In Fig. 6.29, if $ \mathrm{AB}\|\mathrm{CD}, \mathrm{CD}\| \mathrm{EF} $ and $ y: z=3: 7 $, find $ x $.
"
Given:
$AB \parallel CD, CD\parallel EF$ and $y:z=3:7$.
To do:
We have to find $x$.
Solution:
Given,
$AB \parallel CD$ and $CD\parallel EF$
We know that,
When the angles are on the same side of the transversal line the sum of the angles is always $180^o$.
This implies,
$x+y=180^o$
We also know that,
The corresponding interior angles of two parallel lines sum up to $180^o$.
This implies,
$y+z=180^0$
Let us substitute $y=3h$ and $z=7h$ (since $y:z=3:7$)
Therefore,
$3h+7h=180^o$
This implies,
$10=180^o$
$h=18^o$
Therefore,
$y=3\times18^o$
$y=54^o$ and
$z=7\times18^o$
$z=126^o$
Now, by substituting $y$ value in $x+y=180^o$
We get,
$x+54^o=180^o$
$x=180^o-54^o$
$=126^o$
Therefore, $x=126^0$.
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