Question
In figure, ACB is a line such that $\angle\text{DCA}=5\text{x}$ and $\angle\text{DCB}=4\text{x}.$ Find the values of $\angle\text{DCA}$ and $\angle\text{DCB}.$
✓
Answer
It is given that ACB is a line in the figure given below. Thus, $\angle\text{ACD}$ and $\angle\text{BCD}$ from a linear pair. 
Therefore, their sum must be equal to 180°. Or, we can say that$\angle\text{ACD}+\angle\text{BCD}=180^\circ$
Also,$\angle\text{ACD}=4\text{x}$ and $\angle\text{BCD}=5\text{x}.$ This further simplifies to:
4x + 5x = 180 9x = 180$\text{x}=\frac{180}{9}$
x = 20$\therefore\angle\text{DCA}=5\text{x}=5\times20^\circ=100^\circ$
$\angle\text{DCB}=4\text{x}=4\times20^\circ=80^\circ$
Hence, the values of $\angle\text{DCA}$ and $\angle\text{DCB}$ are 100° and 80° respectively.
Therefore, their sum must be equal to 180°. Or, we can say that$\angle\text{ACD}+\angle\text{BCD}=180^\circ$Also,$\angle\text{ACD}=4\text{x}$ and $\angle\text{BCD}=5\text{x}.$ This further simplifies to:
4x + 5x = 180 9x = 180$\text{x}=\frac{180}{9}$
x = 20$\therefore\angle\text{DCA}=5\text{x}=5\times20^\circ=100^\circ$
$\angle\text{DCB}=4\text{x}=4\times20^\circ=80^\circ$
Hence, the values of $\angle\text{DCA}$ and $\angle\text{DCB}$ are 100° and 80° respectively.
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