Mathematics

# Compute the value of $x$ in each of the following figures

$50^{\circ}$

##### SOLUTION
$\Rightarrow$  $\angle ACD+\angle ACB=180^o$             [ Linear pair ]
$\Rightarrow$  $110^o+\angle ACB=180^o$
$\therefore$  $\angle ACB=70^O$
Again,
$\Rightarrow$  $\angle ABE+\angle ABC=180^o$          [ Linear pair ]
$\Rightarrow$  $120^o+\angle ABC=180^o$
$\therefore$  $\angle ABC=60^o$
Now, In $\triangle ABC,$
$\Rightarrow$  $\angle ABC+\angle ACB+\angle BAC=180^o$               [ Sum of angle of property of triangle ]
$\Rightarrow$  $60^o+70^o+x=180^o$
$\Rightarrow$  $130^o+x=180^o$
$\Rightarrow$  $x=50^o$

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