# In Fig. 6.29, if $\mathrm{AB}\|\mathrm{CD}, \mathrm{CD}\| \mathrm{EF}$ and $y: z=3: 7$, find $x$."

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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, $x=126^0$.

Updated on 10-Oct-2022 13:40:32