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Sagot :

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

View image PaightonP659560
View image PaightonP659560
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