In the figure, ACB is a line such that $\angle DCA = 5x$ and $\angle DCB = 4x$. Find the value of $x$.
"
Given:
ACB is a line such that $\angle DCA = 5x$ and $\angle DCB = 4x$.
To do:
We have to find the value of $x$.
Solution:
We know that,
Sum of the angles on a straight line is $180^o$.
Therefore,
From the figure,
$\angle ACD + \angle BCD = 180^o$
$5x + 4x = 180^o$
$9x = 180^o$
$x= \frac{180^o}{9}$
$x = 20^o$
The value of $x$ is $20^o$.
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