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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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Updated on: 10-Oct-2022

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