Mathematics

In Fig., ACB is a line such that $\angle DCA = 5x$ and $\angle DCB = 4x.$ Find the values of $\angle DCA$ and $\angle DCB.$

$\angle DCA = 100^{\circ};\angle DCB = 80^{\circ}$

SOLUTION
Here, $ACB$ is a line.
$\Rightarrow$  $\angle DCA+\angle DCB=180^o$               [ Linear pair ]
$\Rightarrow$  $5x+4x=180^o$
$\Rightarrow$  $9x=180^o$
$\Rightarrow$  $x=\dfrac{180^o}{9}$
$\Rightarrow$  $x=20^o$
$\Rightarrow$  $\angle DCA=5x=5\times 20^o=100^o$
$\Rightarrow$  $\angle DCB=4x=4\times 20^o=80^o$

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