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The answer provided below is on the assumption that |AD| is an angle bisector of <BAC, by Sine rule
[tex]\frac{\sin{\angle BAD}}{24}=\frac{\sin{\angle BDA}}{36} \\ \\ \Rightarrow\sin{\angle BAD}=\frac{24\sin{\angle BDA}}{36}[/tex]
Also,
[tex]\frac{\sin{\angle CAD}}{18}=\frac{\sin{\angle CDA}}{x} \\ \\ \Rightarrow\sin{\angle CAD}=\frac{18\sin{\angle CDA}}{x}[/tex]
But,
[tex]\angle BAD=\angle CAD[/tex]
Thus,
[tex]\frac{24\sin{\angle BDA}}{36}=\frac{18\sin{\angle CDA}}{x} \\ \\ x\sin{\angle BDA}=27\sin{\angle CDA}[/tex]
But, [tex]\angle BDA\ and\ \angle CDA[/tex] are supplementary angles, hence they are equal.
Therefore, x = 27
