 \\ 4x=6 \\ x=4 \end{gathered}[/tex]
The answer would be x = 4 units
