If line AC is the bisector of the angle DCB, therefore angle ACD and angle ACB are equal.
[tex]\begin{gathered} \text{Given } \\ <\text{ACD}=(5x+27)^0 \\ <\text{ACB}=(45+3x)^0 \end{gathered}[/tex][tex]\begin{gathered} \text{Since} \\ <\text{ACD}=<\text{ACB} \\ 5x+27=45+3x \end{gathered}[/tex][tex]\begin{gathered} \text{collect like terms} \\ 5x-3x=45-27 \\ 2x=18 \\ \text{divide through by 2} \\ \frac{2x}{2}=\frac{18}{2} \\ x=9^0 \end{gathered}[/tex]Hence, the value of x is 9°, OPTION A
