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^\circ $. 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^\circ $ and $80^\circ $ respectively.
Therefore, their sum must be equal to $180^\circ $. 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^\circ $ and $80^\circ $ respectively.
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